Question

A thin ring of radius R and mass M rolls without slipping along a level track. It has an initial linear, or translational velocity (of the center of gravity) of 3.50 m/s. The ring rolls to the end of the track, where the track curves upward. The center of gravity of the ring rises to a maximum height h above its initial level. Note that V is the symbol for the linear, or translational velocity (of the center of gravity) in the following questions. Also, you will need to refer Table 7.4 of the text, for the expression for the moment of inertia of a thin ring (same as for a hoop, or hollow cylinder). Remembering that the initial translational velocity of the ring is given by V, the angular velocity of the ring along the level section of the track (i.e., the initial angular velocity) is u^2/R middot u/R R/u u/R The total initial kinetic energy of the ring is (The total kinetic energy is the sum of the KE and RKE of the ring.) 2Mu^2 Mu^2 Mu^2/2 Mu^2 What is the maximum height, h, of the ring? Use conservation of energy for this problem, and enter your answer with exactly three significant figures.

Answer 1

here,

initial linear velocity, u = 3.50 m/s

moment of inertia of ring, I =

Part a:
linear veloicty = angular velocity * radius

angular velocity, w = u/r

option b is correct

Part b:
Total kinetic energy = 0.5*m*u^2 + 0.5*I*w^2
Total kinetic energy = 0.5*m*u^2 + 0.5*mr^2 *(u^2/r^2)
Total kinetic energy = 0.5*m(u^2 + u^2)
Total kinetic energy = mu^2

option d correct

Part c:
From conservation of energy :
Total Kinetic energy = Potential energy gained by ring
m*u^2 = m*g*h

height, h = u^2/g
height, h = 3.50^2/9.81
height, h = 1.249 m