A uniform bar of length 12 cm and mass 20m lies on a smooth horizontal table. Two point masses m and 2m are moving in opposite directions with the same speed v and in the same plane as the bar. The 2m strikes the bar 2 cm from its centre on one side and the m strikes it 4 cm from the centre on the…

Question

A uniform bar of length 12 cm and mass 20m lies on a smooth horizontal table. Two point masses m and 2m are moving in opposite directions with the same speed v and in the same plane as the bar. The 2m strikes the bar 2 cm from its centre on one side and the m strikes it 4 cm from the centre on the other side, and both get stuck to it. After the collision the entire system rotates with angular frequency ω. The ratio of v and ω is

Accepted Answer

(3) 33. First locate the centre of mass of the final system, measuring from one end of the bar. x_cm=(20m(6)+2m(4)+m(10))/(23m)=(120+8+10)/(23)=6 cm — it stays at the centre of the bar, which makes the rest easy. The table is smooth, so angular momentum about that point is conserved through the collision. Before: the 2m is 2 cm from the centre moving one way, the m is 4 cm from it moving the other way, and both turn the system the same way. L=2m v(2)+m v(4)=8mv (in cm units). After: I=((20m)(12)²)/(12)+2m(2)²+m(4)²=240m+8m+16m=264m cm². ω=L/I=(8mv)/(264m)=v/(33). v/ω=33.

Step-by-step solution

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