A solid sphere & a hollow sphere of radius R are rolling down in inclined lane of height h. The ratio of velocities of solid sphere to Hollow sphere on reaching the bottom is

CONCEPT:

Kinetic energy (K.E): 

• The energy possessed by a body by virtue of its motion is called kinetic energy.
• The expression for kinetic energy is

\(KE = \frac{1}{2}m{v^2}\)

Where m = mass of the body and v = velocity of the body

Rotational Kinetic energy (KE):

• The energy, which a body has by virtue of its rotational motion, is called rotational kinetic energy.
• A body rotating about a fixed axis possesses kinetic energy because its constituent particles are in motion, even though the body as a whole remains in place.
• Mathematically rotational kinetic energy can be written as

\(KE = \frac{1}{2}I{\omega ^2}\)

Where I = moment of inertia and ω = angular velocity.

CALCULATION:

• In rolling without slipping through the distance L down the incline, the height of the rolling object changes by "h".
• Hence the gravitational potential energy changes by mgh.

⇒ P.E = K.E_translation + KE_rotational

\(\Rightarrow mgh = \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2} = \frac{1}{2}m{v^2} + \frac{1}{2}I{\left( {\frac{v}{r}} \right)^2}\)

\(\Rightarrow 2mgh = m{v^2}\left( {1 + \frac{I}{{m{r^2}}}} \right)\)

\( ⇒ v = \sqrt {\frac{{2gh}}{{\left( {1 + \frac{I}{{m{r^2}}}} \right)}}}\)

As we know that moment of inertia of solid sphere is

\(\Rightarrow I = \frac{2}{5}m{r^2}\)

∴ The velocity of the solid sphere is 

\(\Rightarrow v_{solid sphere} = \sqrt {\frac{10}{7}gh}\)         ------------ (1)

The moment of inertia of the hollow sphere is

\(\Rightarrow I = \frac{2}{3}m{r^2}\)

∴ The velocity of the hollow sphere is 

\(\Rightarrow v_{hollow sphere} = \sqrt {\frac{6}{5}gh}\)         ------------ (2)

On dividing equation 1 and 2, we get

\(\Rightarrow \frac{{{v_{solid}}}}{{{v_{hollow}}}} = \sqrt {\frac{{25}}{{21}}}\)