Question

A ring of mass 2.4kg, inner radius 6.0cm, and outer radius 8.0cm is rolling (without slipping) up an inclined plane that makes an angle of theta=36.9 with the horizontal. At the moment the ring is x=2.0 m up the plane its speed is 2.8 m/s. the ring continues up the plane for some additional distance and then rolls back down. Assuming that the plane is long enough so that the ring does not roll off the top end, how far up the plane does it go?

Answer 1

Conservation of energy:

(PE + KE)at 2m = (PE + KE)at top

at 2 m

KE = 0.5*m*v^2 = 0.5*2.4*2.8^2 = 9.4 J

PE = mgh = 2.4*9.81*h

h= 2*sin(36.9) = 1.2m

PE = 28.25 J

Total energy = 28.25+9.4 = 37.65J

at the top

KE=0

all the energy is in the form of PE

PE = mgh = 37.65 J

h = 37.65/(2.4*9.81) = 1.6 m

We need to find the distance up the plane thus

d = h/sin (36.9) = 2.66 m