Question

A string wrapped around the pulley as shown in the figure is pulled with a constant force $\overrightarrow{F}$ of magnitude $100\text{N}$. The radius $R$ and moment of inertia $I$ of the pulley are $0.1\text{m}$ and $1.25\times {10}^{-3}{\text{kgm}}^2$ respectively. If the string does not slips, what is the angular velocity of pulley after $1\text{m}$ of string has been unwounded? Assume the pulley starts from rest.

• A. $400\text{rad/sec}$
• B. $200\text{rad/sec}$
• C. $800\text{rad/sec}$
• D. $100\text{rad/sec}$

Answer 1

The correct option is A $400\text{rad/sec}$


From the FBD, force on the bearing of the pulley $\left(\overrightarrow{B}\right)$ and weight of the pulley $\left(Mg\right)$ does not exert a torque around the rotational axis, therefore no work is done by them on the pulley.
As pulley rotates through an angle $\theta ,\overrightarrow{F}$ act through a distance $d$ such that $d=R\theta$.
Since, the torque due to $\overrightarrow{F}$ has magnitude
$\tau =R\times F$
We have, work done, $W=\tau \theta =(R\times F)\times \left(\theta \right)$
$\Rightarrow W=F.d$.
If the force on the string acts through a distance $1\text{m}$, we have, from the work energy theorem,
$W=K{E}_2-K{E}_1$
$\Rightarrow F.d=\frac{1}{2}I{\omega }_2^2-\frac{1}{2}I{\omega }_1^2$ [but ${\omega }_1=0$]
$\Rightarrow \left(100\right)\left(1\right)=\frac{1}{2}(1.25\times {10}^{-3}){\omega }_2^2$
$\Rightarrow {\omega }_2=400\text{rad/sec}$
So, the angular velocity of pulley after $1\text{m}$ of string has been unwounded is $400\text{rad/sec}$