Question 240: Centre of Mass and Rotational Motion

Question

Consider a uniform rod of mass M = 4m and
length $$\ell $$ pivoted about its centre. A mass m
moving with velocity v making angle $$	heta  = {\pi  \over 4}$$ 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 :

Accepted Answer

About hinge(O) net torque $$	au $$ = 0
So angular momentum is conserved about hinge(O),
L
initial
= L
final
m$$\left( {{v \over {\sqrt 2 }}} ight)\left( {{l \over 2}} ight)$$ + 0 = $$\left[ {{{4m{l^2}} \over {12}} + m{{\left( {{l \over 2}} ight)}^2}} ight]\omega $$
$$ \Rightarrow $$ $$\omega $$ = $${{3\sqrt 2 } \over 7}{v \over \ell }$$

Answer

$${{3\sqrt 2 } \over 7}{v \over \ell }$$

Answer

$${3 \over 7}{v \over \ell }$$

Answer

$${3 \over {7\sqrt 2 }}{v \over \ell }$$

Answer

$${4 \over 7}{v \over \ell }$$

Centre of Mass and Rotational Motion

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 :

Choose the correct answer: