Question

A thin uniform rod of length $3\text{m}$ and mass $M$ is held horizontally by two vertical strings attached to the two ends, as shown in figure. One of the strings is cut. Find the angular acceleration of the rod at the moment it is cut. (Take $g=10{\text{m/s}}^2$).

• A. $15{\text{rad/s}}^2$
• B. $10{\text{rad/s}}^2$
• C. $5{\text{rad/s}}^2$
• D. $7.5{\text{rad/s}}^2$

Answer 1

The correct option is C $5{\text{rad/s}}^2$

Given, length of rod $l=3\text{m}$
Just after cutting the string at $\text{B}$, the rod will tend to rotate about its other end i.e $\text{A}$, which is attached with a string.

Moment of inertia of rod about about end $\text{A}$,
$I=\frac{M{l}^2}{3}...\left(i\right)$


From the FBD of rod, since $T$ passes through point $A$, hence ${\tau }_T=0$
For $Mg$ the perpendicular distance from $A$ is $r_{\perp }=\frac{l}{2}$
$\therefore {\tau }_{Mg}=Mg\left(\frac{l}{2}\right)...\left(ii\right)$

Net external torque on rod about $\text{A}$ is
${\tau }_{\text{net}}={\tau }_T+{\tau }_{Mg}={\tau }_{Mg}$
Applying the fundamental equation for torque:
${\tau }_{\text{net}}=I\alpha ...\left(iii\right)$

From Eq $\left(i\right).\left(ii\right),\left(iii\right)$
$Mg\times \left(\frac{l}{2}\right)=\left(\frac{M{l}^2}{3}\right)\alpha$
or $\alpha =\frac{3}{2}\frac{g}{l}$
Substituting $l=3\text{m}$,
$\therefore \alpha =\frac{3\times 10}{2\times 3}=5{\text{rad/s}}^2$