Question

A hollow ball of mass 6.00 kg and radius 0.180 m is rolled up a hill without slipping. If it starts off at the bottom with a linear speed of 7.00 m/s, what vertical height (in m) will it reach?

Answer 1

Using energy conservation

KEi + PEi = KEf + PEf

PEi = 0, since at ground

KEf = 0, since final speed is zero

KEi = PEf

KEt + KEr = PEf

0.5*m*V^2 + 0.5*I*w^2 = m*g*h

I = moment of inertia of hollow sphere = 2*m*r^2/3

w = V/r

0.5*m*V^2 + 0.5*(2*m*r^2/3)*(V/r)^2 = m*g*h

V^2/2 + V^2/3 = g*h

7^2/2 + 7^2/3 = 9.81*h

49/2 + 49/3 = 9.81*h

24.5 + 16.33 = 9.81*h

9.81*h = 40.83

h = 40.83/9.81 = 4.16 m

h = (V^2/g)*(1/2 + 1/3)

Using given values

h = (7^2/9.81)*(5/6)

h = 4.16 m

Please upvote.