91Ó°ÊÓ

The given data can be listed below as:

• The mass of the bowling ball is\(m = 7.25{\rm{ kg}}\).
• The radius of the bowling ball is\(R = 10.8{\rm{ cm}}\).
• The velocity of the bowling ball is \(v = 3.10{\bf{ }}{\rm{m/s}}\).

The bowling ball is rolling from the top of the lane to its bottom.The kinetic energy of the bowling ball is equal to thesum of the rotational and translational kinetic energies.The rotational kinetic energy of the bowling ball can be expressed in terms of Joules.

The rotational kinetic energy of the can be expressed as:

\(K = \frac{1}{2}I{\omega ^2}\) … (i)

Here,\(\omega \)is the angular speed of the bowling ball.

The moment of inertia of the bowling ball can be expressed as:

\(I = \frac{2}{5}M{R^2}\)

Substitute the values in equation (i).

\(\begin{align}K &= \frac{1}{2} \times \frac{2}{5}M{R^2} \times {\omega ^2}\\K &= \frac{1}{5}M\left( {{R^2}{\omega ^2}} \right)\\K &= \frac{1}{5}M{v^2}\end{align}\)

\(K = \frac{1}{5}M{v^2}\) … (ii)

Here,\(v\)is the translational velocity of the bowling ball.

The translational kinetic energy of the bowling ball can be expressed as:

\({K_1} = \frac{1}{2}M{v^2}\) … (iii)

The total kinetic energy is equal to the sum of translational and rotational kinetic energies. Add equations (ii) and (iii) to obtain the total kinetic energy.

\(\begin{align}{K_T} &= K + {K_1}\\{K_T} &= \frac{1}{5}M{v^2} + \frac{1}{2}M{v^2}\end{align}\)

\({K_T} = \frac{7}{{10}}M{v^2}\)

Substitute the values in the above equation.

\(\begin{align}{K_T} &= \frac{7}{{10}} \times 7.25{\rm{ kg}} \times {\left( {3.10{\rm{ m/s}}} \right)^2}\\{K_T} &= 48.77{\bf{ }}{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^2}\left( {\frac{{1{\rm{ J}}}}{{1{\bf{ }}{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^2}}}} \right)\\{K_T} &= 48.77{\rm{ J}}\end{align}\)

Thus, the total kinetic energy of a bowling ball is \({\rm{48}}{\rm{.77 J}}\).