Physics help- unsure where I went wrong

[cpat]

1. A thin, uniform, metal bar, 2.5 m long and weighing 90 N, is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.3 m below the ceiling by a small 4-kg ball, initially traveling horizontally at 14 m/s. The ball rebounds in the opposite direction with a speed of 7 m/s.

Part A. Find the angular speed of the bar just after the collision.

I tried to solve using conservation of angular momentum (it must be a calculation error but I keep getting the same answer):
m1v_0d = -m1vd +(1/3)m2L^2(omega)
(4.0kg)(14m/s)(1.3m) = -(3kg)(7m/s)(1.3m)+(1/3)(90N/9.82m/s)(2.5m)^2(omega)
72.8 = 57.398 + (-27.3)
1001.1 = 57.398
1.7439rad/s = \omega


2. A Ball Rolling Uphill. A bowling ball rolls without slipping up a ramp that slopes upward at an angle beta to the horizontal. Treat the ball as a uniform, solid sphere, ignoring the finger holes.
Part A. What is the acceleration of the center of mass of the ball?

I tried drawing it out, f = mew N, N = wsin(beta)
- (WsinF+ muWcos(beta) = macm
I entered acm = -g(sinF+mucos (beta)
It was incorrect, so I thought I over-did it, and tried just -g(sin(beta)), but that was also incorrect, I'm not sure where to go from here.


Part B. What minimum coefficient of static friction is needed to prevent slipping?
I tried mu = tan (beta)

 

[LowlyPion]

OK

You have

|r X P| = r*m*vi = r*m*(-vf) + I*ω

So r*m*(vi + vf) = 1.3*4*(14 + 7) = 1.3*4*21 = 109.2 = 1/3*90/9.8*L²*ω

So then ...

ω = 109.2*9.8*3/(90*2.5*2.5) = ... ?

 

[cpat]

thanks!- I've got a calculator now (much easier! Mine stopped working last week so I've been borrowing from a friend), 5.70752.

Any tips on the next question?

 

[LowlyPion]

thanks!- I've got a calculator now (much easier! Mine stopped working last week so I've been borrowing from a friend), 5.70752.

Any tips on the next question?

For 2) consider the torque about the center of the solid sphere. That might yield "a" angular acceleration which you can then relate to linear acceleration along the incline.