A force $\overrightarrow{\mathbf{F}}$ is applied to a cylindrical roll of paper of radius R and mass M by pulling on the paper as shown. The acceleration of the centre of mass of the roll of paper (when it rolls without slipping) is:
A) $\frac{1}{2} \frac{F}{M}$ .
B) $\frac{F}{M}$ .
C) $\frac{3}{2} \frac{F}{M}$ .
D) $\frac{4}{3} \frac{F}{M}$ .
E) $\frac{2 F}{M}$ .

Question

A force $\overrightarrow{\mathbf{F}}$ is applied to a cylindrical roll of paper of radius R and mass M by pulling on the paper as shown. The acceleration of the centre of mass of the roll of paper (when it rolls without slipping) is: 
A) $\frac{1}{2} \frac{F}{M}$ . 
B) $\frac{F}{M}$ . 
C) $\frac{3}{2} \frac{F}{M}$ . 
D) $\frac{4}{3} \frac{F}{M}$ . 
E) $\frac{2 F}{M}$ .

Quizplus · Accepted Answer

When the force is applied at the top of the cylinder, it will produce a torque that will cause the cylinder to roll without slipping. The net force acting on the cylinder will be equal to the force applied minus the frictional force (which is equal and opposite to the force required to keep the cylinder from slipping). 
The net force can be found as: F_net = F_applied - F_friction.
The frictional force can be found using the coefficient of static friction μs and the normal force N: F_friction = μs*N.
The normal force can be found using equilibrium of forces in the vertical direction: N = M*g.
Therefore, F_net = F_applied - μs*M*g.
The acceleration of the center of mass can be found using Newton's second law: a = F_net/M.
Substituting F_net, we get:
a = (F_applied - μs*M*g)/M
Substituting the given values, we get:
a = (11eb8b15_74f6_3a68_8952_9b3d049cc852_TB8845_11 - μs*M*g)/M
The coefficient of static friction can be found using the fact that the cylinder is rolling without slipping: 
μs = rω^2/g
where r is the radius of the cylinder and ω is its angular velocity.
Since the cylinder is rolling without slipping, the speed of the center of mass is equal to v = rω. 
Differentiating this expression with respect to time, we get:
a = dv/dt = r*dω/dt = r*α
where α is the angular acceleration of the cylinder.
The torque produced by the applied force is equal to the product of the force and the radius, and this torque causes an angular acceleration of the cylinder. Using Newton's second law for rotation, we get:
τ = I*α
where τ is the torque, and I is the moment of inertia of the cylinder. For a cylinder of mass M and radius R, we have:
I = (1/2)*M*R^2
The torque produced by the applied force is:
τ = F*r = F*R
Therefore, we get:
F*R = I*α = (1/2)*M*R^2*α
α = 2*F/(M*R)
Substituting this expression for α, we get:
a = r*α = 2*F*r/(M*R)
Substituting the given values, we get:
a = 11eb8b15_74f6_888d_8952_bdb4dff49735_TB8845_11. Therefore, the answer is D.

A Force F→\overrightarrow{\mathbf{F}}F Is Applied to a Cylindrical Roll of Paper of Radius

A Force $\overrightarrow{\mathbf{F}}$ Is Applied to a Cylindrical Roll of Paper of Radius