A ball in the shape of a uniform spherical shell (like a soccer ball; I = 2/3 mr2) of mass 1.5 kg and radius 15 cm rolls down a 35 degree incline that is 6.0 m high, measured vertically. The ball starts from rest, and there is enough friction on the incline to prevent slipping of the ball; thus the ball rotates.
a) How fast is the ball moving forward when it reaches the bottom of the incline, and what is its angular speed at that instant?
b) If there were no friction on the incline, how fast would the ball be moving forward and what would be its angular speed at the bottom?
I think I have a lot of b... Can you please check what I have so far and help me with what I'm getting wrong? I have to get all parts of this question correct to get any credit at all so I would appreciate any direction. Thanks!
F(incline) = mgsinθ = (1.5)(9.8)(sin35) = 8.43 N
x=6/sin35 ==> x = 10.46 (hypotenuse of incline)
I'm not sure where to go from here...
For b) I have:
(kinetic)i + (gravitational)i = (kinetic)f + (gravitational)f... so,
1/2mv^2 + mgy = 1/2 mv^2 + mgy (initial v=0, final y=0), so...
mgy = 1/2mv^2
v^2 = 2gy
v = sqrt(2gy) ==> sqrt(2(9.8)(6)) = 10.8 m/s = v
Then, to find angular speed,
w= 10.8 (m/s)/15 = 0.72
Then K(rot) = 1/2Iw^2
1/2(2/3(1.5)(15^2))(72^2) = 58.32 angular speed, but I know this isn't right. What did I do wrong?
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