Question

Two identical masses are connected to a horizontal thin massless rod as shown in the figure. When their distance from the pivot is $x$, a torque produces an angular acceleration ${\alpha }_1$. If the masses are now repositioned so that they are at distance $2x$ each from the pivot, the same torque will produce an angular acceleration ${\alpha }_2$ such that :

• A. ${\alpha }_2=4{\alpha }_1$
• B. ${\alpha }_2={\alpha }_1$
• C. ${\alpha }_2=\frac{{\alpha }_1}{2}$
• D. ${\alpha }_2=\frac{{\alpha }_1}{4}$

Answer 1

The correct option is D ${\alpha }_2=\frac{{\alpha }_1}{4}$

Torque $\tau$ acting when distance is x
$\tau =\left(2m{x}^2\right){\alpha }_1$
When the distance is increased to 2x we get
$\tau =(2\times m\times 4x^2){\alpha }_2$
Equating we get
${\alpha }_2=\frac{{\alpha }_1}{4}$