JEE MAIN - Physics (2020 - 8th January Morning Slot - No. 6)
Consider a uniform rod of mass M = 4m and
length $$\ell $$ pivoted about its centre. A mass m
moving with velocity v making angle $$\theta = {\pi \over 4}$$ to
the rod's long axis collides with one end of the
rod and sticks to it. The angular speed of the
rod-mass system just after the collision is :
$${{3\sqrt 2 } \over 7}{v \over \ell }$$
$${3 \over 7}{v \over \ell }$$
$${3 \over {7\sqrt 2 }}{v \over \ell }$$
$${4 \over 7}{v \over \ell }$$
Explanation
_8th_January_Morning_Slot_en_6_1.png)
About hinge(O) net torque $$\tau $$ = 0
So angular momentum is conserved about hinge(O),
Linitial = Lfinal
m$$\left( {{v \over {\sqrt 2 }}} \right)\left( {{l \over 2}} \right)$$ + 0 = $$\left[ {{{4m{l^2}} \over {12}} + m{{\left( {{l \over 2}} \right)}^2}} \right]\omega $$
$$ \Rightarrow $$ $$\omega $$ = $${{3\sqrt 2 } \over 7}{v \over \ell }$$
Comments (0)
