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 frictional force on the disc is \(\\{\mathrm{A}\\}[(\mathrm{Mg} \sin \theta) / 3]\) \(\\{\mathrm{B}\\}[(2 \mathrm{Mg} \sin \theta) / 3]\) \(\\{\mathrm{C}\\} \mathrm{Mg} \sin \theta\) \(\\{\mathrm{D}\\}\) None

Short answer

The frictional force on the disc rolling without slipping down an inclined plane at an angle θ with the horizontal is given by \(f = \frac{2Mg\sin\theta}{3}\), which corresponds to option B.

Step-by-step solution

Draw a free body diagram

Draw a free body diagram for the disc while it's rolling down the incline showing all the forces acting on it: gravitational force (Mg), frictional force (f), and normal force (N).

Determine the equations for the horizontal and vertical forces

Resolve the forces in the horizontal (x) and vertical (y) directions, and describe the acceleration and torque acting on the disc: Sum of the forces in the x-direction: \(f - Mg\sin\theta = Ma\) Sum of the forces in the y-direction: \(N - Mg\cos\theta = 0\)

Determine the rotational equation

Since the disc is rolling without slipping, we need to consider a rotational equation using the equation of torque about the center of the disc: Torque = I * angular acceleration \(fR = \frac{1}{2}MR^2α\) Where I is the moment of inertia of the disc (\(\frac{1}{2}MR^2\)) and α is the angular acceleration.

Determine the No-slip condition equation

A no-slip condition means that when the disc is rolling, the point of contact with the inclined plane has zero velocity. Using the relationship between linear acceleration and angular acceleration: \( a = Rα \)

Solve the equations for frictional force

Using Steps 2, 3, and 4, we can solve the system of equations to find the frictional force (f): From Step 4, \(α = \frac{a}{R}\). Plugging this value into Step 3 equation: \(fR = \frac{1}{2}MR^2 (\frac{a}{R}) \) Solving for f: \(f = \frac{1}{2}Ma\) Now, lets substitute \(f = \frac{1}{2}Ma\) into the Step 2 equation of horizontal forces: \(\frac{1}{2}Ma - Mg\sin\theta = Ma\) Solving for a: \(a = \frac{2}{3}g\sin\theta\) Finally, substituting the value of a back to frictional force equation: \(f = \frac{1}{2}M(\frac{2}{3}g\sin\theta)\)

Choose the correct answer

Comparing this result with the given options: \(f = \frac{2Mg\sin\theta}{3}\), which corresponds to option B.