Question

The density of a rod $AB$ increases continuously from $A$ to $B.$ Is it easier to set it in rotation by clamping it at $A$ and applying a perpendicular force at $B$ or by clamping it at $B$ and applying the force at $A?$
Explain your answer.

Answer 1

Solution

REF.Image

It will be easier when clamping

at B. Because we know that,

$T=\left(I\right)\alpha$

If same force (Torque)

is applied in both cases.

In case - I,

$T=\left(I_1\right){\alpha }_1$

$I_1=I_{COM}+M{a}^2$ [parallel axis theor.]

$\Rightarrow \frac{I}{I_{COM}+M{a}^2}$

Similarly in case - II

$\frac{T}{I_{COM}+M{b}^2}={\alpha }_2$

As the mass distributed is increasing

towards B, $a<b$. so,

$\frac{T}{I_{COM}+M{a}^2}<\frac{T}{I_{COM}+M{b}^2}$

$\Rightarrow {\alpha }_1<{\alpha }_2$

By applying same force, we are

getting better angular acceleration