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)

Key concepts

Inclined Plane

An inclined plane is a flat surface tilted at an angle, different from being horizontal or vertical. It's one of the basic classical machines, often used to make loads easier to move upwards by extending the distance over which they are applied.
When dealing with physics problems involving inclined planes, gravitational potential energy becomes a key factor.
The potential energy an object holds at a height is converted into kinetic energy as it moves downward. This makes inclined planes excellent for studying the conservation of energy.

• Angle of Inclination: The greater the angle, the more rapid the increase in speed.
• Friction: Frictional forces can add complexity by opposing motion.

Understanding the geometry and friction involved helps solve problems efficiently by applying the principles of energy conservation.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion.
It depends directly on both the mass and the velocity of the object, calculated using the formula:\[KE = \frac{1}{2} \times m \times v^{2}\]where \(m\) is mass and \(v\) is velocity.

In the context of a sliding particle on an inclined plane, all the gravitational potential energy at the top converts into kinetic energy at the bottom. This principle of energy conversion is crucial:

• Direct Proportionality: Both the mass and the square of velocity directly affect kinetic energy.
• Speed Increase: A larger velocity results in exponentially higher energy.

Understanding kinetic energy helps explain how objects accelerate down an inclined plane and reach different velocities depending on their initial potential energy.

Moment of Inertia

The moment of inertia is an object's resistance to changes in its rotational motion.

This is especially important for objects rolling down an incline, like the ring in our example. It's calculated depending on the mass distribution relative to the axis of rotation:

For a ring, the moment of inertia \(I\) is given by:\[I = m \times R^2\]where \(m\) is mass and \(R\) is radius.

• Distribution of Mass: The further the mass is from the center, the higher the inertia.
• Rotational Effect: A higher moment of inertia means more energy is needed to change the object's rotational speed.

Understanding the moment of inertia explains why different shapes roll differently down an incline, impacting their kinetic energy distribution between rotational and translational movement.

Rolling Motion

Rolling motion occurs when an object rotates while moving forward, such as a ring or wheel rolling down an incline without slipping.

This combines both translational movement and rotational movement. The crucial calculation here involves relating linear velocity and angular velocity:

For a ring rolling without slipping, the relationship is:\[v = R \times \omega\]where \(v\) is the linear velocity, \(R\) is the radius, and \(\omega\) is the angular velocity.

• Conservation of Energy: Both forms of motion are considered in energy calculations.
• Non-Slipping Condition: Ensures energy split between rotation and translation.

This concept helps to analyze how much faster or slower rolling objects move compared to sliding, influenced by their shapes and moments of inertia.