Question

A uniform rod of mass $15\text{kg}$ and length $3\text{m}$ hinged at point $\text{O}$, is held stationary initially with the help of a light string (connected at point $\text{A}$) as shown in figure. At $t=0$, if the string breaks, find the hinge force acting on the rod at that instant. (Take $g=10{\text{m/s}}^2)$

• A. $122\text{N}$
• B. $82.5\text{N}$
• C. $11.5\text{N}$
• D. $15\text{N}$

Answer 1

The correct option is A $122\text{N}$

Given, mass of the rod $\left(m\right)=15\text{kg}$,
length of the rod $l=3\text{m}$
MOI of the rod about $\text{A},I=\frac{m{l}^2}{3}$

Torque about hinge $\text{O}$ is due to weight of the rod alone.
$\tau =I\alpha$
$\Rightarrow mg\times \frac{l}{2}\sin \theta =\frac{m{l}^2}{3}\alpha$
$\Rightarrow \alpha =\frac{3g}{2l}\sin \theta =\frac{3\times 10}{2\times 3}\sin {37}^{\circ }=3{\text{rad/s}}^2$

FBD of the rod just after it is released is given below.


Applying force equations for the COM of rod at $t=0$ :
Net radial force $=0$
($\because$ angular velocity of rod just after it's released is zero, hence centripetal acceleration of COM is also zero)
$\Rightarrow F_r=mg\cos {37}^{\circ }$
$\Rightarrow F_r=15g\times \frac{4}{5}=120\text{N}$

Net tangential force $=mg\sin {37}^{\circ }-F_t=m{a}_t$
$\Rightarrow F_t=mg\sin {37}^{\circ }-m{a}_t$

We know that tangential acceleration of COM
$a_t=r\alpha =\left(\frac{l}{2}\right)\alpha =4.5{\text{m/s}}^2$
$\Rightarrow F_t=15g\times \frac{3}{5}-15\times 4.5=22.5\text{N}$

Therefore, total hinge force $F=\sqrt{F_r^2+F_t^2}$
$=122\text{N}$