A small block (mass m) slides down a circle path (radius R) which is cut into a

Question

A small block (mass m) slides down a circle path (radius R) which is cut into a large block (mass M), as shown. M rests on a table and both blocks slide without friction. The blocks are initially at rest, and m starts at the top of the path. Determine the velocity of m when it loses constant with the large block. Hand in (Room 1312) a derivation of the formula: v=[2MgR/(M+m)]1/2 [Hint: The center of mass point does not move.] Calculate v, for the following parameter values. Data: M=3.1 kg; m=1.4 kg; R=0.8 m.

Explanation / Answer

(a) According to conservation of energy,the initial system of energy is equals to the final system of energy. MgR + 1/2mv^2 = 1/2(M+m)V^2 MgR = 1/2(M+m)V^2    (since block m is at rest, v = 0) V = (2MgR/M+m)^1/2 the velocity of the block when it loses the contact is (2MgR/M+m)^1/2 . (b)the velocity V = (2*3.1*9.8*0.8/4.5)^1/2       V = 3.286 m/s

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