Question

1)A ball with an initial velocity of 9.6 m/s rolls up a hill without slipping. a)Treating the ball as a spherical shell, calculate the vertical height it reaches in m. b) Repeat the calculation for the same ball if it slides up the hill without rolling in m.

2) Suppose we want to calculate the moment of inertia of a 56.5 kg skater, relative to a vertical axis through their center of mass. Calculate the moment of inertia in (kg*m^2) when the skater has their arms pulled inward assuming they are cylinder of radius 0.125m

Answer 1

The MI of a spherical shell is J = (2/3)mr^2 where r is radius.
The initial KE of the sphere is (1/2)mv^2 + (1/2)Jw^2 where w = v/r is the angular velocity (assuming there is no sliding).
Using conservation of energy, PE gained = KE lost so
mgh = (1/2)mv^2+(1/2)Jw^2
= (1/2)mv^2+(1/2)(2/3)mr^2(v/r)^2
= (1/2)mv^2+(1/3)mv^2
mgh = (5/6)mv^2
h = 5v^2/6g = (5*9.6^2)/(6*9.8) = 7.84 m

If instead the ball slides without ROLLING (not 'slipping' - sliding is slipping) then w=0 so
mgh = (1/2)mv^2
h = v^2/2g = (9.6^2)/(2*9.8) = 4.702 m