A thin cylindrical rod of length 2.40 m and mass 1.3 kg has two spheres attached

Question

A thin cylindrical rod of length 2.40 m and mass 1.3 kg has two spheres attached on its ends. The centers of the spheres are 1.20 m from the center of the rod. The mass of each sphere is M = 0.78 kg. The rod and spheres are initially at rest on a frictionless icy surface; the figure below is drawn looking down on that icy surface, with north towards the top of the page. A small mass m = 0.19 kg, sliding across the frictionless ice with a speed of v = 1.5 m/s and in the Eastward direction as shown, strikes the southernmost sphere and sticks to it. At the same instant, an identical mass having twice the speed, and also traveling to the East, strikes the northernmost sphere and also sticks. As a result of the simultaneous collisions, the entire system starts to translate and also to rotate. (a) Determine the speed of the center of mass of the rod after the simultaneous collisions. (b) Determine the magnitude and direction of the angular velocity of the system after the simultaneous collisions. You may treat the two moving masses and the two spheres at the ends of the rod as if they were point particles.

http://www.webassign.net/webassignalgphys1/10-p-060.gif

Explanation / Answer

the formula you are using is correct ie., Vcm = m1 v1 + m2 v2 / m1 + m2 = 65.103 x 0.64 + 92.103 x 1.1 / 65.103 + 92.103 = 0.91 m/sec ( nearly ) and this velocity of centre of mass will be the same before and after collision

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