Sledgehammer falling on pivot. physics question. Where are my workings wrong?
A sledgehammer with a mass of 2.80 kg 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=0.560 m, and the moment of inertia about the center of mass is I_cm = 0.0410 kgm2. If the hammer is released from rest at an angle of theta=45.0 degrees such that H=0.396 m, what is the speed of the center of mass when it passes through horizontal?
OK, so using parallel axis theorem:
I = Icm + Md^2 = 0.91908
gives me moment of inertia at the pivot
plug that into the energy equations
gravitational potential energy (initial) = Kinetic energy (final)
mgH = (1.2)(I)(w^2 )
finding for w = 23.67
So now that I have omega I can find linear speed by using v=rw where r is the distance to the CM/
v = 13.3 m/s
When I enter this answer in the assignment however it's wrong. Can anyone tell me where I messed up? Thanks!
