Question 100: Centre of Mass and Rotational Motion

Question

A uniform rod of mass $m$ and length $l$ suspended by means of two identical inextensible light strings as shown in figure. Tension in one string immediately after the other string is cut, is $\_\_\_\_$ . $(g$ acceleration due to gravity)

Accepted Answer

Immediately after one string is cut, the only remaining upward force is the Tension (T) at end A and the downward force is the Weight $(\mathrm{mg})$ acting at the centre of mass (distance $1 / 2$ from each end).
Let a be the downward linear acceleration of the center of mass and $\alpha$ be the angular acceleration of the rod about its centre of mass.
For translational motion
Using Newton's Second Law ( $\mathrm{F}_{	ext {net }}=\mathrm{ma}$ ):
$$ \mathrm{mg}-\mathrm{T}=\mathrm{ma} $$ .....(i)
For rotational motion
The torque $(	au)$ about the centre of mass is provided by the tension T at distance $\frac{1}{2}$ :
$$ 	au=\mathrm{T} \cdot \frac{1}{2} $$
Also, $	au=\mathrm{I} \alpha$, where I is the moment of inertia of a rod about its center $\left(\mathrm{I}=\frac{1}{12} \mathrm{ml}^2ight)$ :
$$ \mathrm{T} \cdot \frac{\mathrm{l}}{2}=\left(\frac{1}{12} \mathrm{ml}^2ight) \alpha $$
Immediately after the cut, end A is effectively the pivot point. The relationship between linear acceleration (a) of the center and angular acceleration ( $\alpha$ ) is:
$$ a=\alpha \cdot \frac{l}{2} \Rightarrow \alpha=\frac{2 a}{l} \ldots (ii)$$
From relations (ii) and (iii)
$$ \mathrm{T} \cdot \frac{\mathrm{l}}{2}=\frac{1}{12} \mathrm{ml}^2\left(\frac{2 \mathrm{a}}{\mathrm{l}}ight) $$
$$\Rightarrow $$ $\mathrm{T}=\frac{4 \mathrm{ma}}{12}=\frac{\mathrm{ma}}{3}$
$$ \Rightarrow \mathrm{ma}=3 \mathrm{~T} $$
From relation (i),
$$ \mathrm{mg}-\mathrm{T}=3 \mathrm{~T} $$
$$\Rightarrow $$ $\mathrm{mg}=4 \mathrm{~T}$
$$\Rightarrow $$ $$ \mathrm{T}=\frac{\mathrm{mg}}{4} $$
Therefore, immediately after one string is cut, the tension in the remaining string $\frac{\mathrm{mg}}{4}$.
Hence, the correct option is (B).

Answer

$\mathrm{mg} / \mathrm{s}$

Answer

$m g / 4$

Answer

$m g$

Answer

$m g / 2$

Centre of Mass and Rotational Motion

A uniform rod of mass m and length l suspended by means of two identical inextensible light strings as shown in figure. Tension in one string immediately after the other string is cut, is \_\_\_\_ . (g acceleration due to gravity)

Choose the correct answer: