Problem 643 - JEE Physics

Question

A uniform disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) rolls without slipping down a plane inclined at an angle \(\theta\) with the horizontal. The magnitude of torque acting on the disc is \(\\{\mathrm{A}\\} \mathrm{MgR}\) \(\\{\mathrm{B}\\} \mathrm{MgR} \sin \theta\) \(\\{\mathrm{C}\\}[(2 \mathrm{MgR} \sin \theta) / 3]\) \(\\{\mathrm{D}\\}[(\mathrm{MgR} \sin \theta) / 3]\)

Step-by-Step Solution

Answer

In conclusion, the torque acting on the uniform disc rolling without slipping down a plane inclined at an angle θ with the horizontal is given by \( Torque = (\frac{1}{2} MgR \sin \theta) \). The given alternatives do not provide the correct option for the calculated torque.

Step 1: 1. Identify the forces acting on the disc

The first step is to identify all the forces acting on the disc. In this case, there are three main forces acting: a. Gravitational force (Mg) - Acting vertically downwards b. Normal force (N) - Acting perpendicular to the inclined surface c. Frictional force (f) - Acting parallel to the inclined surface in the direction opposing motion

Step 2: 2. Resolve gravitational force along and perpendicular to the incline

To consider the forces acting along and perpendicular to the inclined plane, we need to project the gravitational force into components. Let's find the components of gravitational force Mg as: a. Component parallel to incline: Mg sin(θ) b. Component perpendicular to incline: Mg cos(θ)

Step 3: 3. Apply Newton's Second Law in the vertical direction

As the disc is rolling without slipping, there is no vertical acceleration. So the net force in the vertical direction should be zero. We have: Normal force (N) - Mg cos(θ) = 0 Normal force (N) = Mg cos(θ)

Step 4: 4. Apply Newton's Second Law in the horizontal direction along the incline

The disc is rolling down the inclined plane, so it has a horizontal acceleration a along the incline. The net force acting along the incline can be found by Newton's Second Law: f = M * a

Step 5: 5. Calculate the torque due to frictional force

The frictional force provides the required torque for the rolling motion of the disc. The magnitude of this torque can be found using the formula: Torque = force * radius Torque = f * R

Step 6: 6. Determine angular acceleration and relate it to linear acceleration

As the disc rolls without slipping, it undergoes angular acceleration (α). According to the relationship between the linear acceleration (a) and angular acceleration (α) for rolling without slipping: a = α * R

Step 7: 7. Calculate the torque due to gravitational force down the incline

By equating the above equations, we can calculate the torque due to the gravitational force acting down the incline. Torque = f * R = M * a * R = M * (α * R) * R Since the disc is a uniform disc, we can relate the torque to the moment of inertia (I) and angular acceleration (α) using the formula: I * α = M * (α * R) * R For a uniform solid disk, the moment of inertia (I) = 0.5 * M * R^2. So, α = (2 * Torque) / (M * R^2) Now, a = αR = (2 * Torque) /M. Torque = a * (M/2) = (M * g * sin(θ))/ 2 The torque acting on the uniform disc is given by \( Torque = (\frac{1}{2} MgR \sin \theta) \) This option is not provided in the given alternatives, so there might be a mistake in the problem or in the choices. In conclusion, the torque acting on the disc rolling without slipping down a plane inclined at an angle θ with the horizontal can be calculated by following these steps. In this particular case, the answer provided by the alternatives is not matching our calculated torque.

Key Concepts

Rolling without Slipping

When a disc rolls without slipping down an inclined plane, it means there is no relative motion between the point of contact of the disc and the plane. The disc moves such that its rotational or angular motion is synchronized with its linear or translational motion. This condition ensures that each point on the rim of the disc moves linearly as much as it rotates around the center of the disc.

The mathematical relationship for rolling without slipping is given by:

• Linear velocity at the point of contact \( v = \omega \, R \)
• Linear acceleration \( a = \alpha \, R \)

Where \( \omega \) is the angular velocity, \( \alpha \) is the angular acceleration, and \( R \) is the radius of the disc. Understanding this relationship helps in solving problems involving motion on inclined planes.

Inclined Plane Dynamics

The dynamics of an object moving down an inclined plane involves analyzing forces and how they result in motion. In our scenario, a disc rolls down an incline, and it's important to identify all forces and their respective contributions.

• Gravitational Force: Always acts vertically downward with a magnitude \( Mg \), where \( M \) is the mass of the disc and \( g \) is the acceleration due to gravity.
• Normal Force: Acts perpendicular to the surface of the incline and balances the perpendicular component of gravitational force.
• Frictional Force: Acts parallel to the incline and provides the torque necessary for rolling.

Recognizing these forces and their components allows us to apply Newton's laws effectively, resulting in accurate predictions of the disc's motion on the slope.

Gravitational Force Components

The gravitational force acting on an object on an inclined plane has two essential components: parallel and perpendicular to the incline.

• Parallel Component: This component causes the object to accelerate down the incline. It is given by \( Mg \sin(\theta) \), where \( \theta \) is the angle of inclination.
• Perpendicular Component: This component is balanced by the normal force and is given by \( Mg \cos(\theta) \).

Understanding these components is crucial for analyzing forces on inclined planes. The gravitational force parallel to the plane provides the necessary force to keep the disc moving, while the perpendicular component helps in determining the normal force from the surface.

Newton's Laws of Motion

Newton's laws are fundamental when tackling problems in mechanics, including motion on inclined planes. In the context of our rolling disc, these laws explain why the disc moves as it does.

• First Law (Inertia): The disc will continue its state of motion unless an unbalanced force acts on it. Rolling on an incline typically involves unbalanced forces that necessitate motion.
• Second Law (F=ma): Relates the net force acting on the disc to its mass and acceleration. For rolling motion, we consider both linear and angular acceleration.
• Third Law (Action-Reaction): Explains the interaction between the disc and the inclined plane. The disc exerts a force on the incline, and the incline exerts an equal but opposite force on the disc, mostly visible via the normal and frictional forces.

Together, these laws provide a comprehensive framework to analyze and solve problems of motion like rolling on inclined planes, allowing students to predict and understand the behavior of rolling objects.