Learning Objectives:

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

Kinetic Energy due to Translational Motion:

Kinetic Energy due to Translational Motion (in x-y plane):

Kinetic Energy due to Rotational Motion:

Kinetic Energy about Instantaneous Center:

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:

A. $\dfrac{1}{2} \times I \times \omega$

B. $\dfrac{1}{2} \times I^2 \times \omega$

C. $\dfrac{1}{2} \times I \times \omega^2$

D. $\dfrac{1}{2} \times I^2 \times \omega^2$

Correct Answer: C

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?

A. $\sqrt{(3L/g)}$

B. $\sqrt{(3gL)}$

C. $\sqrt{(2g/L)}$

D. $\sqrt{(3g/L)}$

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

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.

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

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:

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

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

Correct Answer: B

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:

A. 5 m/s

B. 10 m/s

C. 10$\sqrt{3}$ m/s

D. 20 m/s

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

Correct Answer: B

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:

A. $\geq \sqrt{gh}$

B. $\geq \sqrt{\dfrac{10}{7}gh}$

C. $\geq \sqrt{2gh}$

D. $\geq \dfrac{10}{7}gh$

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*} \[v \geq \sqrt{\dfrac{10}{7}}gh\]

Correct Answer: A

Solved Example: 36-1-05

An object of mass 2000 g possesses 100 J kinetic energy. The object must be moving with a speed of:

A. 10.0 m/s

B. 11.1 m/s

C. 11.2 m/s

D. 12.1 m/s

\begin{align*} KE &= \dfrac{1}{2}mv^2\\ 100 &= \dfrac{1}{2}(2)v^2\\ 100 &= v^2\\ 10\ \mathrm{m/s} &= v \end{align*}

Correct Answer: A

Solved Example: 36-1-06

A thin disc and a thin ring, both have mass M and radius R. Both rotate about axes through their centre of mass and are perpendicular to their surfaces at the same angular velocity. Which of the following is true?

A. The ring has higher kinetic energy

B. The disc has higher kinetic energy

C. The ring and the disc have the same kinetic energy

D. Kinetic energies of both the bodies are zero since they are not in linear motion

Correct Answer: A

Solved Example: 36-1-07

The expression $\dfrac{I\omega^2}{2}$ represents Rotational ___________:

A. Kinetic energy

B. Angular momentum

C. Torque

D. Power

Correct Answer: A