Question

Figure shows a spool with thread wound on it placed on a smooth plane inclined at angle from horizontal. The spool has mass $m$, edge radius $R$, and is wound up to a radius $r$, its moment of inertia about its own axis is $I$. The free end of the thread is attached as shown in the figure. So that the thread is parallel to the inclined plane. $T$ is the tension in the thread. Which of the following is correct?

• A. The linear acceleration of the spool axis down the slope is $\frac{mgsin\theta -T}{m}$.
• B. Angular acceleration is $\frac{Tr}{2I}$.
• C. The linear acceleration of the spool axis down the plane is $\frac{T{r}^2}{I}$.
• D. The acceleration of the spool axis down the slope is $\frac{gsin\theta }{1+\frac{I}{m{r}^2}}$.

Answer 1

Answer: (A), (C), (D)

The correct options are
A The linear acceleration of the spool axis down the slope is $\frac{mgsin\theta -T}{m}$.
C The linear acceleration of the spool axis down the plane is $\frac{T{r}^2}{I}$.
D The acceleration of the spool axis down the slope is $\frac{gsin\theta }{1+\frac{I}{m{r}^2}}$.

Linear acceleration of point P: $a_{linear}=\frac{mgsin\theta -T}{m}$
Angular acceleration at point P:
$\alpha =\frac{Torque\left(Tr\right)}{Momentofinertia\left(I\right)}$

Acceleration due to rotation:
$a_{rotational}=r\alpha =r\left[\frac{Tr}{I}\right]=\frac{T{r}^2}{I}$

As the point P is fixed to a rigid support, its net acceleration should be zero.

So, the magnitudes of both accelerations must match,
$\therefore \frac{T{r}^2}{I}=\frac{mgsin\theta -T}{m}$
or $\frac{Tm{r}^2}{l}+T=mgsin\theta$
or $T[1+\frac{m{r}^2}{I}]=mgsin\theta$ or $T=\frac{mgsin\theta }{1+\frac{m{r}^2}{l}}$

So, acceleration of the spool down the plane :
$a_{linear}=\frac{mgsin\theta -T}{m}$
$a_{linear}=gsin\theta -\frac{T}{m}$
$a_{linear}=gsin\theta -\frac{gsin\theta }{1+\frac{m{r}^2}{l}}$
$\Rightarrow a_{linear}=\frac{gsin\theta }{1+\frac{I}{m{r}^2}}$