### Kinetic Energy of a Rigid Body

• Obtain an expression for kinetic energy of a rigid body in pure rotation.

Consider a rigid body rotating about an axis with the angular velocity $\omega$. Let the axis of rotation passes through the CG of the body. In case of pure rotation CG of the body doesn’t have any linear velocity and hence the body has only angular velocity.
Suppose that the rigid body is composed of large no. of elementary particles having masses m$_1$, m$_2$, m$_3$…….m$_n$ at the distances r$_1$, r$_2$, r$_3$………………..r$_n$ respectively from the axis of rotation. Here by the definition of rigid body, each particle has the same angular velocity equal to that of the rigid body. But the linear velocity is different for different particles due to their positions. i.e. $v_1=r_1\omega,v_2=r_2\omega,v_3=r_3\omega,..........,v_n=r_n\omega$ So that, total kinetic energy of the body is given by, $T=\dfrac{m_1v_1^2}{2}+\dfrac{m_2v_2^2}{2}+.......+ \dfrac{m_nv_n^2}{2}$ $T=\dfrac{\omega^2}{2}(m_1r_1^2+m_2r_2^2+......+m_nr_n^2)$

### Solved Example:

#### 36-1-01

In linear motion, energy is given by $\dfrac{1}{2}$ mv$^2$. Similarly, in rotational motion, rotational energy is given by:

### Solved Example:

#### 36-1-02

A thin straight uniform rod AB of length L and mass M, held vertically with the end A on horizontal floor, is released from rest and is allowed to fall. Assuming that the end A (of the rod) on the floor does not slip, what will be the linear velocity of the end B when it strikes the floor?

Solution:

When the rod falls, its gravitational potential energy gets converted into rotational kinetic energy. Therefore we have

$\dfrac{MgL}{2} = \dfrac{1}{2} I \omega^2$

where I is the moment of inertia of the rod about a normal axis passing through its end and $\omega$ is the angular velocity of the rod when it strikes the floor.

[Note that initially the centre of gravity of the rod is at a height L/2 and that’s why the initial potential energy is MgL/2]

The moment of inertia of the rod about the normal axis through its end is given by

$I = \dfrac{ML^2}{3}$

The moment of inertia of the rod about an axis through its centre and perpendicular to its length is ML$^2$/12. On applying parallel axes theorem, the moment of inertia about a parallel axis through the end is:

$\dfrac{ML^2}{12} + M \left(\dfrac{L}{2}\right)^2 = \dfrac{ML^2}{3}$ Substituting for I in Eq.(i), we have $\dfrac{MgL}{2} = \dfrac{1}{2} (\dfrac{ML^2}{3}) \omega^2$

Therefore $\omega = \sqrt{(3g/L)}$

The linear velocity v of the end B of the rod is given by, $v = \omega L = \sqrt{(3gL)}$

### Solved Example:

#### 36-1-03

A block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 m/s$^2$). The speed of the block when it reaches the point Q is: (JEE Advanced 2013)

Solution:
Kinetic energy of block at Q $KE = mgR \sin 30^\circ - 150\ J$ $\dfrac{1}{2}mv^2 = 1 \times 10 \times 40 \times \dfrac{1}{2} - 150 = 50\ J$ $v = 10\ m/s$

### Solved Example:

#### 36-1-04

A solid sphere is rolling on a frictionless surface, shown in figure with a translational velocity v m/s. If it is to climb the inclined surface then v should be:

Solution:
Applying law of conservation of energy for rotating body, \begin{align*} \dfrac{1}{2}mv^{2} &+\dfrac{1}{2}I \omega^{2} &\geq mgh\\ \dfrac{1}{2}mv^{2} &+\dfrac{1}{2}\left(\dfrac{2}{5}mr^{2} \right) \left( \dfrac{v}{r}\right) ^{2} &\geq mgh\\ \dfrac{v^{2}}{2}&+\dfrac{2v^{2}}{10} &\geq gh\\ \end{align*} $v \geq \sqrt{\dfrac{10}{7}}gh$