Problem 651 - JEE Physics

Question

A circular disc of mass \(\mathrm{m}\) and radius \(\mathrm{r}\) is rolling on a smooth horizontal surface with a constant speed v. Its kinetic energy is \(\\{\mathrm{A}\\}(1 / 4) \mathrm{mv}^{2}\) \(\\{\mathrm{B}\\}(1 / 2) \mathrm{mv}^{2}\) \(\\{\mathrm{C}\\}(3 / 4) \mathrm{mv}^{2}\) \(\\{\mathrm{D}\\} \mathrm{mv}^{2}\)

Step-by-Step Solution

Answer

The total kinetic energy of the circular disc is \(\frac{3}{4}mv^2\). Hence, the correct answer is \(\boxed{\text{C} (3 / 4) \mathrm{mv}^{2}}\).

Step 1: Identify the given variables.

In this problem, we are given the following variables: 1. Mass of the circular disc (m) 2. Radius of the circular disc (r) 3. Constant speed of the circular disc (v)

Step 2: Calculate the translational kinetic energy.

The translational kinetic energy of the disc, moving with a constant speed v, is given by the formula: Translational Kinetic Energy = \( \frac{1}{2} * m * v^2 \)

Step 3: Calculate the rotational kinetic energy.

For a rolling object, the moment of inertia (I) for a circular disc about its center of mass is given by the formula: Moment of Inertia (I) = \( \frac{1}{2} * m * r^2 \) The rotational kinetic energy is given by the formula: Rotational Kinetic Energy = \( \frac{1}{2} * I * ω^2 \) Since we are given the constant speed (v) and we know that the object is rolling without slipping, we can find the angular velocity (ω) using the formula: \( ω = \frac{v}{r} \) Now we plug the values into the rotational kinetic energy formula: Rotational Kinetic Energy = \( \frac{1}{2} * (\frac{1}{2} * m * r^2) * (\frac{v}{r})^2 \) Simplify the equation: Rotational Kinetic Energy = \( \frac{1}{4} * m * v^2 \)

Step 4: Calculate the total kinetic energy.

Now we can find the total kinetic energy by summing both the translational and rotational kinetic energies: Total Kinetic Energy = Translational Kinetic Energy + Rotational Kinetic Energy Total Kinetic Energy = \( (\frac{1}{2} * m * v^2) + (\frac{1}{4} * m * v^2) \) Total Kinetic Energy = \( (\frac{3}{4} * m * v^2) \)

Step 5: Find the correct answer choice.

Now we have the total kinetic energy of the circular disc as \( (\frac{3}{4} * m * v^2) \). Looking at the given options, the correct answer is: \(\boxed{\text{C} (3 / 4) \mathrm{mv}^{2}}\)

Key Concepts

Rolling without Slipping

Imagine a wheel moving along a surface without sliding. This is what we call "rolling without slipping." It's like a perfect blend between rotational and translational motions. When a disc rolls without slipping, every point on the rim comes into contact with the surface exactly once per cycle. This means the speed of the contact point is zero relative to the ground. This condition links the translational speed of the disc to its angular speed in a simple way.\
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Here is the important relationship for rolling without slipping: the linear velocity at the disc's center equals the product of the angular velocity and the radius of the disc. In formula terms, that's expressed as:

• \(v = \omega \times r\)

Where \(v\) is the linear velocity, \(\omega\) is the angular velocity, and \(r\) is the radius of the disc. This relationship helps us understand how rolling without slipping connects rotational motion with linear motion.

Translational Kinetic Energy

Translational kinetic energy is the energy related to the motion of the center of mass of an object moving through space. Imagine how a car rolls down a hill solely because of its forward motion. The translational kinetic energy depends on two things: the mass of the object and its velocity.\
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The formula to calculate translational kinetic energy is quite straightforward:

• \(KE_{\text{trans}} = \frac{1}{2} \times m \times v^2\)

Where \(m\) is the mass of the object, and \(v\) is the linear velocity. This formula shows that as the mass or velocity of an object increases, its translational kinetic energy increases too. In the case of the rolling disc, the translational kinetic energy provides insight into how much energy the disc has due to its motion along the surface.

Rotational Kinetic Energy

When an object spins, its parts move with both speeds and possess energy because of that rotation. This is known as rotational kinetic energy. For a rotating object like a disc, not only its whole body translates, but also every little piece rotates around the center.\
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The formula to determine the rotational kinetic energy is analogous to that for translational motion:

• \(KE_{\text{rot}} = \frac{1}{2} \times I \times \omega^2\)

Where \(I\) is the moment of inertia, and \(\omega\) is the angular velocity. The moment of inertia is like the rotational equivalent of mass and depends on how the object's mass is distributed around the center. This energy contributes significantly to the total kinetic energy when a disc rolls, providing a more complete picture of its energy dynamics.\

Moment of Inertia

The moment of inertia tells us how much resistance an object has to changes in its rotational motion. Think of it as rotational "mass." It depends not only on the object's mass but also on how far each part of the object is from the axis of rotation.\
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For simple shapes, common moment of inertia formulas are well known. For a disc rotating about its center, the moment of inertia is:

• \(I = \frac{1}{2} \times m \times r^2\)

Here, \(m\) is the mass, and \(r\) is the radius. Recognizing the concept of moment of inertia helps understand parts of the kinetic energy due to rotation and why different objects with the same mass might roll differently.

Angular Velocity

Angular velocity measures how fast something is spinning. It's like the speed for rotational movement. The angular velocity gives you an idea of how many turns an object makes in a certain time period. In rolling motion, it links directly to translational speed via the radius.\
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The relationship can be expressed as:

• \(\omega = \frac{v}{r}\)

Where \(\omega\) is the angular velocity, \(v\) is the linear velocity, and \(r\) is the radius. This formula shows that a higher speed or a smaller radius results in a quicker spin. Understanding angular velocity is key to grasping more complex concepts in rotating systems like a rolling disc.\