Question

A particle of mass \(\mathrm{m}\) slides down on inclined plane and reaches the bottom with linear velocity \(\mathrm{V}\). If the same mass is in the form of ring and rolls without slipping down the same inclined plane. Its velocity will be \(\\{\mathrm{A}\\} \mathrm{V}\) \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~V})}\) \(\\{\mathrm{C}\\}(\mathrm{V} / \sqrt{2})\) \(\\{\mathrm{D}\\} 2 \mathrm{~V}\)

Short answer

The velocity of the rolling ring without slipping is given by \(v_{r}\) = \(\frac{V}{\sqrt{(1 + R^2)}}\). Assuming R = 1, the correct answer is \(\mathrm{C}\) (\(\mathrm{V} / \sqrt{2}\)).

Step-by-step solution

Calculate the particle's kinetic energy at the bottom of the inclined plane

A particle of mass m slides down the inclined plane and reaches the bottom with a linear velocity V. At the bottom of the inclined plane, the particle has only kinetic energy, which can be calculated using the formula: Kinetic Energy = \(\frac{1}{2} \times m \times V^{2}\)

Calculate the gravitational potential energy at the top of the inclined plane

Let h be the vertical height of the inclined plane. The gravitational potential energy (GPE) of the particle at the top of the inclined plane can be calculated using the formula: GPE = m × g × h

Apply conservation of mechanical energy

As the particle slides down the inclined plane without any external forces, the total mechanical energy remains constant. Thus, the GPE at the top of the inclined plane is equal to the kinetic energy at the bottom. So, we have: m × g × h = \(\frac{1}{2} \times m \times V^{2}\)

Determine the moment of inertia of the rolling ring

A ring of mass m and radius R has a moment of inertia I given by: I = m × R^2

Calculate the kinetic energy of the rolling ring

As the ring rolls without slipping, it has both translational and rotational kinetic energy. The total kinetic energy of the rolling ring at the bottom of the inclined plane is given by: Total Kinetic Energy = \(\frac{1}{2} \times m \times v_{r}^{2}\) + \(\frac{1}{2} \times I \times \omega^{2}\) Since the ring is rolling without slipping, \(\omega = \frac{v_{r}}{R}\), where \(\omega\) is the angular velocity of the ring.

Apply conservation of mechanical energy for the rolling ring

The GPE at the top of the inclined plane is equal to the total kinetic energy of the rolling ring at the bottom. So, we have: m × g × h = \(\frac{1}{2} \times m \times v_{r}^{2}\) + \(\frac{1}{2} \times (m \times R^2) \times (\frac{v_{r}}{R})^{2}\)

Solve for the velocity of the rolling ring

Now, we can solve the equation from Step 6 for the velocity of the rolling ring, v_r, and then compare it to the given options to find the correct one. m × g × h = \(\frac{1}{2} \times m \times v_{r}^{2}\) + \(\frac{1}{2} \times (m \times R^2) \times (\frac{v_{r}}{R})^{2}\) From the conservation of mechanical energy equation for the particle, we already know that m × g × h = \(\frac{1}{2} \times m \times V^{2}\), so: \(\frac{1}{2} \times m \times V^{2}\) = \(\frac{1}{2} \times m \times v_{r}^{2}\) + \(\frac{1}{2} \times (m \times R^2) \times (\frac{v_{r}}{R})^{2}\) Cancel out the mass term m and divide both sides by \(\frac{1}{2}\) to get: \(V^{2}\) = \(v_{r}^{2}\) + \(R^2 \times (\frac{v_{r}}{R})^{2}\) Now, we can factor out \(v_{r}^{2}\) from the equation: \(V^{2}\) = \(v_{r}^{2}\) (1 + R^2) Divide both sides by (1 + R^2) and take the square root to solve for \(v_{r}\): \(v_{r}\) = \(\frac{V}{\sqrt{(1 + R^2)}}\) Comparing our result with the given options, we find that the correct answer is: \(\\{\mathrm{C}\\}(\mathrm{V} / \sqrt{2})\) (assuming R = 1 in this case)