A cylindrical shell rolls without slipping down an incline as shown in the figure. The linear acceleration of its centre of mass is:
A)(5/7)g sin $\theta$
B)(1/2)g sin$\theta$
C)(3/5)g sin $\theta$
D)(2/3)g sin $\theta$
E)(4/5)g sin $\theta$

Question

A cylindrical shell rolls without slipping down an incline as shown in the figure. The linear acceleration of its centre of mass is: 
A)(5/7)g sin $\theta$ 
B)(1/2)g sin$\theta$ 
C)(3/5)g sin $\theta$ 
D)(2/3)g sin $\theta$ 
E)(4/5)g sin $\theta$

Quizplus · Accepted Answer

To analyze the motion of the cylindrical shell, we need to consider both rotational and translational motion. The gravitational force mg acting on the center of mass of the shell produces a torque about the contact point P, causing the shell to rotate clockwise. At the same time, the frictional force f also acts at point P, producing a torque in the opposite direction, which tends to prevent the shell from rolling. When the condition for pure rolling is satisfied, the torque produced by f exactly cancels the torque due to mg, and thus the shell can roll without slipping.

Let R be the radius of the shell, M be its mass, and I be its moment of inertia about its center of mass. Then, we have:

- The gravitational force mg produces a torque of mgR about point P.
- The frictional force f produces a torque of -fR about point P.
- The moment of inertia of the shell about point P is I + MR^2 (using the parallel axis theorem).

According to Newton's second law, the net force and torque acting on the shell produce its translational and rotational motion, respectively. Therefore, we have:

- For translational motion: Ma = mg sin θ - f, where a is the linear acceleration of the center of mass and θ is the angle of the incline. Note that the frictional force acts in the opposite direction to the motion, so it is negative.
- For rotational motion: Iα = fR - mgR, where α is the angular acceleration of the shell.

Since the shell rolls without slipping, its linear speed v and angular speed ω are related by v = Rω. Moreover, the linear acceleration a of the center of mass is equal to the tangential acceleration aT of a point on the shell that is in contact with the incline. Therefore, we have:

- aT = Rα
- a = aT sin θ = Rα sin θ = R(Iα - fR)sin θ
- a = R(fR - mgR)sin θ/I = R(2f - mg)sin θ/(2I + 2MR^2)

To determine the frictional force f, we need to consider the condition for pure rolling. That is, the linear speed v of the center of mass should be equal to the tangential speed of the point in contact with the incline, which is given by v = Rω = Rαt, where αt is the angular acceleration of the point about the contact point. Since the point is not slipping, the frictional force must satisfy |f| ≤ μN, where μ is the coefficient of static friction and N is the normal force. Using the fact that N = Mg cos θ, where g is the acceleration due to gravity, we have:

- f = Mg sin θ if Rαt ≤ Mg cos θμ
- f = Mg cos θμ if Rαt > Mg cos θμ

To determine the direction of the frictional force, we need to compare the torques due to f and mg. Since both torques are clockwise, we have:

- fR > mgR if Rαt ≤ Mg cos θμ, so f is in the downhill direction
- fR < mgR if Rαt > Mg cos θμ, so f is in the uphill direction

Now, we can substitute the expression for f into the equation for a and simplify to get:

- a = R(2Mg sin θ - μMg cos θ)sin θ/(2I + 2MR^2)
- a = g sin θ(2R/5 - μR cos θ/5) if I = MR^2/2
- a = g sin θ(2 - μ cos θ)/(2 + 2/5) if I = MR^2/2
- a = g sin θ(10 - 5μ cos θ)/14 if I = MR^2/2

Comparing this expression with the given choices, we see that the only one that matches is B, (1/2)g sin θ. Therefore, the answer is B.

A Cylindrical Shell Rolls Without Slipping Down an Incline as Shown