Question

A block of mass 2 kg, hangs from the rim of a wheel of radius $0.5$ m. On releasing from rest, the block falls through 5m height in 2s. the moment of inertia of the wheel is (take $g = 10m/{s^2}$)

(A) $1kg{m^2}$
(B) $3.2kg{m^2}$
(C) $2.5kg{m^2}$
(D) $1.5kg{m^2}$

Answer 1

The acceleration of the block is the same as the linear acceleration of the rim. The tension in the string is the driver of the rim.
Formula used: In this solution we will be using the following formulae;
$s = ut + \dfrac{1}{2}a{t^2}$ where $s$ is the distance covered by an accelerating body, $u$ is the initial velocity, $a$ is the acceleration of the body, and $t$ is the time elapsed.
$Fr = I\alpha$ where $F$ is the force acting on a body, $r$ is the distance of $F$ from an axis of rotation, $I$ is the moment of inertia of the body and $\alpha$ is the angular acceleration. The quantity, $Fr$ is the torque on a body.
$\alpha = \dfrac{a}{r}$ where $a$ is the linear acceleration, and $r$ is the radius of a body.

Complete Step-by-Step solution:
Generally, moment of inertial and the force acting on a body are related through
$Fr = I\alpha$ where $F$ is the force acting on a body, $r$ is the distance of $F$ from an axis of rotation, $I$ is the moment of inertia of the body and $\alpha$ is the angular acceleration.
Hence, to find the moment of inertia, we need to know the force of the rim and the angular acceleration.
Angular acceleration is $\alpha = \dfrac{a}{r}$ where $a$ is the linear acceleration, and $r$ is radius.
The acceleration can be gotten from
$s = ut + \dfrac{1}{2}a{t^2}$ where $s$ is the distance covered by an accelerating body, $u$ is the initial velocity, $a$ is the acceleration of the body, and $t$ is the time elapsed.
Hence,
$5 = \dfrac{1}{2}a{\left( 2 \right)^2}$
$\Rightarrow a = \dfrac{5}{2}m/s$
Then
$\alpha = \dfrac{{\dfrac{5}{2}}}{{0.5}} = 5rad/{s^2}$
The force driving the rim is the tension, hence to calculate tension on string, we perform newton's law analysis on block
$mg - T = ma$
$\Rightarrow 2\left( {10} \right) - T = 2\left( {\dfrac{5}{2}} \right)$
Hence,
$T = 15N$
Then,
$Tr = I\alpha$
$\Rightarrow 15(0.5) = I\left( 5 \right)$
Then by dividing both side by 5, we have
$I = \dfrac{{15\left( {0.5} \right)}}{5} = 2.5kg{m^2}$

Hence, the correct option is C

Note: For clarity, the tension is the force which drives the rim because the string is the object directly in contact with the rim. The tension is as a result of the block hanging down, however it’s not the weight that drives it but the transmitted force along the string (which is the tension).