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## Homework Statement

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

*O*through angle

*θ*before momentarily stopping. Find

*θ*.

## Homework Equations

L=I*w

P=mv

v=wr

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?