Question

Two masses are attached to a rod end to end. If torque is applied, they rotate with angular acceleration $\alpha$.If their distances are doubled and same torque is applied, then the move with angular acceleration:

• A. 4$\alpha$
• B. $\alpha$
• C. 3$\alpha$
• D. $\alpha$/4

Answer 1

Answer: (D)

$\alpha$/4

Suppose the masses $m_1$and $m_2$ are attached to the ends of a massless rod. $C_1$ is the axis of rotation which is at a distance R from each of the masses.

.'. The moment of inertia of the system,

$I=m_1{R}^2+m_2{R}^2=(m_1+m_2)R^2$

Now if the distance between the masses is doubled, then the moment of inertia,

$I^{\prime }=m_1(2R{)}^2+m_2(2R{)}^2=(m_1+m_2)4R^2$

$\tau =I\alpha =I^{\prime }{\alpha }^{\prime }$

Since the same torque is applied in both the cases,

So New angular acceleration ${\alpha }^{\prime }=\frac{I^{\prime }\alpha }{I}=\frac{(m_1+m_2)R^2\alpha }{(m_1+m_2)4R^2}=\frac{\alpha }{4}$