Answer to Question #212539 in Mechanics | Relativity for Gar12

Question

A uniform solid cylinder with mass M and radius 2R rests on a horizontal tabletop. A string is 

attached by a yoke to a frictionless axle through the center of the cylinder so that the cylinder 

can rotate about the axle. The string runs over a disk-shaped pulley with mass M and radius R 

that is mounted on a frictionless axle through its center. A block of mass M is suspended from 

the free end of the string (Fig. P10.75). The string does not slip over the pulley surface, and the cylinder rolls without slipping on the tabletop. Find the magnitude of the acceleration of the block after the system is released from rest.

Answer 1

Explanations & Calculations

• The question sounds to be a picture like the attached one.

What happens is,

• The hanging mass accelerates downwards.
• The solid cylinder rolls on the table with the help of the friction, therefore, attain a translational as well as a rotational motion.
• The pully has mass thus it also rotates with some rotational momentum.

Apply $\small \Sigma F=m a$, $\small \tau= I\alpha\,\,\&\,\,a=r\alpha$ where applicable and find the acceleration of the hanging mass.

• This acceleration is the linear acceleration of both the cylinder & the pully as the same thread passes on them.

Then,

• For the hanging mass,

$\qquad\qquad \begin{aligned} \small \downarrow F&=\small ma\\ \small Mg-T_1&=\small Ma\cdots(1) \end{aligned}$

• For the pully,

$\qquad\qquad \begin{aligned} \small \tau&-=\small I\alpha\\ \small T_1R-T_2R&=\small \frac{MR^2}{2}.\frac{a}{R}\\ \small T_1-T_2&=\small \frac{Ma}{2}\cdots(2) \end{aligned}$

• For the cylinder,

$\qquad\qquad \begin{aligned} \small F&=\small ma\\ \small T_2-f&=\small Ma\cdots(3)\\\\ \small \tau&=\small I\alpha\\ \small f.(2R)&=\small \frac{M.(2R)^2}{2}.\frac{a}{2R}\\ \small f&=\small \frac{Ma}{2}\cdots(4)\\\\ &\small\text{By (3) \& (4)},\\\\ \small T_2&=\small \frac{3Ma}{2}\cdots(5) \end{aligned}$

• Substituting for $\small T_1\,\, \& \,\,T_2$ in (2) by (1) and (5), the linear accelration could be found.

$\qquad\qquad \begin{aligned} \small (Mg-Ma)-(\frac{3Ma}{2})&=\small \frac{Ma}{2}\\ \small g-a -\frac{3a}{2}&=\small \frac{a}{2}\\ \small a&=\small \frac{g}{3} \end{aligned}$

$\qquad\qquad \begin{aligned} \small \end{aligned}$