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 \(\\{\mathrm{A}\\} \mathrm{MgR}\) \(\\{\mathrm{B}\\} \mathrm{MgR} \sin \theta\) \(\\{\mathrm{C}\\}[(2 \mathrm{MgR} \sin \theta) / 3]\) \(\\{\mathrm{D}\\}[(\mathrm{MgR} \sin \theta) / 3]\)

Short 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-by-step solution

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

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)

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)

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

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

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

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.