Finding Torque, Angular Velocity & Potential Energy of a Uniform Rod

[grouchy]

A uniform rod has length 87 m and mass 8 kg. A mass of 8 kg is attached at one end. The other end of the rod is pivoted about the horizontal axis.

a) determine the torque immediately after the rod-plus-mass system is released from the horizontal position.

b) After the rod-plus-mass system is released, it rotates freely. Determine its angular velocity as the rod passes through the vertical direction. Answer in units of s^-1

c) The angle between the instantaneous position of the rod and the initial horizontal direction is 67 degrees. How much potential energy is released in going from 0 degrees to 67 degrees?

 

[grouchy]

A uniform rod has length 87 m and mass 8 kg. A mass of 8 kg is attached at one end. The other end of the rod is pivoted about the horizontal axis.

a) determine the torque immediately after the rod-plus-mass system is released from the horizontal position.

b) After the rod-plus-mass system is released, it rotates freely. Determine its angular velocity as the rod passes through the vertical direction. Answer in units of s^-1

c) The angle between the instantaneous position of the rod and the initial horizontal direction is 67 degrees. How much potential energy is released in going from 0 degrees to 67 degrees?

I found how to do part a

T = m(rod)gl +m(mass)gr
T = 8(9.8)(87) + 8(9.8)(43.5)
T = 10231.2 kgm^2/s^2

still got no clue how to attack b or c.

 

[grouchy]

A uniform rod has length 87 m and mass 8 kg. A mass of 8 kg is attached at one end. The other end of the rod is pivoted about the horizontal axis.

a) determine the torque immediately after the rod-plus-mass system is released from the horizontal position.

b) After the rod-plus-mass system is released, it rotates freely. Determine its angular velocity as the rod passes through the vertical direction. Answer in units of s^-1

c) The angle between the instantaneous position of the rod and the initial horizontal direction is 67 degrees. How much potential energy is released in going from 0 degrees to 67 degrees?

Found b...
Στ=I*α

I=8*87^2/3+8*87^2
I=80736

α=0.127 rad/s^2

θ=π/2

using energy
m*g*h=.5*m*v^2=.5*I*ω^2

9.81*(8*43.5+8*87)=.5*80736*ω^2
0.50 rad/s

P.S. Any help with c?