ROTATORY MOTION (PART-2)

Vertical Circular Motion (with Variable Speed).
Motion in Vertical Circle (Condition for Looping a Loop)
Uniform Vertical Circular Motion
Torque
1. Moment of Force or Torque
2. Torque on a Rigid Body About an Axis
Couple and Moment of Couple
Rigid Body
1. Mechanical Equilibrium of a Rigid Body
Banking of Roads
Overturning of Vehicle on an Unbanked Path
Moment of Inertia
Radius of Gyration (K)

11. Principle of Parallel and Perpendicular Axes
i) Parallel Axis Theorem
ii) Perpendicular Axis Theorem

Vertical Circular Motion (with Variable Speed)

Consider a body of mass m tied at one end of a string of length r and whirled in a vertical circle by fixing the other end at O. Let V₁ be the velocity of the body at the lowest point.

Velocity of the body at any point on the vertical circle: Let V_θ be the velocity of the body when the string makes an angle θ with the vertical as shown in the figure.

According to the law of conservation of mechanical energy,

E_A = E_P

K_A + U_A = K_P + U_P

K_P = K_A − (U_P − U_A)

K_P = 1 2 mV_θ²

K_A = 1 2 mV₁² & U_P − U_A = mgh

1 2 mV_θ² = 1 2 mV₁² − mgh

V_θ² = V₁² − 2gh

From the triangle OPI,

h = r(1 − cos θ)

V_θ² = V₁² − 2gr(1 − cos θ)

V_θ = √[V₁² − 2gr(1 − cos θ)]

If V₂ is the velocity of the body at the highest point (θ = 180°),

V₂ = √[V₁² − 2gr(1 + 1)]

V₂² = V₁² − 4gr

Tension in the String at Any Point

Let T_θ be the tension in the string and V_θ be the velocity of the body when the string makes an angle θ with the vertical.

At point P, the net force towards the centre is T_θ − mg cos θ. This force provides the centripetal force at this point.

T_θ − mg cos θ = mV_θ² r

T_θ = mV_θ² r + mg cos θ

At the lowest point, θ = 0° and V_θ = V₁. The tension in the string is

T_L = mV₁² r + mg (maximum)

At the highest point, θ = 180° and V_θ = V₂. The tension in the string is

T_H = mV₂² r − mg (minimum)

When the string is horizontal, θ = 90°, V_θ = V_hor. The tension in the string at this position is

T_(hor) = mV_hor² r

Note 1: The difference in maximum and minimum tension in the string is

T_max − T_min = mV₁² r + mg − mV₂² r + mg

= m r (V₁² − V₂²) + 2mg

= m r (4gr) + 2mg = 4mg + 2mg = 6mg

Note 2: Ratio of maximum tension to minimum tension in the string is

T_max T_min = mV₁² r + mg mV₂² r − mg

= V₁² + rg V₂² − rg (V₂² = V₁² − 4gr)

Note 3:

When the body is at a point P as shown in the figure:

(a) Tangential force acting on the particle

\[ F_t = mg\sin\theta \]

Tangential acceleration is

\[ a_t = g\sin\theta \]

(b) Centripetal force acting on the particle

\[ F_c = \frac{mV_\theta^2}{r} = T_\theta – mg\cos\theta \]

Centripetal acceleration is

\[ a_c = \frac{V_\theta^2}{r} \]

(c) Net force acting on the particle at point P

\[ F = \sqrt{F_t^2 + F_c^2} \]

Net acceleration of the particle at point P is

\[ a = \sqrt{a_t^2 + a_c^2} \]

The angle made by the net force or net acceleration with the centripetal component is \(\phi\), and

\[ \tan\phi = \frac{F_t}{F_c} = \frac{a_t}{a_c} \]

Note 4:

In a vertical circular motion, if the velocity at the highest point is \(\sqrt{ngr}\), then the velocity and tension at different positions will be:

\[ T = mg(n-1) \]

At the highest point,

\[ V_2 = \sqrt{ngr} \]

At the lowest point,

\[ V_1 = \sqrt{(n+4)gr} \]

The tension at the lowest point is

\[ T_1 = mg(n+5) \]

At an arbitrary point on the vertical circle, making an angle \(\theta\) with the downward vertical diameter,

\[ V_\theta = \sqrt{ngr+2gr(1+\cos\theta)} \]

The tension at that point is

\[ T_\theta = mg(n+2+3\cos\theta) \]

Therefore, the ratio of velocity at the lowest and highest positions is

\[ \frac{V_1}{V_2} = \sqrt{\frac{n+4}{n}} \]

The ratio of tensions at the lowest and highest positions is

\[ \frac{T_{\max}}{T_{\min}} = \frac{n+5}{n-1} \]

Motion in Vertical Circle (Condition for Looping a Loop)

Consider a body of mass m tied at the end of a string and whirled in a vertical circle of radius r. Let v₁ and v₂ be the velocities of the body and T₁ and T₂ be the tensions in the string at the lowest point A and highest point B, respectively.

The velocity of the body at any point of the vertical circle is directed along the tangent to the circular path, while the tension in the string is always directed towards the centre of rotation O.

Forces acting on the body at the lowest point A

When the body is at the lowest point A, the forces acting on it are:

Tension T₁ acting in the upward direction.
Weight mg of the body acting in the downward direction.

The net force towards the centre supplies the necessary centripetal force. Therefore,

\[ T_1-mg=\frac{mv_1^2}{r} \]

Hence,

\[ T_1=\frac{mv_1^2}{r}+mg \tag{1} \]

Forces acting on the body at the highest point B

When the body is at the highest point B, both the tension T₂ and the weight mg act towards the centre of the circle.

Therefore, the net force acting towards the centre is T₂ + mg.

\[ T_2+mg=\frac{mv_2^2}{r} \]

Hence,

\[ T_2=\frac{mv_2^2}{r}-mg \tag{2} \]

Condition for completing the vertical circle

The body will complete the vertical circular path only if the minimum tension at the highest point is not negative. Therefore,

\[ T_{2(\min)}\geq0 \]

From equation (2),

\[ \frac{mv_2^2}{r}-mg\geq0 \]

\[ \frac{mv_2^2}{r}\geq mg \]

\[ v_2^2\geq gr \]

\[ v_2\geq\sqrt{gr} \]

Therefore, the minimum speed at the highest point B to just complete the vertical circle is

\[ v_{2(\min)}=\sqrt{gr} \]

This minimum speed is also called the critical velocity.

If the velocity of the body at the highest point is less than \(\sqrt{gr}\), the string becomes loose \((T_2<0)\), and the body falls down without continuing along the circumference of the vertical circle.

Relation between velocities at the lowest and highest points

According to the law of conservation of mechanical energy,

\[ E_A=E_B \]

\[ K_A+U_A=K_B+U_B \]

Taking the lowest point A as the reference level for gravitational potential energy,

\[ K_A=K_B+U_B-U_A \]

The kinetic energies at the lowest and highest points are

\[ K_A=\frac{1}{2}mv_1^2 \]

\[ K_B=\frac{1}{2}mv_2^2 \]

Since the highest point is at a height of \(2r\) above the lowest point,

\[ U_B-U_A=2mgr \]

Therefore,

\[ \frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2+2mgr \]

On simplifying,

\[ v_1^2=v_2^2+4gr \tag{3} \]

Minimum velocity at the lowest point

To complete the vertical circle, the velocity of the body at the highest point must not be less than the critical velocity \(\sqrt{gr}\).

Therefore, for the body to just complete the vertical circle, we substitute

\[ v_2=\sqrt{gr} \]

in equation (3):

\[ v_{1(\min)}^2=(\sqrt{gr})^2+4gr \]

\[ v_{1(\min)}^2=gr+4gr \]

\[ \boxed{v_{1(\min)}=\sqrt{5gr}} \]

Therefore, for a body to move in a complete vertical circular path, its minimum velocity at the lowest point A must be

\[ \boxed{(v_1)_{\min}=\sqrt{5gr}} \]

Minimum tension at the lowest point

The minimum tension that must be maintained in the string at the lowest point to just complete the vertical circle is obtained from equation (1):

\[ T_{1(\min)} = \frac{m(v_{1(\min)})^2}{r}+mg \]

Since \[ (v_{1(\min)})^2=5gr, \] we get

\[ T_{1(\min)} = \frac{m(5gr)}{r}+mg \]

\[ \boxed{T_{1(\min)}=6mg} \]

Application 1

While looping an aircraft in a vertical circular path or driving a motorcycle in a vertical circular path inside a circus globe, the minimum velocity for the safety of the rider or exhibitor at the lowest point of the circular path must be

\[ \boxed{\sqrt{5gr}} \]

Application 2

If some water is taken in a bucket and whirled with the help of a string in the vertical plane, the water will not spill out from the bucket even when it is at the highest point of its circular path if a minimum velocity of \(\sqrt{gr}\) is maintained at the highest point.

Application 3

A motorcyclist moves in a vertical circle inside a spherical globe without falling at the topmost point if he maintains a minimum speed of \(\sqrt{gr}\) at the highest point.

Note 1: Critical velocity at any point of the vertical circle

The critical velocity of the body at any point P in a vertical circle to just complete the loop is

\[ V_{\theta(\min)} = \sqrt{ V_{1(\min)}^2-2gr(1-\cos\theta) } \]

Since \[ V_{1(\min)}=\sqrt{5gr}, \] we have

\[ V_{\theta(\min)} = \sqrt{ 5gr-2gr(1-\cos\theta) } \]

On simplifying,

\[ \boxed{ V_{\theta(\min)} = \sqrt{gr(3+2\cos\theta)} } \]

Note 2: Minimum tension at any point of the vertical circle

The minimum tension to be maintained in the string when the string makes an angle \(\theta\) with the vertical, to just complete the loop, is

\[ T_{\theta(\min)} = \frac{mV_{\theta(\min)}^2}{r} + mg\cos\theta \]

But,

\[ V_{\theta(\min)} = \sqrt{gr(3+2\cos\theta)} \]

Therefore,

\[ T_{\theta(\min)} = \frac{m\left(gr(3+2\cos\theta)\right)}{r} + mg\cos\theta \]

\[ \boxed{ T_{\theta(\min)} = 3mg(1+\cos\theta) } \]

Note 3: Velocity and tension at different positions

When a body is moving in a vertical circle with critical speed, the velocities and tensions in the string at different positions are as shown below.

At the highest point B,

\[ V_2=\sqrt{gr}, \qquad T_2=0 \]

At point E,

\[ V=\sqrt{2gr}, \qquad T=\frac{3mg}{2} \]

At the horizontal point C,

\[ V_3=\sqrt{3gr}, \qquad T_3=3mg \]

At point D,

\[ V=\sqrt{4gr}, \qquad T=\frac{9mg}{2} \]

At the lowest point A,

\[ V_1=\sqrt{5gr}, \qquad T_1=6mg \]

Hence, in a vertical circle with critical speed, the ratio of the minimum speeds to be maintained at the highest, horizontal and lowest positions is

\[ \boxed{ 1:\sqrt{3}:\sqrt{5} } \]

The respective kinetic energies at these points are in the ratio

\[ \boxed{1:3:5} \]

Note 4: Non-uniform circular motion in a vertical plane

In the case of non-uniform circular motion in a vertical plane, if the velocity of the body at the lowest point is less than \(\sqrt{5gr}\), the particle will not complete the circle in the vertical plane. It can either oscillate about the lowest point or leave the circle without looping.

Case 1: Condition for oscillating about the lowest position

Let \(V_L\) be the minimum velocity given to a body at the lowest position to reach the horizontal position.

According to the law of conservation of mechanical energy,

\[ T.E._A=T.E._B \]

\[ KE_A+U_A=KE_B+U_B \]

At the lowest point \(A\), the kinetic energy is \(\frac{1}{2}mV_L^2\), while at the horizontal point \(B\), the velocity is zero and the height reached is \(r\). Therefore,

\[ \frac{1}{2}mV_L^2+0=0+mgr \]

Hence,

\[ \boxed{V_L=\sqrt{2gr}} \]

If \[ V_L=\sqrt{2gr}, \] the body oscillates with an angular amplitude of \(90^\circ\), and the tension in the string when the string is in the horizontal position is \(T=0\).

If the velocity of the body at the lowest position is less than \(\sqrt{2gr}\), it oscillates with an angular amplitude \(\theta<90^\circ\).

Tension at an intermediate position during oscillation

Consider the body at position \(P\), where the string makes an angle \(\theta\) with the vertical. At this extreme position, the velocity of the body becomes zero:

\[ V_P=0 \]

Therefore, the centripetal force at this position is zero and the tension in the string is

\[ T = mg\cos\theta \qquad \left[ \frac{mV_P^2}{r}=0 \right] \]

Note 5: Maximum height and angle for oscillation

If the velocity of the body at the lowest point \(V_L\) is less than \(\sqrt{2gr}\), then the maximum height reached by the body just before its velocity becomes zero is given by

\[ \boxed{ h_1=\frac{V_L^2}{2g} } \]

The angle made by the string with the vertical when its velocity becomes zero is given by

\[ \boxed{ \cos\theta=1-\frac{V_L^2}{2gr} } \]

Case 2: Condition for leaving the circular path without looping

When the velocity given to the body at the lowest point is equal to \(\sqrt{5gr}\), then it just completes the vertical circle.

If the velocity given to the body at the lowest position is greater than \(\sqrt{2gr}\) and less than \(\sqrt{5gr}\), then the body crosses the horizontal position but cannot reach the highest position.

At a certain point, the tension in the string becomes zero and the body possesses some velocity. Now the only real force acting on the body at that position is the gravitational force. Therefore, the body follows its own parabolic path from that point.

Let \(V_L\) be the velocity of the body at the lowest position. The tension at any point \(P\) is given by

\[ T_P = \frac{mV_\theta^2}{r} + mg\cos\theta \]

When the tension becomes zero,

\[ T_P=0 \]

Therefore,

\[ \frac{mV_\theta^2}{r} + mg\cos\theta = 0 \]

Let the body rise through a vertical height \(h\) from the lowest point. From the conservation of mechanical energy,

\[ \frac{1}{2}mV_L^2 = \frac{1}{2}mV_\theta^2+mgh \]

Therefore,

\[ V_\theta^2=V_L^2-2gh \]

At the point where the string just becomes loose, substituting this value in the condition \(T_P=0\),

\[ \frac{m}{r}(V_L^2-2gh)+mg\left(\frac{r-h}{r}\right)=0 \]

Dividing by \(m\) and multiplying by \(r\),

\[ V_L^2-2gh+gr-gh=0 \]

Hence,

\[ 3gh=V_L^2+gr \]

\[ \boxed{ h=\frac{V_L^2+gr}{3g} } \]

If the string makes an angle \(\theta\) with the vertical at this position, then

\[ h=r(1-\cos\theta) \]

Therefore,

\[ r(1-\cos\theta) = \frac{V_L^2+gr}{3g} \]

Rearranging,

\[ \boxed{ \cos\theta = \frac{2}{3} – \frac{V_L^2}{3gr} } \]

Uniform Vertical Circular Motion

Consider a body of mass m tied at one end of a string and whirled in a vertical circle of radius r with uniform speed v by fixing the other end at O.

Motion in a vertical circle

Let T_θ be the tension in the string when it makes an angle θ with the vertical. At point P, the net force towards the centre is T_θ − mg cos θ. This net force provides the centripetal force at this point.

\(T_\theta-mg\cos\theta=\frac{mv^2}{r}\)

\[ \boxed{T_\theta=\frac{mv^2}{r}+mg\cos\theta} \]

When the body is at the lowest point (θ = 0°), the tension in the string is maximum.

\[ T_L=\frac{mv^2}{r}+mg \qquad (\text{maximum}) \]

When the body is at the highest point (θ = 180°), the tension in the string is minimum.

\[ T_H=\frac{mv^2}{r}-mg \qquad (\text{minimum}) \]

Note 1

The ratio of tensions at the lowest and highest positions is

\[ \frac{T_{\max}}{T_{\min}} = \frac{\frac{mv^2}{r}+mg} {\frac{mv^2}{r}-mg} \]

\[ \boxed{ \frac{T_{\max}}{T_{\min}} = \frac{v^2+gr}{v^2-gr} } \]

Note 2

The difference of tensions at the lowest and highest positions of the body is given by

\[ T_{\max}-T_{\min} = \left(\frac{mv^2}{r}+mg\right) – \left(\frac{mv^2}{r}-mg\right) \]

\[ \boxed{T_{\max}-T_{\min}=2mg} \]

Application 1: Bucket containing water moving in a vertical circle

A bucket containing water is made to revolve in a vertical circle, such that water does not fall even in the inverted position at the highest point of the path. If r is the radius of the circle, then the maximum time period of revolution of the bucket is

\[ \boxed{T=2\pi\sqrt{\frac{r}{g}}} \]

At the highest point of the path, if the velocity of the bucket \(V\geq\sqrt{gr}\), then water does not fall.

\[ r\omega\geq\sqrt{gr} \]

\[ \omega\geq\sqrt{\frac{g}{r}} \]

\[ T=\frac{2\pi}{\omega} \leq 2\pi\sqrt{\frac{r}{g}} \]

Hence, the maximum time period of revolution of the bucket is

\[ \boxed{T_{\max}=2\pi\sqrt{\frac{r}{g}}} \]

Application 2: Reaction of Road on Car

The reaction of the road on a car moving along a curved road depends upon whether the car is moving on a concave bridge or a convex bridge.

1) When car moves on a concave bridge

For a car moving on a concave bridge, the centripetal force is

\[ N-mg\cos\theta=\frac{mv^2}{r} \]

Therefore, the reaction of the road on the car is

\[ \boxed{ N=mg\cos\theta+\frac{mv^2}{r} } \]

2) When car moves on a convex bridge

For a car moving on a convex bridge, the centripetal force is

\[ mg\cos\theta-N=\frac{mv^2}{r} \]

Therefore, the reaction of the road on the car is

\[ \boxed{ N=mg\cos\theta-\frac{mv^2}{r} } \]

Torque

Definition: Torque is the turning effect of a force about a fixed point or axis. It is produced when a force tends to rotate a body about an axis.

Symbol: Torque is represented by τ (Greek letter tau).

Explanation

Consider a rigid body capable of rotating about a fixed point O. A force F is applied at point P, at a distance r from the axis of rotation. The turning effect of the force depends on the magnitude of the force, the distance from the axis, and the angle between the position vector and the force.

The magnitude of torque is

τ = rF sin θ

where r sin θ represents the perpendicular distance (moment arm) of the line of action of the force from the axis of rotation.

τ = F × perpendicular distance

If the force acts perpendicular to the radius, θ = 90°, and hence

τ = rF

Thus, the torque is maximum when the force acts perpendicular to the radius.

SI Unit

The SI unit of torque is newton metre (N m).

1 N m = 1 kg m² s⁻²

Dimensional Formula

From

τ = rF

we have

[τ] = [L][MLT⁻²]

Therefore,

[τ] = [ML²T⁻²]

Nature of Physical Quantity

Torque is a vector quantity.

Its direction is perpendicular to the plane containing the position vector r and force F, and is determined using the right-hand thumb rule.

Important Point: Torque depends not only on the magnitude of the applied force but also on its perpendicular distance from the axis of rotation. A larger perpendicular distance produces a greater turning effect for the same force.

Moment of Force or Torque

Moment of force or torque is defined as the turning effect of a force about a fixed point and is measured as the product of the magnitude of the force and the perpendicular distance of the line of action of the force from the fixed point.

Let a force F act on a particle at a point P in the XOY plane. Let the position vector of the point P with respect to the origin O be r.

Let θ be the angle between the position vector r and the force F.

The magnitude of the moment of the force or torque of the force F about the point O is given by:

\( \tau = F r_{\perp} \)

From the geometry of the figure,

\( \sin\theta = \frac{r_{\perp}}{r} \)

Therefore,

\( r_{\perp}=r\sin\theta \)

Hence,

\( \boxed{\tau=rF\sin\theta} \)

Meaning of the Symbols

r is the magnitude of the position vector.
F is the magnitude of the force F.
θ is the angle between r and F.
r_⊥ is the perpendicular distance from the point of rotation to the line of action of the force.

Torque in Vector Form

Torque of the force F about a point O in vector form is given by:

\( \boxed{\vec{\tau}=\vec{r}\times\vec{F}} \)

The direction of torque is perpendicular to the plane containing both r and F. Its direction can be determined by using the right-hand thumb rule.

Dimensions and SI Unit

Moment of force has dimensions:

\( \boxed{ML^2T^{-2}} \)

Its dimensions are the same as those of work or energy. However, moment of force is a very different physical quantity from work.

Moment of a force is a vector, while work is a scalar.

The SI unit of moment of force is newton metre (N m).

Torque by Resolving the Force into Components

The torque of a force acting at point P with respect to the origin O can also be determined by resolving the force into two components:

F cos θ along r
F sin θ perpendicular to r

The component F cos θ acts along the position vector r. Therefore, its line of action passes through O and produces no torque about O.

\( \tau_1=0(F\cos\theta)=0 \)

The component F sin θ acts perpendicular to r. Therefore, its torque about O is:

\( \tau_2=r(F\sin\theta) \)

Hence, the total torque of the force F about O is:

\( \tau=\tau_1+\tau_2 \)

\( \tau=0+r(F\sin\theta) \)

\( \boxed{\tau=r(F\sin\theta)} \)

Note: The magnitude of torque τ of a force F acting at a point P with respect to O depends on r, F and θ.

Maximum Torque

For given values of r and F, torque τ is maximum when:

\( \theta=90^\circ \)

Therefore,

\( \boxed{\tau_{\max}=rF} \)

Thus, torque is maximum when the force is perpendicular to the position vector.

Zero Torque

For given values of r and F, torque τ is zero when:

\( \theta=0^\circ \)

That is, the line of action of the force passes through the point O.

\( \boxed{\tau=0} \)

Important Note: In the formula for torque about a point,

\( \vec{\tau}=\vec{r}\times\vec{F} \)

r is the position vector of the point of application of the force with respect to the point about which we want to determine the torque.

Example: Torque About Different Points

Consider a force F acting at point A.

The torque of force F acting at point A with respect to O is given by:

\( \boxed{\vec{\tau}=\vec{r}_A\times\vec{F}} \)

Therefore, the magnitude of torque is:

\( \tau=r_AF\sin\theta \)

Here, θ is the angle between r_A and F.

Now consider the torque of the same force F acting at point A with respect to point B.

\( \boxed{\vec{\tau}=(\vec{r}_A-\vec{r}_B)\times\vec{F}} \)

Hence, the magnitude of torque is:

\( \tau=|\vec{r}_A-\vec{r}_B|F\sin\alpha \)

Here, α is the angle between (r_A − r_B) and the force F.

Torque on a Rigid Body About an Axis

In case of a single particle, if a single force or number of forces acting on the particle, then torque due to the force (or net force) about a point is determined by using the formula

\[ \vec{\tau}=\vec{r}\times\vec{F}_{net} \]

In case of a rigid body, if a single force or number of forces acting on it at different positions, then the torque of the force or forces acting on the body about the axis of rotation is determined by using

\[ \vec{\tau} = (\vec{r}_1\times\vec{F}_1) + (\vec{r}_2\times\vec{F}_2) +\cdots \]

Consider a rigid body. Let \(\vec{F}\) be the force acting on the body at point P whose position vector is \(\vec{r}\) with reference to the point O on the axis of rotation.

The torque acting on the body about the axis of rotation is

\[ \vec{\tau}=\vec{r}\times\vec{F} \]

If a number of forces are acting on the body at different positions of the body, then the net torque about the axis of rotation is the vector sum of torques due to all forces.

\[ \vec{\tau}_{net} = \vec{\tau}_1+\vec{\tau}_2+\vec{\tau}_3+\cdots \]

Note 1: Torque produced by a force about an axis can be zero in the following cases.

(i) If the force vector passes through the axis of rotation.

(ii) If the force vector is parallel to the axis of rotation.

Note 2: Moment of force or torque which tends to produce rotation in anticlockwise direction is taken as positive, whereas the moment of force which tends to produce rotation in clockwise direction is taken as negative.

Couple and Moment of Couple

Two equal, unlike parallel forces whose lines of action are not the same, constitute a couple.

The moment (or) torque of a couple is the product of either force and the perpendicular distance between the forces.

Moment of the couple is the vector sum of the torques produced by the two forces individually about a fixed point.

Figure: A couple consists of two equal and opposite parallel forces separated by a perpendicular distance d.

The effect of a couple on a rigid body is to produce rotation.

Let F and −F be the forces acting on a body simultaneously at points A and B as shown in the figure. As the lines of action of these forces do not coincide, we say that a couple is acting on the body.

The net force acting on the body is

\[ \vec{F}_{net}=\vec{F}-\vec{F}=\vec{0} \]

i.e., the net force exerted by a couple is zero.

Let \(\vec{r}_1\) and \(\vec{r}_2\) be the position vectors of points A and B about an axis passing through a fixed point O.

The torque produced by the force at A about O is

\[ \vec{\tau}_1=\vec{r}_1\times\vec{F} \]

Its turning effect is in the anticlockwise (positive) direction and the direction of torque is normal to the plane and outwards.

The torque produced by the force at B about O is

\[ \vec{\tau}_2=\vec{r}_2\times\vec{F} \]

Its turning effect is clockwise (negative) direction and its direction is normal to the plane and inwards.

Net torque produced by the couple about an axis passing through the point O is

\[ \vec{\tau}=\vec{\tau}_1+\vec{\tau}_2 \]

\[ \vec{\tau}=\vec{r}_1\times\vec{F}-\vec{r}_2\times\vec{F} \]

\[ \vec{\tau}=(\vec{r}_1-\vec{r}_2)\times\vec{F} \]

The magnitude of the torque is

\[ \tau=Fd \]

where \(\left|\vec{r}_1-\vec{r}_2\right|=d\)

Note 1: The moment of a given couple is constant about any point.

Examples for Couple

When we turn a door knob, we produce a couple on the knob.
When we wind a watch, we produce a couple to the winding key with our fingers.
When we use a screw driver, we produce a couple on its head.
When we turn a tap, we produce a couple on its head.
While spinning a ball or a top, we produce a couple.

Rigid Body

A rigid body is an ideal object whose shape and size can never be changed, whatever may be the external influence.

In all possible motions of a rigid body as a whole, the separation between any two parts of the rigid body must remain constant. Thus, the distance between any two points of a rigid body does not change even when the body undergoes motion.

All the points in a rigid body rotate through the same angle and in the same sense, except the particles lying on the axis of rotation. Therefore, the instantaneous angular velocities of all the particles in a rigid body must be the same.

A rigid body is an idealized concept. In reality, every material body undergoes some amount of deformation when an external force is applied, but for many physical problems this deformation is so small that the body can be treated as perfectly rigid.

Mechanical Equilibrium of a Rigid Body

Mechanical equilibrium: A rigid body is said to be in mechanical equilibrium if it has both translational and rotational equilibrium.

Consider a rigid body as shown in the figure. Let the forces acting on the body be \(\vec{F}_1,\vec{F}_2,\vec{F}_3,\vec{F}_4,\ldots,\vec{F}_n\) at the points A, B, C, D, …. Let \(PQ\) be the axis of rotation of the body and let \(\vec{r}_1,\vec{r}_2,\vec{r}_3,\vec{r}_4,\ldots,\vec{r}_n\) be the position vectors of the points A, B, C, D, …. with respect to the point O on the axis of rotation.

Condition for translational equilibrium: If the vector sum of all the forces acting on a body is zero, then the body is in translational equilibrium.

\(\vec{F}_1+\vec{F}_2+\vec{F}_3+\vec{F}_4+\cdots+\vec{F}_n=\sum_{i=1}^{n}\vec{F}_i=\vec{0}\)

According to Newton’s second law,

\(\vec{F}=\frac{d\vec{p}}{dt}\)

\(\vec{0}=\frac{d\vec{p}}{dt}\)

\(\therefore\vec{p}=\text{constant}\)

i.e., when the body is in translational equilibrium, the total linear momentum of the body is constant.

Condition for rotational equilibrium: If the vector sum of all the torques about the given axis of rotation of the body is zero, then the body is in rotational equilibrium.

The torques \(\vec{\tau}_1,\vec{\tau}_2,\vec{\tau}_3,\vec{\tau}_4,\ldots,\vec{\tau}_n\) are the torques produced by the forces acting on the body at different positions about the given axis of rotation.

\(\vec{\tau}_1+\vec{\tau}_2+\vec{\tau}_3+\vec{\tau}_4+\cdots+\vec{\tau}_n=\sum_{i=1}^{n}\vec{\tau}_i=\vec{0}\)

We know that

\(\vec{\tau}=\frac{d\vec{L}}{dt}\)

\(\vec{0}=\frac{d\vec{L}}{dt}\)

\(\therefore\vec{L}=\text{constant}\)

i.e., in rotational equilibrium, the angular momentum of the body is constant.

Banking of Roads

Banking of roads is the process of raising the outer edge of a curved road above the inner edge so that the road surface is inclined at an angle to the horizontal. This inclination helps provide the necessary centripetal force for a vehicle to move safely along the curved path.

When a vehicle moves along a circular road, it requires a centripetal force directed towards the centre of the circular path. On an ordinary level road, this force is mainly provided by friction between the tyres and the road.

On a banked road, the normal reaction of the road has a component directed towards the centre of the circular path. This component helps provide the required centripetal force. Therefore, the vehicle can negotiate the curve more safely and with less dependence on friction.

For a vehicle of mass m moving with speed v on a road of radius r, banked at an angle θ, the condition for ideal banking, when friction is neglected, is

\(N\cos\theta=mg\)

\(N\sin\theta=\frac{mv^2}{r}\)

Dividing the two equations,

\(\tan\theta=\frac{v^2}{rg}\)

\(v=\sqrt{rg\tan\theta}\)

Thus, for a given radius of the road and a given banking angle, there is a particular speed at which a vehicle can move around the curve without requiring friction to provide the centripetal force.

Importance of Banking of Roads

Banking reduces the dependence on friction between the tyres and the road while a vehicle is taking a turn. It provides a component of the normal reaction towards the centre of the circular path and thereby helps the vehicle negotiate the curve safely.

Applications of Banking of Roads

1. Curved highways: Roads are banked at bends to help vehicles negotiate curves safely at appropriate speeds.

2. Railway tracks: At a curved railway track, the outer rail is raised above the inner rail. This is the railway equivalent of banking and helps provide the required centripetal force.

3. Race tracks: Curved portions of racing tracks are highly banked so that vehicles can negotiate curves at high speeds.

4. Circular tracks: Banking is also used in specially designed circular tracks where vehicles or cyclists have to move along curved paths.

Overturning of Vehicle on an Unbanked Path

When a vehicle negotiates a sharp turn on a level road, its driver should be careful about two possible mishaps:

1) Skidding and 2) Overturning.

The cause of skidding and the speed limit imposed for its avoidance have already been discussed earlier. Here, we study why a vehicle overturns while moving along a curved path and how it can be prevented.

Suppose the radius of the curved path is r, the speed of the vehicle is v and the mass of the vehicle is m. The centre of mass C of the vehicle is at a height h above the ground. The distance between the inner and outer wheels of the vehicle is 2ℓ.

The forces acting on the vehicle are:

1) The force of gravity mg acting vertically downwards through C.

2) The normal reactions N₁ and N₂ acting vertically upwards through the inner and outer wheels respectively.

3) The forces of static friction f₁ and f₂ acting horizontally towards the centre of the curved path at the bottom of the inner and outer wheels respectively.

Here the two frictional forces provide the centripetal force. Therefore,

f₁ + f₂ = mv² r

Since the vehicle has no vertical acceleration,

N₁ + N₂ = mg

The various torques acting on the vehicle about the centre of mass are:

i) Torque due to N₁:

τ₁ = N₁ℓ (clockwise)

ii) Torque due to N₂:

τ₂ = N₂ℓ (anticlockwise)

iii) Torque due to weight mg:

τ₃ = 0

The weight acts through the centre of mass.

iv) Torque due to f₁:

τ₄ = f₁h (clockwise)

v) Torque due to f₂:

τ₅ = f₂h (clockwise)

For the vehicle not to overturn, the net torque about the centre of mass C is zero.

i.e. the sum of anticlockwise torques = the sum of clockwise torques.

τ₂ = τ₁ + τ₄ + τ₅

N₂ℓ = N₁ℓ + (f₁ + f₂)h

N₂ − N₁ = (f₁ + f₂)h ℓ

On solving the equations, we get

N₁ = m 2 (g − v²h rℓ ) and N₂ = m 2 (g + v²h rℓ )

Hence, N₂ is always positive, but N₁ may become zero (or negative if v is sufficiently large). In that event, the inner wheels leave the ground and the car tends to overturn towards the outer edge of the road.

At the limiting condition, N₁ = 0. Therefore,

m 2 (g − v²h rℓ ) = 0

V = √ grℓ / h

which should be the speed for the limiting case.

Precautions to Avoid Overturning at a Curved Path

1) The speed v must be reduced before entering the curved path.

2) The road must be banked with the suitable angle θ.

3) The height of the centre of gravity h above the ground should be lowered.

4) The wheel base 2ℓ should be large.

5) The radius of curvature must be large.

Note 1: In case of overturning of a vehicle on a level road, the inner wheels leave the ground first.

Moment of Inertia (M.I.)

In translatory motion, the property of a body by virtue of which it opposes any change of state of rest or of uniform motion along a straight line is inertia. In translatory motion, mass is the measure of inertia.

Similarly, in rotational motion, the property of a rigid body by virtue of which it opposes any change in its state of rest or of uniform circular motion about the axis of rotation is rotational inertia. A measure of rotational inertia is called moment of inertia. Moment of inertia is the rotational analogue of inertia in translatory motion.

Analytically, for a particle of mass m rotating in a circle of radius r, the moment of inertia about the axis of rotation is given by

I = mr²

The moment of inertia of a rigid body about an axis is defined as the sum of the products of the masses of its different particles and the squares of their respective perpendicular distances from the axis of rotation.

Consider a rigid body rotating about an axis. Suppose the body is composed of a large number of small particles having masses m₁, m₂, …, m_n, at perpendicular distances r₁, r₂, …, r_n respectively from the axis of rotation.

The moment of inertia of the particle of mass m₁ about the axis is m₁r₁². Similarly, the moments of inertia of the other particles are m₂r₂², m₃r₃², and so on.

Therefore, the moment of inertia of the whole body is equal to the sum of the moments of inertia of all its constituent particles.

I = m₁r₁² + m₂r₂² + m₃r₃² + \cdots + m_nr_n²

I = \sum_i=1ⁿ m_ir_i²

where m_i is the mass of the ith particle and r_i is its perpendicular distance from the axis of rotation.

For a body having a continuous mass distribution, the summation is replaced by integration:

I = ∫ r² dm

If the density of the body is ρ and dV is an infinitesimal volume element, then

I = ∫ r²ρ dV

SI Unit of Moment of Inertia

The SI unit of moment of inertia is kilogram metre squared.

SI unit = kg m²

Dimensional Formula of Moment of Inertia

From the relation

I = mr²

the dimensional formula is

[M L² T⁰]

Nature of Moment of Inertia

Moment of inertia is a tensor quantity. It is not a vector quantity because it does not have a direction in the same sense as a vector. It is also not simply a scalar quantity because its value depends on the chosen axis of rotation.

For a particular body, the moment of inertia can have different values about different axes.

Properties and Characteristics of Moment of Inertia

1. Dependence on the axis of rotation: The moment of inertia of a given body depends on the position and orientation of the axis of rotation.

2. Dependence on mass distribution: For a given axis, the moment of inertia depends on how the mass of the body is distributed about that axis. The farther the mass is distributed from the axis, the greater the moment of inertia.

3. Dependence on shape and size: For a given mass and axis, the moment of inertia depends on the shape and size of the body. Thus, two bodies having the same mass can have different moments of inertia if their shapes or mass distributions are different.

4. Effect of density: If a steel sphere and a wooden sphere have the same mass, the moment of inertia about their diameters is greater for the wooden sphere because the wooden sphere has a larger size and its mass is distributed farther from the axis.

5. Effect of mass: If a steel sphere and a wooden sphere have the same radius, the moment of inertia about their diameters is greater for the steel sphere because the steel sphere has greater mass for the same size.

6. Effect of shape: If a steel sphere and a steel plate have the same mass, the moment of inertia about a normal axis through the centre of the plate can be greater than that of the sphere about its diameter because the mass of the plate is distributed differently with respect to the axis.

7. Effect of distribution of mass: For a given shape, size, mass and axis, the moment of inertia depends on the distribution of mass within the body. Moving more mass farther from the axis increases the moment of inertia.

For example, when a person rotating about an axis through the centre of mass stretches the arms outward, the moment of inertia increases because the mass of the arms is farther from the axis.

8. Effect of temperature: When the temperature of a body increases, its dimensions generally increase due to thermal expansion. Consequently, its moment of inertia about a fixed corresponding axis also increases.

For small temperature changes, the change in moment of inertia may be written as

ΔI = 2IαΔT

where α is the coefficient of linear expansion and ΔT is the change in temperature.

9. Independence from rotational motion: For a rigid body of fixed mass distribution and fixed axis, the moment of inertia is independent of angular velocity, frequency and time period of rotation.

10. Rotational analogue of mass: Moment of inertia plays the same role in rotational motion as mass plays in translational motion. A body having a large moment of inertia is more difficult to start rotating or to stop once it is rotating.

A large moment of inertia also helps to maintain uniform rotational motion. This is one reason why flywheels are used in stationary engines.

Physical Significance of Moment of Inertia

Moment of inertia represents the resistance offered by a body to a change in its rotational motion. It depends not only on the total mass of the body but also on how far that mass is distributed from the axis of rotation.

Thus, for the same mass, a body with more of its mass concentrated farther from the axis has a larger moment of inertia.

Radius of Gyration (K)

Radius of gyration of a rigid body about an axis of rotation is the distance between the axis of rotation and a point at which the whole mass of the body can be supposed to be concentrated so that its moment of inertia would be the same as with the actual distribution of mass.

It is the distance whose square when multiplied with the mass of the body gives the moment of inertia of the body about the given axis.

Consider a rigid body of mass M rotating about an axis of rotation OO′. We can imagine that the total mass M is concentrated at point P. Its perpendicular distance from the axis of rotation is radius of gyration, denoted by K.

The moment of inertia of the body about the axis OO′ is MK². This must be equal to the moment of inertia of the body with the actual distribution of mass.

\[ I = MK^2 = \sum_{i=1}^{n} m_i r_i^2 \]

\[ K = \sqrt{\frac{I}{M}} \]

\[ K = \sqrt{ \frac{ \displaystyle\sum_{i=1}^{n} m_i r_i^2 }{M} } \]

The SI unit of radius of gyration is metre (m) and its CGS unit is centimetre (cm).

Since radius of gyration is a distance, its dimensional formula is

\[ [K] = [M^0L^1T^0] \]

Note:

The radius of gyration depends on the following factors:

1. The position of the axis of rotation.

2. The distribution of mass about the axis of rotation.

3. Temperature: The radius of gyration increases with increase in temperature.

4. The radius of gyration is independent of angular velocity, frequency and time period of rotation.

Principle of Parallel and Perpendicular Axes

There are two general theorems or principles of moment of inertia which sometimes enable us to determine the moment of inertia of a body about an axis when the moment of inertia about some other axis is known.

Parallel Axis Theorem

Statement: The moment of inertia of a rigid body about any axis is equal to the moment of inertia of the body about a parallel axis passing through its centre of mass, plus the product of the total mass of the body and the square of the perpendicular distance between the two parallel axes.

\[ \boxed{I=I_{\mathrm{CM}}+Md^2} \]

Here, \(I\) is the moment of inertia of the body about the given axis, \(I_{\mathrm{CM}}\) is the moment of inertia about the parallel axis passing through the centre of mass, \(M\) is the total mass of the body, and \(d\) is the perpendicular distance between the two parallel axes.

Geometrical Representation

Figure: Geometrical representation of the Parallel Axis Theorem

Proof

Consider a rigid body of total mass \(M\). Let \(Z’\) be an axis passing through the centre of mass of the body and let \(Z\) be another axis parallel to \(Z’\), separated from it by a perpendicular distance \(d\).

Consider a small element of mass \(dm\) situated at a perpendicular distance \(r’\) from the axis \(Z’\). Let \(r\) be its perpendicular distance from the parallel axis \(Z\).

The position of the mass element relative to the axis \(Z\) can be expressed in terms of the position relative to the axis \(Z’\) and the displacement between the two parallel axes.

\[ \vec r=\vec r’+\vec d \]

Taking the square of the magnitude of both sides,

\[ r^2=(\vec r’+\vec d)\cdot(\vec r’+\vec d) \]

Expanding the scalar product,

\[ r^2=\vec r’\cdot\vec r’ +\vec d\cdot\vec d +2\vec r’\cdot\vec d \]

Since

\[ \vec r’\cdot\vec r’=r’^2 \]

and

\[ \vec d\cdot\vec d=d^2, \]

while the scalar product of two vectors is given by

\[ \vec r’\cdot\vec d=r’d\cos\theta, \]

we obtain

\[ \boxed{r^2=r’^2+d^2+2r’d\cos\theta} \]

Thus, the square of the perpendicular distance \(r\) from the mass element to the axis \(Z\) is related to \(r’\), \(d\), and the angle \(\theta\) by the above expression.

The moment of inertia of the body about the axis \(Z\) is

\[ I=\int r^2\,dm \]

Substituting the expression for \(r^2\),

\[ I=\int \left(r’^2+d^2+2r’d\cos\theta\right)dm \]

Separating the terms of the integral,

\[ I= \int r’^2\,dm + \int d^2\,dm + 2d\int r’\cos\theta\,dm \]

Since \(d\) is constant for all mass elements of the body,

\[ I= \int r’^2\,dm + d^2\int dm + 2d\int r’\cos\theta\,dm \]

The first integral represents the moment of inertia of the body about the axis \(Z’\), which passes through the centre of mass. Therefore,

\[ \int r’^2\,dm=I_{\mathrm{CM}} \]

The second integral gives the total mass of the body:

\[ \int dm=M \]

Now, \(r’\cos\theta\) represents the component of the position vector \(r’\) along the direction of the displacement \(d\). Since \(Z’\) passes through the centre of mass, the first moment of the mass distribution about the centre of mass is zero. Hence,

\[ \int r’\cos\theta\,dm=0 \]

Therefore, the cross term vanishes and we get

\[ I=I_{\mathrm{CM}}+Md^2 \]

Hence, the Parallel Axis Theorem is proved.

\[ \boxed{\displaystyle I=I_{\mathrm{CM}}+Md^2} \]

Meaning of the Theorem

The theorem shows that when the axis of rotation is shifted from an axis passing through the centre of mass to another parallel axis at a distance \(d\), the moment of inertia increases by \(Md^2\).

Therefore, if the moment of inertia about an axis through the centre of mass is known, the moment of inertia about any parallel axis can be calculated directly using

\[ \boxed{I=I_{\mathrm{CM}}+Md^2} \]

Perpendicular Axis Theorem

Statement: For a plane lamina, the moment of inertia about an axis perpendicular to the plane of the lamina and passing through a point is equal to the sum of the moments of inertia about two mutually perpendicular axes lying in the plane of the lamina and passing through the same point.

\[ \boxed{I_Z=I_X+I_Y} \]

Here, \(I_Z\) is the moment of inertia about the axis \(Z\) perpendicular to the plane of the lamina, while \(I_X\) and \(I_Y\) are the moments of inertia about two mutually perpendicular axes \(X\) and \(Y\) lying in the plane of the lamina.

Geometrical Representation

Figure: Perpendicular axes through a point of a plane lamina

Proof

Consider a plane lamina lying in the \(XY\)-plane. Let the \(X\)-axis and \(Y\)-axis lie in the plane of the lamina and intersect at the point \(O\). Let the \(Z\)-axis pass through \(O\) perpendicular to the plane of the lamina.

Consider a small element of mass \(dm\) situated at the point \(P\), having coordinates \((x,y)\) with respect to the \(X\)- and \(Y\)-axes.

Let \(r\) be the perpendicular distance of the mass element \(dm\) from the \(Z\)-axis.

Geometrical Relation

Since the \(X\)- and \(Y\)-axes are mutually perpendicular, the coordinates \(x\), \(y\), and the distance \(r\) form a right-angled triangle.

Therefore, by applying the Pythagorean theorem,

\[ r^2=x^2+y^2 \]

This relation is the geometrical basis of the Perpendicular Axis Theorem.

Moment of Inertia About the \(Z\)-Axis

The moment of inertia of the lamina about the \(Z\)-axis is

\[ I_Z=\int r^2\,dm \]

Substituting the geometrical relation \(r^2=x^2+y^2\),

\[ I_Z=\int (x^2+y^2)\,dm \]

Expanding the integral,

\[ I_Z=\int x^2\,dm+\int y^2\,dm \]

Identifying the Moments of Inertia About the \(X\)- and \(Y\)-Axes

The perpendicular distance of the mass element \(dm\) from the \(X\)-axis is \(y\). Therefore, the moment of inertia about the \(X\)-axis is

\[ I_X=\int y^2\,dm \]

Similarly, the perpendicular distance of the mass element \(dm\) from the \(Y\)-axis is \(x\). Therefore, the moment of inertia about the \(Y\)-axis is

\[ I_Y=\int x^2\,dm \]

Substituting these expressions into the equation for \(I_Z\),

\[ I_Z=I_Y+I_X \]

Hence,

\[ \boxed{I_Z=I_X+I_Y} \]

Therefore, the Perpendicular Axis Theorem is proved.

Final Result

\[ \boxed{I_Z=I_X+I_Y} \]

Thus, the moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two mutually perpendicular axes lying in the plane of the lamina and passing through the same point.

Conditions for Applying the Theorem

1. The body must be a plane lamina or a body whose mass distribution is confined to a plane.

2. The \(X\)- and \(Y\)-axes must lie in the plane of the lamina.

3. The \(X\)- and \(Y\)-axes must be mutually perpendicular.

4. The \(Z\)-axis must pass through their point of intersection and be perpendicular to the plane of the lamina.

5. The three axes must therefore intersect at the same point.

Important Note

The Perpendicular Axis Theorem is applicable to plane laminae. It cannot, in general, be applied directly to an arbitrary three-dimensional body.

Table of Contents

Vertical Circular Motion (with Variable Speed)

Tension in the String at Any Point

Note 3:

Note 4:

Motion in Vertical Circle (Condition for Looping a Loop)

Forces acting on the body at the lowest point A

Forces acting on the body at the highest point B

Condition for completing the vertical circle

Relation between velocities at the lowest and highest points

Minimum velocity at the lowest point

Minimum tension at the lowest point

Application 1

Application 2

Application 3

Note 1: Critical velocity at any point of the vertical circle

Note 2: Minimum tension at any point of the vertical circle

Note 3: Velocity and tension at different positions

Note 4: Non-uniform circular motion in a vertical plane

Case 1: Condition for oscillating about the lowest position

Tension at an intermediate position during oscillation

Note 5: Maximum height and angle for oscillation

Case 2: Condition for leaving the circular path without looping

Uniform Vertical Circular Motion

Note 1

Note 2

Application 1: Bucket containing water moving in a vertical circle

Application 2: Reaction of Road on Car

1) When car moves on a concave bridge

2) When car moves on a convex bridge

Torque

Explanation

SI Unit

Dimensional Formula

Nature of Physical Quantity

Moment of Force or Torque

Meaning of the Symbols

Torque in Vector Form

Dimensions and SI Unit

Torque by Resolving the Force into Components

Maximum Torque

Zero Torque

Example: Torque About Different Points

Torque on a Rigid Body About an Axis

Couple and Moment of Couple

Examples for Couple

Rigid Body

Mechanical Equilibrium of a Rigid Body

Banking of Roads

Importance of Banking of Roads

Applications of Banking of Roads

Overturning of Vehicle on an Unbanked Path

Precautions to Avoid Overturning at a Curved Path

Moment of Inertia (M.I.)

SI Unit of Moment of Inertia

Dimensional Formula of Moment of Inertia

Nature of Moment of Inertia

Properties and Characteristics of Moment of Inertia

Physical Significance of Moment of Inertia

Radius of Gyration (K)

Principle of Parallel and Perpendicular Axes

Parallel Axis Theorem

Geometrical Representation

Proof

Meaning of the Theorem

Perpendicular Axis Theorem

Geometrical Representation

Proof

Geometrical Relation

Moment of Inertia About the \(Z\)-Axis

Identifying the Moments of Inertia About the \(X\)- and \(Y\)-Axes

Final Result

Conditions for Applying the Theorem

Important Note