Question

A spool of inner radius $R$ and outer radius $3R$ has a moment of inertia $=M{R}^2$ about an axis passing through its geometric centre, where $M$ is the mass of the spool. A thread wound on the inner surface of the spool is pulled horizontally with a constant force $=Mg$. Find the acceleration of the point on the thread which is being pulled assuming that the spool rolls purely on the floor.

• A. $40m/s^2$
• B. $8.4m/s^2$
• C. $3.33m/s^2$
• D. $16m/s^2$

Answer 1

The correct option is D $16m/s^2$

We will equate the torques acting on the spool about the instantaneous center of rotation.
$Mg\left(4R\right)=I\alpha$
I about the instantaneous center is $M{R}^2+M(3R{)}^2=10M{R}^2$(use parallel axis theorem)
thus we get
$4MgR=10M{R}^2\alpha$
or
$4g=10R\alpha$
or
$4g=10R\frac{a}{3R}$
or
$4g=\frac{10}{3}a$
or
$a=\frac{12}{10}g$
This a is the acceleration of the center of mass.
Thus the acceleration of the thread $a_t$ is given as the acceleration of point at a distance $R$ from the center or $4R$ from the instantaneous axis.
Thus we get $a_t=4R\alpha =4R\frac{a}{3R}=\frac{4}{3}a$
or
$a_t=\frac{4}{3}\frac{12}{10}g=\frac{16}{10}g\approx 16m/s^2$