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

Question

A small block (mass m) slides down a circular 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 contact with the large block. (a) Hand in a derivation of the formula: v=[2MgR/(M+m)]1/2 [HINT: The center of mass point does not move.] (b) Calculate v, for the following parameter values. Data: M = 3.0 kg; m = 1.6 kg; R = 1.0 m.

Explanation / Answer

We can do this as following. From conservation of momentum 0 = MV_M + mV_m or from center of mass we get the same result as above. So, V_M = -mV_m/M Using energy equation mg*R = .5*(M*V_M^2 + m*V_m^2) substituting the value of V_M mgR = .5*(M*V_m^2*m^2/M^2 + m*V_m^2) on rearranging we get 2MgR = V_m^2 (M+m) V_m = sqrt(2MgR/(M+m)) which is required to prove b) substituting the values. M = 3.0 kg; m = 1.6 kg; R = 1.0 m. V_m = sqrt(2*3*9.8*1/4.6) = 3.575m/s

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