In the figure, a small 0.124 kg block slides down a frictionless surface through

Question

In the figure, a small 0.124 kg block slides down a frictionless surface through height h = 0.545 m and then sticks to a uniform vertical rod of mass M = 0.569 kg and length d = 2.27 m. The rod pivots about point O through angle theta before momentarily stopping. Find theta.

In the figure, a small 0.124 kg block slides down a frictionless surface through height h = 0.545 m and then sticks to a uniform vertical rod of mass M = 0.569 kg and length d = 2.27 m. The rod pivots about point O through angle theta before momentarily stopping. Find theta.

Explanation / Answer

0.5mv^2= mgh

v = sqrt(2*9.8*0.545)=3.26

now by angular mementum conservation

mvr = (I/r +mr)u

mv = (M/3 +m)u

u =0.124*3.26/(0.545 /3+0.124) =1.32

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