Question

A solid cylinder of mass M and radius R
rotates about its axis with angular speed w.
Its rotational kinetic energy is
​

Answer 1

Answer:

Correct option is

B

K

rot

′

=

3

1

K

rot

Moment of inertia of the solid cylinder I=

2

1

mr

2

Thus initial kinetic energy of the system K

rot

=

2

1

Iw

2

=

4

1

mr

2

w

2

Final moment of inertia of the system I

′

=

2

1

mr

2

+

2

1

(2m)r

2

=

2

3

mr

2

Let the final angular velocity of the system be w

′

.

Using conservation of angular momentum : Iw=I

′

w

′

∴

2

1

mr

2

w=

2

3

mr

2

w

′

⟹w

′

=

3

w

Thus final kinetic energy of the system K

rot

′

=

2

1

I

′

(w

′

)

2

=

12

1

mr

2

w

2

⟹ K

rot

′

=

3

K

rot

Answer 2

Rotational Kinetic Energy :

• Cylinder has mass M and radius R, rotates about the geometrical axis with an angular velocity of $\omega$.

$KE = \dfrac{1}{2} \times I \times { \omega}^{2}$

$\implies KE = \dfrac{1}{2} \times \dfrac{M{R}^{2} }{2} \times { \omega}^{2}$

$\implies KE = \dfrac{M{R}^{2} { \omega}^{2} }{4}$

So, final answer is :

$\boxed{ KE = \dfrac{M{R}^{2} { \omega}^{2} }{4} }$