Question

A system consists of two identical small balls of mass $2\text{kg}$ each connected to the two ends of a $1\text{m}$ long light rod. The system is rotating about a fixed axis through the centre of the rod and perpendicular to it at an angular speed of $9\text{rad/s}$. An impulsive force of average magnitude $10\text{N}$ acts on one of the masses in the direction of its velocity for $0.20\text{s}$. Calculate the new angular velocity of the system.

• A. $9\text{rad/s}$
• B. $8\text{rad/s}$
• C. $10\text{rad/s}$
• D. $5\text{rad/s}$

Answer 1

The correct option is C $10\text{rad/s}$

Given,
Mass of identical balls, $m=2\text{kg}$
Length of rod $\left(l\right)=1\text{m}$
Angular speed of rod $w=9\text{rad/s}$
Impulsive force $F=10\text{N}$
As we know, torque
$\overrightarrow{\tau }=\frac{d\overrightarrow{L}}{dt}$
where $\tau =\overrightarrow{F}\times \overrightarrow{d}$
i.e $\overrightarrow{\tau }.dt=d\overrightarrow{L}$
or $\tau .\Delta t=\Delta L$
or $(F.d).\Delta t=(L_f-L_i)---\left(1\right)$

As we know, $L=I\omega$
Initial angular momentum $L_i=I{\omega }_1$
$I$ = $m{\left(\frac{l}{2}\right)}^2+m{\left(\frac{l}{2}\right)}^2=\frac{m{l}^2}{2}$
$=\frac{2\times 1^2}{2}=1{\text{kg - m}}^2$
Thus, $L_i=1\times 9=9\text{kg}\frac{m^2}{s}$
Final angular momentum $L_f=I{\omega }_2$
$L_f=1\times {\omega }_2$
Now, from eq $\left(1\right)$:
$(F.d)\Delta t=L_f-L_i$
$10\times \left(\frac{1}{2}\right)\times 0.2={\omega }_2-9$
${\omega }_2=10\frac{\text{rad}}{s}$