Question

A circular disc rolls down an inclined plane. The fraction of its total energy associated with its rotational motion is
1/4
1/3
1/2
1

Answer 1

Given :

A circular disc rolls down an inclined plane.

To Find :

The fraction of its rotational kinetic energy associated with total kinetic energy.

Solution :

★ Rotational kinetic energy associated with rotating body is given by

• $\underline{\boxed{\bf{\orange{K_R=\dfrac{1}{2}mv^2\bigg(\dfrac{k^2}{R^2}\bigg)}}}}$

★ Total kinetic energy associated with rotating body is given by

• $\underline{\boxed{\bf{\blue{K_T=\dfrac{1}{2}mv^2\bigg(1+\dfrac{k^2}{R^2}\bigg)}}}}$

m denotes mass of objectt

v denotes linear velocity

k denotes radius of gyration

R denotes radius of object

⧪ Moment of inertia of a circular disc rotating about an axis passing through its centre and perpendicular to the plane is given by

➠ I = mR²/2

We know that, I = mk²

➠ mk² = mR²/2

➠ k²/R² = 1/2

By taking ratio of rotational kinetic energy and total kinetic energy;

$:\implies\sf\:\dfrac{K_R}{K_T}=\dfrac{\dfrac{1}{2}mv^2\bigg(\dfrac{k^2}{R^2}\bigg)}{\dfrac{1}{2}mv^2\bigg(1+\dfrac{k^2}{R^2}\bigg)}$

$:\implies\sf\:\dfrac{K_R}{K_T}=\dfrac{\bigg(\dfrac{1}{2}\bigg)}{\bigg(1+\dfrac{1}{2}\bigg)}$

$:\implies\sf\:\dfrac{K_R}{K_T}=\dfrac{\bigg(\dfrac{1}{2}\bigg)}{\bigg(\dfrac{3}{2}\bigg)}$

$:\implies\:\underline{\boxed{\bf{\gray{\dfrac{K_R}{K_T}=\dfrac{1}{3}}}}}$

∴ (B) is the correct answer!