Question

A light thread is wound four times around a pulley disc of diameter $30cm$ and mass $5kg$. We want to unwind the thread by pulling it in s. The minimum constant force require to do so is
$\begin{align}
  & A)12.25N \\
 & B)18.85N \\
 & C)24.5N \\
 & D)37.7N \\
\end{align}$

Answer 1

Hint: The work done to unwind the pulley will increase the kinetic energy of the pulley. Equating these two we can find out the required force. Here we will assume the pulley as a cylinder and will use its formula of moment of inertia to find its kinetic energy. So let us start with the solution.
Formula used:
  $I=\dfrac{1}{2}M{{R}^{2}}$

Complete answer:
Let $F$ be the required force and $s$ be the displacement of the tip of the thread while unwinding. Thus the total work done by the force is $Fs$.
Now the change in kinetic energy of the pulley is given by
\[\vartriangle K=\dfrac{1}{2}I{{\omega }^{2}}\]………..$(1)$, where $I$ is the moment of inertia and$\omega $ is the angular velocity. Now the value of $I$ for a cylinder about the central axis is given by
$I=\dfrac{1}{2}M{{R}^{2}}$, where $M$ is the mass and $R$ is the radius of the cylinder. Again if $v$ is the linear velocity of the thread then we have
$\begin{align}
  & v=wR \\
 & or w=\dfrac{v}{R} \\
\end{align}$
Putting these values in (1) we get
$\vartriangle K=\dfrac{1}{2}\times \dfrac{1}{2}M{{R}^{2}}\times {{(\dfrac{v}{R})}^{2}}=\dfrac{1}{4}M{{v}^{2}}$. Thus
$Fs=\dfrac{1}{4}M{{v}^{2}}$ ……………$(2)$
Now if $a$ be the acceleration of the thread then
$\begin{align}
  & s=0\times 1+\dfrac{1}{2}a\times {{1}^{2}} \\
 & or s=\dfrac{a}{2} \\
\end{align}$
And
$\begin{align}
  & v=0+a\times 1 \\
 & \Rightarrow v=a \\
\end{align}$
Thus we get
$v=2s$.
Putting this value in $(2)$ we get
\[Fs=\dfrac{1}{4}M{{(2s)}^{2}}\]
$F=Ms$ …….$(3)$
Now the displacement of the tip is given by
$s=2\pi Rn$, where $n$ is the number of wounds and it is $4$ here. So putting the values of all the quantities we get
$s=2\times 3.14\times 0.15\times 4=3.768m$.
Putting this value and the value of $M$ in $(3)$ we get,
$F=5\times 3.768N=18.84N$

So, the correct answer is “Option B”.

Note:
Here we have applied conservation of energy theorem.We have to eliminate the acceleration of the thread from the given conditions. We also have to use the formula for the moment of inertia to get the change in kinetic energy of the cylinder. We have to change the diameter of the pulley in the SI system.