Question

Consider a rigid body that rotates but whose center of mass is at rest. True or false: the rotational kinetic energy of the entire object is equivalent to the sum of the translational kinetic energy of each small piece of the object.

Answer 1

Answer:

True

Explanation:

Let us consider a rigid body which contains tiny particles of mass dm each. and the angular velocity ω for each of the particle is same and teh radius of rotation is r.

the linear velocity, v = r ω

The total kinetic energy is given by

$K=\int \frac{1}{2}dm v^{2}$

$K=\int \frac{1}{2}dm r^{2}\omega^{2}$

$K=\frac{1}{2}\times \omega^{2}\int dm r^{2}$

K = 1/2 Iω^2

where I is the moment of inertia.

Answer 2

The rotational kinetic energy should be equal to the sum of translational kinetic energy for every small piece of the object. The given statement is true.

Kinetic energy:

Since the rigid body shoud be made of small particles so the mass should contain the similar angular velocity $\omega$ and radius r.

So, the linear velocity should be $v=r\omega$

Now

the kinetic energy of the particle should be $dK=\frac{1}{2} dm\cdot (r \omega)^2$

And,

total kinetic energy of the system K should be$=\int \frac{1}{2} dm\cdot (r \omega)^2 K=\frac{1}{2}\omega ^{2}\int dm\cdot (r)^2$

The quantity $\int dm\cdot (r)^2$ is called as the moment of inertia of the rigid body.

Also, $K=\frac{1}{2}I\omega ^{2}$ called as the total rotational energy of the system.

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