Physics question

Question

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?

Answer 1

(a)

 

Apply conservation of energy.

 

mv²/2 + Iω²/2 = mgh

 

ω = v/r

 

mv²/2 + I(v/r)²/2 = mgh

 

v²[m/2 + I/(2r²)] = mgh

 

v² = mgh / [m/2 + I/(2r²)]

 

I = 2mr²/3 ⇒

 

v² = gh / (1/2 + 1/3) = gh/(5/6) = (6/5)gh

 

v = √[(6/5)gh]

 

Plug in g = 9.8 m/s² and h = 6.0 m, and get v in m/s.

 

ω = v/r

 

Plug in the result for v above and r = 0.15 m, and get ω in rad/s.

 

(b)

 

If there is no friction, the ball slides down without rolling.  Its angular speed is zero.

 

mv²/2 = mgh

 

v = √(2gh)

 

Plug in g and h.