Angular Collision. A uniform rod of mass M and length d is initially at rest on a horizontal and frictionless table in the x-y plane. the figure attached is a top view, with gravity pointing into the page. The rod is free to rotate about an axis perpendicular to the plane and is passing through the pivot point at a distance d/3 measured from one of its ends as shown. A small point particle of mass m = M/4, which is moving with speed $v_0$, hits the rod and sticks to it at the point of impact at a distance d/3 from the pivot.

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$a)$ Find the magnitude of the angular velocity of the rod-and-mass system after the collision.

$b)$ Using again $M = 4m$, find the speed of the centre of mass of the rod right after the collision.


I'm assuming you start to answer this question by using the equation of angular momentum $L = I\omega$ Really confused by this if someone could help me understand the question. Thanks in advance.


There is no external force acting on the system of ball and the rod except for gravity which acts along the $z$ axis. Hence no external torque is produced on the system. We know that $$\tau=\frac {d\mathbf L}{dt}$$ $$0=\frac {d\mathbf L}{dt}$$. Thus the angular momentum is conserved. . Hence by conserving angular momentum about the pivot point we get $$\frac {mv_0d}{3}=\frac {(M+m)d^2\omega}{9}$$ Hence we shall get

$$\omega= \frac {3mv_0}{(M+m)d}$$ While for the second part, by conserving linear momentum we get $$mv_0= (m+4m)v_{cm}$$ $$\Rightarrow v_{cm}= \frac {v_0}{5}$$

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  • $\begingroup$ Hi, thanks for the reply, could you just break down the first and second line of the solution. Thanks @manthanein $\endgroup$ – Ben Jones Jan 14 '18 at 19:18
  • $\begingroup$ What do you want me to explain exactly? $\endgroup$ – Rohan Shinde Jan 15 '18 at 2:21
  • $\begingroup$ @downvoter Why the downvote? $\endgroup$ – Rohan Shinde Jan 15 '18 at 3:28

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