Question

A uniform meter rod consists of half wood and the remaining steel as shown in the figure. $a_1$ is the angular acceleration of the rod about O when force is applied as shown in figure A, whereas $a_2$ is the angular acceleration about $O^1$ when the force is applied as shown in figure B. The following choice is true:

• A. $a_1=a_2$
• B. $a_1<a_2$
• C. $a_1>a_2$
• D. None of these

Answer 1

The correct option is B $a_1<a_2$

A uniform meter rod of length $1m$, consists of half wood and the remaining steel as shown in the figure.

Let suppose the mass of half part of wood be $m_{w}kg$ and that of steel part be $m_{s}kg$.

Torque $\left(\tau \right)$ which is applied by the force $F$ is same in fig $A$ and fig $B$

$\because$ Both lengths of the rod and the applied force are the same.

$\Rightarrow$ $\tau =R\times F$ (where $R=1m$ is the length of rod)

In fig $A$, the pivot point is $O$ and moment of inertia about $O$ is given as;

$I_o={\sum }_i{m}_i{r_i}^2=m_w{\left(\frac{1}{4}\right)}^2+m_s{(\frac{1}{2}+\frac{1}{4})}^2=\frac{m_w+9m_s}{16}$

In fig $B$, the pivot point is $O^1$ and moment of inertia about $O^1$ is given as;

$I_{o^1}={\sum }_i{m}_i{r_i}^2=m_s{\left(\frac{1}{4}\right)}^2+m_w{(\frac{1}{2}+\frac{1}{4})}^2=\frac{m_s+9m_w}{16}$

Since, $m_s>m_w$ therefore $I_o>I_{o^1}$

Now, $Torque\left(\tau \right)=I\alpha =R\times F$

Here, $\tau$ is constant $\Rightarrow$ $I\propto \frac{1}{\alpha }$

Hence, if $I_o>I_{o^1}$ then ${\alpha }_1<{\alpha }_2$