JEE MAIN - Physics (2019 - 10th January Evening Slot - No. 7)
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 -
_10th_January_Evening_Slot_en_7_1.png)
$${g \over {13l}}$$
$${g \over {2l}}$$
$${g \over {3l}}$$
$${7g \over {3l}}$$
Explanation
_10th_January_Evening_Slot_en_7_2.png)
Applying torque equation about point P.
2M0 (2l) $$-$$ 5 M0 gl = I$$\alpha $$
I = 2M0 (2l)2 + 5M0 l2 = 13 M0l2d
$$ \therefore $$ $$\alpha $$ = $$-$$ $${{{M_0}gl} \over {13{M_0}{\ell ^2}}}$$ $$ \Rightarrow $$ $$\alpha $$ = $$-$$ $${g \over {13\ell }}$$
$$ \therefore $$ $$\alpha $$ = $${g \over {13\ell }}$$ anticlockwise
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