Question 86: Centre of Mass and Rotational Motion

Question

Two masses 400 g and 350 g are suspended from the ends of a light string passing over a heavy pulley of radius 2 cm . When released from rest the heavier mass is observed to fall 81 cm in 9 s . The rotational inertia of the pulley is $\_\_\_\_$ $\mathrm{kg} \cdot \mathrm{m}^2$. $\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right)$

Accepted Answer

The heavier mass starts from rest ( $u=0$ ) and falls a certain distance ( $h$ ) in time ( $t$ ).
Using the kinematic equation :
$$ \mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2 $$
$$\Rightarrow $$ $$ \mathrm{h}=\frac{1}{2} \mathrm{at}^2 \Rightarrow \mathrm{a}=\frac{2 \mathrm{~h}}{\mathrm{t}^2} $$
The given distance $\mathrm{h}=81 \mathrm{~cm}=0.81 \mathrm{~m}$
The time taken is $\mathrm{t}=9 \mathrm{~s}$
$$ a=\frac{2 	imes 0.81}{9^2}=\frac{1.62}{81} $$
$$\Rightarrow $$ $$ \mathrm{a}=0.02 \mathrm{~m} / \mathrm{s}^2 $$
Since the pulley has mass (and inertia), the tension is not the same on both sides. Let $\mathrm{T}_1$ be the tension on the heavier mass side ( $\mathrm{m}_1$ ) and $\mathrm{T}_2$ be the tension on the lighter mass side ( $\mathrm{m}_2$ ).
For heavier mass ( $\mathrm{m}_1=0.4 \mathrm{~kg}$ ) which is moving down.
$$ \mathrm{m}_1 \mathrm{~g}-\mathrm{T}_1=\mathrm{m}_1 \mathrm{a} $$
$$\Rightarrow $$ $$ \mathrm{T}_1=\mathrm{m}_1(\mathrm{~g}-\mathrm{a}) \ldots(\mathrm{i}) $$
For lighter mass ( $\mathrm{m}_2=0.35 \mathrm{~kg}$ ) which is moving up.
$$ \mathrm{T}_2-\mathrm{m}_2 \mathrm{~g}=\mathrm{m}_2 \mathrm{a} $$
$$\Rightarrow $$ $$ \mathrm{T}_2=\mathrm{m}_2(\mathrm{~g}+\mathrm{a}) \ldots 	ext { (ii) } $$
The pulley rotates due to the difference in tensions. The net torque $(	au)$ causes angular acceleration $(\alpha)$.
$$ 	au=\left(\mathrm{T}_1-\mathrm{T}_2ight) \mathrm{R} $$
Also, from Newton's Second Law for rotation:
$$ 	au=\mathrm{I} \alpha $$
Therefore :
$$ \left(\mathrm{T}_1-\mathrm{T}_2ight) \mathrm{R}=\mathrm{I} \alpha $$
Assuming the string does not slip, the linear acceleration of the string
$$ \mathrm{a}=\mathrm{R} \alpha \Rightarrow \alpha=\frac{\mathrm{a}}{\mathrm{R}} $$
Substituting in the torque equation :
$$ \left(\mathrm{T}_1-\mathrm{T}_2ight) \mathrm{R}=\mathrm{I}\left(\frac{\mathrm{a}}{\mathrm{R}}ight) $$
$$ \mathrm{I}=\frac{\left(\mathrm{T}_1-\mathrm{T}_2ight) \mathrm{R}^2}{\mathrm{a}} \ldots (iii)$$
On subtracting (ii) from (i)
$$ \mathrm{T}_1-\mathrm{T}_2=\left[\mathrm{m}_1(\mathrm{~g}-\mathrm{a})ight]-\left[\mathrm{m}_2(\mathrm{~g}+\mathrm{a})ight] $$
$$\Rightarrow $$ $\mathrm{T}_1-\mathrm{T}_2=\left(\mathrm{m}_1 \mathrm{~g}-\mathrm{m}_1 \mathrm{a}ight)-\left(\mathrm{m}_2 \mathrm{~g}+\mathrm{m}_2 \mathrm{a}ight)$
$$\Rightarrow $$ $$ \mathrm{T}_1-\mathrm{T}_2=\left(\mathrm{m}_1-\mathrm{m}_2ight) \mathrm{g}-\left(\mathrm{m}_1+\mathrm{m}_2ight) \mathrm{a} $$
Putting the values,
$$ \mathrm{T}_1-\mathrm{T}_2=(0.400-0.350) 	imes 9.8-(0.400+0.350) 	imes 0.02 $$
$$\Rightarrow $$ $\mathrm{T}_1-\mathrm{T}_2=(0.05 	imes 9.8)-(0.75 	imes 0.02)$
$$\Rightarrow $$ $$ \mathrm{T}_1-\mathrm{T}_2=0.475 \mathrm{~N} $$
So, putting in equation (iii),
$$ \mathrm{I}=\frac{(0.475) 	imes \mathrm{R}^2}{\mathrm{a}} $$
Given radius $\mathrm{R}=2 \mathrm{~cm}=0.02 \mathrm{~m}$.
$$ I=\frac{0.475 	imes(0.02)^2}{0.02}=0.475 	imes 0.02=9.5 	imes 10^{-3} \mathrm{kgm}^2 $$
Therefore, the rotational inertia of the pulley is $9.5 	imes 10^{-3} \mathrm{kgm}^2$.
Hence, the correct option is (D).

Answer

$8.3 	imes 10^{-3}$

Answer

$4.75 	imes 10^{-3}$

Answer

$1.86 	imes 10^{-2}$

Answer

$9.5 	imes 10^{-3}$

Centre of Mass and Rotational Motion

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