In the figure, a small 0.235 kg block slides down a frictionless surface through height h = 0.471 m and then sticks to a uniform vertical rod of mass M = 0.470 kg and length d = 2.36 m. The rod pivots about point Othrough angle θ before momentarily stopping. Find θ.
Ki=Kf -----> 0.5mv^2=mgh , and this gives expression for 'v'
w^2=w0^2 + 2*alpha*(delta ø) , for when alpha(angular acceleration) is constant
The Attempt at a Solution
in effort to find ø, alpha and w are needed. I established that m*v=I*w (not sure if this is correct)
by substituting relevant equations, i got w0=[sqrt(2gh)]/d^2
and w, is according to the question, 0.
now i can't seem to find an expression for alpha.
could u point out what i missed out?
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