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

[bocobuff]

Homework Statement

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*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?

Homework Equations

Thin rod, about end I=1/3M*r_cm=0.529 kg*m²
I=I_cm+Md²
Then some torque equations I don't know which to use, if any
τ=r*F*sinϕ
τ=m*r²*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

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_cm² 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

 

[Saladsamurai]

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?

 

[marlon]

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?

That's good advice

 

[Saladsamurai]

You know that the initial velocity is=0 so KE1 is 0. It has only PE1.

At the horizontal, PE2=0, so you can find the velocity.
(If you do not understand why PE2=0, please ask and I will explain)