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A block of mass M carrying a smooth semicircular groove of radius R sits on a frictionless horizontal floor. A small uniform cylinder of radius r and mass m is let go from rest at the upper edge A of the groove and slides without friction down to the lowest point B. Through what horizontal distance does the block shift, and what is its speed, at the instant the cylinder arrives at B?
- Block shifts m(R-r)/(M+m); block speed = m*sqrt(2g(R-r)/(M+m)) / sqrt(M+m)... i.e. v_M = m*sqrt(2g(R-r)/(M(M+m)))
- Block shifts (R-r); block speed = sqrt(2g(R-r))
- Block shifts M(R-r)/(M+m); block speed = M*sqrt(2g(R-r)/(M(M+m)))
- Block shifts m(R-r)/M; block speed = sqrt(2g(R-r))
Correct answer: Block shifts m(R-r)/(M+m); block speed = m*sqrt(2g(R-r)/(M+m)) / sqrt(M+m)... i.e. v_M = m*sqrt(2g(R-r)/(M(M+m)))
Solution
Because the floor is frictionless, the system's horizontal momentum and centre of mass are conserved. The cylinder descends (R-r) vertically while the block recoils so the COM stays put.
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