A solid cylinder rolls without slipping down an incline as shown in the figure. The linear acceleration of its center of mass is
A) (5/7)g sin θ
B) (1/2)g sin θ
C) (2/3)g sin θ
D) (3/5)g sin θ
E) (4/5)g sin θ

Question

A solid cylinder rolls without slipping down an incline as shown in the figure. The linear acceleration of its center of mass is 
A) (5/7)g sin θ 
B) (1/2)g sin θ 
C) (2/3)g sin θ 
D) (3/5)g sin θ 
E) (4/5)g sin θ

Quizplus · Accepted Answer

The linear acceleration of the center of mass of a rolling cylinder is given by:
$a_{cm}=\frac{1}{2} \alpha R$
where $R$ is the radius of the cylinder and $\alpha$ is the angular acceleration.

Since the cylinder is rolling without slipping, the angular acceleration can be related to the linear acceleration by:
$a_{cm}=\alpha R$
or
$\alpha=\frac{a_{cm}}{R}$

The torque about the center of mass of the cylinder is due to gravity and is given by:
$	au=mg\sin	heta R$

Using the equation of motion for rotational motion, we can relate the torque and angular acceleration by:
$	au=I\alpha$
where $I$ is the moment of inertia of the cylinder about its center of mass.

For a solid cylinder rotating about its central axis perpendicular to its length, the moment of inertia is given by:
$I=\frac{1}{2}MR^2$

Substituting the values, we get:
$mg\sin	heta R=\frac{1}{2}MR^2 \frac{a_{cm}}{R}$

Simplifying, we get:
$a_{cm}=\frac{2}{3}g\sin	heta$

Thus, the correct choice is C) $(2/3)g\sin	heta$.

A Solid Cylinder Rolls Without Slipping Down an Incline as Shown