Question

A solid cylinder of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) rolls down an inclined plane of height \(\mathrm{h}\). The angular velocity of the cylinder when it reaches the bottom of the plane will be. \(\\{\mathrm{A}\\}(2 / \mathrm{R}) \sqrt{(\mathrm{gh})}\) \(\\{\mathrm{B}\\}(2 / \mathrm{R}) \sqrt{(\mathrm{gh} / 2)}\) \(\\{\mathrm{C}\\}(2 / \mathrm{R}) \sqrt{(\mathrm{gh} / 3)}\) \(\\{\mathrm{D}\\}(1 / 2 \mathrm{R}) \sqrt{(\mathrm{gh})}\)

Short answer

The short answer is: The angular velocity of the cylinder when it reaches the bottom of the plane is \(\frac{2}{R}\sqrt{\frac{gh}{3}}\).

Step-by-step solution

Find the total potential energy at the incline's top.

At the top of the inclined plane, the cylinder has a potential energy given by: \(PE = mgh\) Where: - \(PE\) = potential energy of the cylinder - \(m\) = mass of the cylinder (\(M\)) - \(g\) = acceleration due to gravity - \(h\) = height of the inclined plane

Find the rotational and translational kinetic energies at the bottom of the incline.

At the bottom of the inclined plane, the cylinder has both translational kinetic energy (due to its linear motion) and rotational kinetic energy (due to its rotation). These can be given as follows: \(KE_t = \frac{1}{2} mv^2\) \(KE_r = \frac{1}{2} I \omega^2\) Where: - \(KE_t\) = translational kinetic energy - \(KE_r\) = rotational kinetic energy - \(v\) = linear velocity at the bottom of the incline - \(I\) = moment of inertia of the cylinder - \(\omega\) = angular velocity of the cylinder at the bottom of the incline The moment of inertia for a solid cylinder is \(I = \frac{1}{2} MR^2\).

Apply the conservation of energy.

By the conservation of energy, the potential energy at the top of the incline is equal to the sum of the translational and rotational kinetic energies at the bottom. Thus, \(mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2\) Substituting the moment of inertia for the solid cylinder, we get: \(mgh = \frac{1}{2} mv^2 + \frac{1}{4} MR^2 \omega^2\)

Relate linear velocity and angular velocity.

For a rolling object without slipping, the linear velocity \(v\) and the angular velocity \(\omega\) are related as follows: \(v = \omega R\) We can substitute this into the conservation of energy equation: \(mgh = \frac{1}{2} m(\omega R)^2 + \frac{1}{4} MR^2 \omega^2\)

Solve for angular velocity.

Now we can solve for the angular velocity \(\omega\): \(\omega^2(R^2 + \frac{1}{2}R^2) = 2gh\) Solving for \(\omega\), we get: \(\omega = \sqrt{\frac{2gh}{\frac{3}{2}R^2}}\) \(\omega = \sqrt{\frac{4gh}{3R^2}}\) \(\omega = \frac{2}{R}\sqrt{\frac{gh}{3}}\) Thus, the angular velocity of the cylinder when it reaches the bottom of the plane is \(\frac{2}{R}\sqrt{\frac{gh}{3}}\), which corresponds to option C.