(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A sledgehammer with a mass of 2.60kg is connected to a frictionless pivot at the tip of its handle. The distance from the pivot to the center of mass is r_{cm}=0.610m, and the moment of inertia about the center of mass is I_{cm}=0.04kg*m^{2}. If the hammer is released from rest at an angle of Θ=53.0° such that H=0.487m, what is the speed of the center of mass when it passes through horizontal?

2. Relevant equations

Thin rod, about end I=1/3M*r_{cm}=0.529 kg*m^{2}

I=I_{cm}+Md^{2}

Then some torque equations I don't know which to use, if any

τ=r*F*sinϕ

τ=m*r^{2}*α

τ_{grav}=-M*g*r_{cm}

and α=τ_{net}/I

And eventually I'll have to use either

v_{tan}=ω*r or v=√a_{rad}*r

360°=2π radians

3. The attempt at a solution

I think F_{grav}runs along H. ϕ=143° so F_{tan}=F_{grav}*sin(-143°) where F_{grav}=m*g.

I can also get I=I_{cm}+M*r_{cm}^{2}but then I get lost and don't know where to go with all the info. If I get α, I don't know how I would use rotational kinematics since there is no time given. Any clues to get going would help. Thanks

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# Sledgehammer on pivot, using inertia, center of mass, finding speed

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