(I) Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.20 m high. Assume it starts from rest and rolls without slipping.

The given data can be listed below as:

• The height of the cylinder is\(h = 7.20{\rm{ m}}\).
• The initial velocity of the cylinder is\({v_i} = 0\).
• The acceleration due to gravity is\(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

The cylinder starts rolling from the top of the inclined plane to the bottom of the inclined plane. The energy on top of the incline is in the form of potential energy.The energy conversion takes place on the inclined plane, moving from top to bottom.

The total potential energy is converted into total kinetic energy. The sum of rotational and translational kinetic energies is stored in the cylinder at the bottom.

The cylinder is at rest on top of the incline. The rotational kinetic energy of the cylinder can be expressed as:

\(K = \frac{1}{2}I{\omega ^2}\) … (i)

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

The moment of inertia of the solid cylinder can be expressed as:

\(I = \frac{1}{2}M{R^2}\)

Substitute the values in equation (i).

\(\begin{align}K &= \frac{1}{2} \times \frac{1}{2}M{R^2} \times {\omega ^2}\\ &= \frac{1}{4}M{R^2}{\omega ^2}\end{align}\)

\(K = \frac{1}{4}M{v^2}\) … (ii)

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

The translational kinetic energy of the rotor can be expressed as:

\({K_1} = \frac{1}{2}M{v^2}\) … (iii)

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

\(\begin{align}{K_T} &= K + {K_1}\\{K_T} &= \frac{1}{4}M{v^2} + \frac{1}{2}M{v^2}\end{align}\)

\({K_T} = \frac{3}{4}M{v^2}\) … (iv)

The potential energy of the cylinder can be expressed as:

\(P.E = Mgh\) … (v)

The potential energy is getting converted into kinetic energy and is equal to the total kinetic energy. Equate equations (iv) and (v). This can be expressed as:

\(\begin{align}P.E &= {K_T}\\Mgh &= \frac{3}{4}M{v^2}\\{v^2} &= \frac{{4Mgh}}{{3M}}\\v &= \sqrt {\frac{{4gh}}{3}} \end{align}\)

Substitute the values in the above equation.

\(\begin{align}v &= \sqrt {\frac{{4 \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times 7.20{\rm{ m}}}}{3}} \\ &= \sqrt {94.176} {\rm{ m/s}}\\ &= {\rm{9}}{\rm{.7 m/s}}\end{align}\)

Thus, the translational speed of the cylinder is 9.7 m/s.