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

1. Nov 10, 2008

### bocobuff

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 rcm=0.610m, and the moment of inertia about the center of mass is Icm=0.04kg*m2. 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*rcm=0.529 kg*m2
I=Icm+Md2
Then some torque equations I don't know which to use, if any
τ=r*F*sinϕ
τ=m*r2
τgrav=-M*g*rcm
and α=τnet/I
And eventually I'll have to use either

3. The attempt at a solution
I think Fgrav runs along H. ϕ=143° so Ftan=Fgrav*sin(-143°) where Fgrav=m*g.
I can also get I=Icm+M*rcm2 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

2. Nov 10, 2008

How about conservation of energy here? That way you do not have to deal with any forces.

PE1+KE1=PE2+KE2

Also, why did you calculate the mass moment of inertia if it was given to you in the problem statement?

3. Nov 10, 2008

### marlon

4. Nov 10, 2008