Question

A ball with an initial velocity of 4.29 m/s rolls up a hill without slipping.
(a)
Treating the ball as a spherical shell, calculate the vertical height (in m) it reaches.
(b)
Repeat the calculation (in m) for the same ball if it slides up the hill without rolling.

Answer 1

Given v = 4.29 m/s and g = 9.80 m/s^2.

(a) For a thin spherical shell, I = (2/3)mR^2. With rolling without slipping (ω = v/R):
total initial energy = (1/2)mv^2 + (1/2)Iω^2
= (1/2)mv^2 + (1/2)(2/3)mR^2 (v^2/R^2)
= (1/2 + 1/3) m v^2 = (5/6) m v^2.
Set mgh = (5/6) m v^2 ⇒ h = (5/6) v^2 / g.

h = (5/6)(4.29^2) / 9.80 = (5/6)(18.4041) / 9.80 ≈ 1.57 m.

(b) If it slides (no rotation), (1/2) m v^2 = m g h ⇒ h = v^2/(2g):

h = 4.29^2 / (2·9.80) = 18.4041 / 19.6 ≈ 0.939 m.

Answers:
(a) 1.57 m
(b) 0.939 m