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.


closed as off-topic by Bill N, stafusa, knzhou, Jon Custer, JMac Jan 15 '18 at 17:28

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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 $$mv_0= (m+4m)v_{cm}$$ $$\Rightarrow v_{cm}= \frac {v_0}{5}$$

  • $\begingroup$ 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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