Question 296: Centre of Mass and Rotational Motion

Question

A rigid massless rod of length 3l has two masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal axis (see figure). When released from initial horizontal position, its
instantaneous angular acceleration will be -

Accepted Answer

Applying torque equation about point P.
2M
0
(2l) $$-$$ 5 M
0
gl = I$$\alpha $$
I = 2M
0
(2l)
2
+ 5M
0
l
2
= 13 M
0
l
2
d
$$ 	herefore $$  $$\alpha $$ = $$-$$ $${{{M_0}gl} \over {13{M_0}{\ell ^2}}}$$ $$ \Rightarrow $$ $$\alpha $$ = $$-$$ $${g \over {13\ell }}$$
$$ 	herefore $$  $$\alpha $$ = $${g \over {13\ell }}$$ anticlockwise

Answer

$${g \over {13l}}$$

Answer

$${g \over {2l}}$$

Answer

$${g \over {3l}}$$

Answer

$${7g \over {3l}}$$

Centre of Mass and Rotational Motion

A rigid massless rod of length 3l has two masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal axis (see figure). When released from initial horizontal position, its instantaneous angular acceleration will be -

Choose the correct answer: