Question

A uniform rod or length I m and mass 4 kg can rotate freely in a vertical plane about its end A. The rod is initially held in a horizontal position and then released. At the time the rod makes an angle ${30}^{\circ }$ With the vertical. then

• A. its angular acceleration is 7.5 $\mathrm{rad}/s^2$
• B. its angular velocity is 5.1 $\mathrm{rad}/s^2$
• C. its total acceleration vector of tip p is $\left[{\omega }^2\ell \right(\sin \theta )+\alpha \ell \cos \theta .](-\hat{i})+\left[{\omega }^{2}e\right(\cos \theta )-\alpha \ell \sin \theta .]\left(\hat{j}\right)$
• D. None of these

Answer 1

Answer: (A), (C)

The correct options are
A its angular acceleration is 7.5 $\mathrm{rad}/s^2$
C its total acceleration vector of tip p is $\left[{\omega }^2\ell \right(\sin \theta )+\alpha \ell \cos \theta .](-\hat{i})+\left[{\omega }^{2}e\right(\cos \theta )-\alpha \ell \sin \theta .]\left(\hat{j}\right)$

$\begin{array}{c}mg\frac{l}{2}\sin \left(\frac{\pi }{6}\right)=\frac{M{l}^2}{3}.\propto \\ \Rightarrow \frac{3}{2}\frac{g}{l}.\sin \frac{\pi }{6}=\propto \\ \Rightarrow \propto =\frac{3g}{4l}\\ =\frac{3\times 10}{4\times 1}=\frac{30}{4}=7.5rad/s^2\end{array}$

and total ace of tip:

$a=-\left(w^{2}l\sin \theta +al\cos \theta \right)\hat{i}+\left(w^{2}l\cos \theta +al\sin \theta \right)\hat{j}$

$\therefore$ Option $A,C$ is correct.