Question

A thin rod $AB$ is held horizontally so that it can freely rotate in a vertical plane about the end $A$ as shown in the figure. The potential energy of the rod when it hangs vertically is taken to be zero. The end $B$ of the rod is released from rest from a horizontal position. At the instant the rod makes an angle $\theta$ with the horizontal:

• A. the speed of end $B$ is proportional to $\sqrt{\sin \theta }$
• B. the potential energy is proportional to $(1-\cos \theta )$
• C. the angular acceleration is proportional to $\cos \theta$
• D. the torque about $A$ remains the same as its initial value

Answer 1

Answer: (A), (C)

the speed of end $B$ is proportional to $\sqrt{\sin \theta }$
the angular acceleration is proportional to $\cos \theta$

Let the mass and length of the rod be $m$ and $L$ respectively.

Points $C,C^{\prime }{C}^{\prime \prime }$ represent the center of mass of the rod.

From geometry, we get $O{C}^{\prime }=\frac{L}{2}cos\theta$ and $OA=\frac{L}{2}sin\theta$

$⟹$ $OC=AC-OA=\frac{L}{2}(1-sin\theta )$

Potential energy of the rod $P.E=mg\left(OC\right)=mg\frac{L}{2}(1-sin\theta )$

$⟹$ $P.E\propto (1-sin\theta )$

Applying conservation of energy : $P.E_i+K.E_i=P.E_f+K.E_f$

$\therefore$ $mg\left(AC\right)+0=mg\left(OC\right)+\frac{1}{2}I{w}^2$

$\therefore$ $mg\frac{L}{2}+0=mg\frac{L}{2}(1-sin\theta )+\frac{1}{2}\times \frac{1}{3}m{L}^2{w}^2$ $⟹w=\sqrt{\frac{3g}{L}sin\theta }$

Speed of end B $v_B=wL=\sqrt{3gLsin\theta }$

$⟹$ $v\propto \sqrt{sin\theta }$

Torque about point A $\tau =mg\left(O{C}^{\prime }\right)=mg\frac{L}{2}cos\theta$

$⟹\tau \propto cos\theta$

From $\tau =I\alpha$, we get $mg\frac{L}{2}cos\theta =I\alpha$

$⟹$ $\alpha \propto cos\theta$