Question

Question No. 39
If rotational kinetic energy is 40% of total K.E. for a rolling body then find out shape of body :-
O Ring
O Hollow sphere
O Solid sphere
O Solid cylinder​

Answer 1

Answer:

Hollow sphere

Explanation:

Given :

The rotational kinetic energy is 40% of total K.E. for a rolling body.

To find :

the shape of body

Solution :

The total Kinetic Energy of a rolling body is equal to the sum of the translational Kinetic Energy and Rotational Kinetic Energy.

  translational kinetic energy =  $\sf \dfrac{1}{2}mv^2$

  rotational kinetic energy =  $\sf \dfrac{1}{2}mv^2 \bigg(\dfrac{K^2}{R^2} \bigg)$

The total Kinetic energy  

     =  $\sf \dfrac{1}{2}mv^2+\dfrac{1}{2}mv^2 \bigg(\dfrac{K^2}{R^2} \bigg)$

    =  $\sf \dfrac{1}{2}mv^2 \bigg(1+\dfrac{K^2}{R^2} \bigg)$

where

v is the velocity of the center of mass

m is the mass of the rolling body

K is the radius of the gyration

R is the radius of the body

As given,

   $\sf \dfrac{1}{2}mv^2 \bigg(\dfrac{K^2}{R^2} \bigg) = 40 \% \ of \ \sf \dfrac{1}{2}mv^2 \bigg(1+\dfrac{K^2}{R^2} \bigg) \\\\ \sf \dfrac{1}{2}mv^2 \bigg(\dfrac{K^2}{R^2} \bigg) = \dfrac{40}{100} \times \dfrac{1}{2}mv^2 \bigg(1+\dfrac{K^2}{R^2} \bigg) \\\\ \sf \dfrac{K^2}{R^2}=\dfrac{2}{5} \bigg(1+\dfrac{K^2}{R^2} \bigg) \\\\ \sf \dfrac{K^2}{R^2}=\dfrac{2}{5}+\dfrac{2K^2}{5R^2} \\\\ \sf \dfrac{K^2}{R^2}-\dfrac{2K^2}{5R^2}=\dfrac{2}{5} \\\\ \sf \dfrac{3K^2}{5R^2}=\dfrac{2}{5}$

  $\longrightarrow \boxed{\sf \dfrac{K^2}{R^2}=\dfrac{2}{3}}$

Now, let's look at options

• Ring

Moment of Inertia, I = MR² = MK²

R² = K²

K²/R² = 1

• Hollow sphere

Moment of Inertia, I = 2MR²/3 = MK²

2R²/3 = K²

K²/R² = 2/3

• Solid sphere

Moment of Inertia, I = 2MR²/5 = MK²

2R²/5 = K²

K²/R² = 2/5

• Solid cylinder

Moment of Inertia, I = MR²/2 = MK²

R²/2 = K²

K²/R² = 1/2

As we got K²/R² = 2/3, the shape of the rolling body is hollow sphere.