Question

A hoop with a mass of 2.95 kg starts from rest at the top of a ramp. The ramp is 4.6 m long and 1.9 m high. What is the rotational kinetic energy of the hoop after it has rolled without slipping to the bottom?

Answer 1

Using Energy Conservation for hoop:

KEi + PEi = KEf + PEf

KEi = Initial kinetic energy = (1/2)*M*V^2 + (1/2)*I*w^2,

Given that initial speed of hoop(V) = 0

KEi = 0

PEi = Initial potential energy = m*g*h

KEf = final kinetic energy = (1/2)*M*V^2 + (1/2)*I*w^2,

final speed of hoop = Suppose V

I = moment of inertia of Hoop = M*R^2

w = angular velocity of ball at bottom of slope = V/R

Which gives

KEf = TKE + RKE = (1/2)*m*V^2 + (1/2)*(m*R^2)*(V/R)^2 = (1/2)*m*V^2 + (1/2)*m*V^2 = m*V^2

PEf = final potential energy = 0

h = height = 1.9 m

Now Using these values

0 + m*g*h = 0 + m*V^2

V = sqrt(g*h) = sqrt(9.81*1.9)

V = 4.32 m/s

now, RKE = rotational kinetic energy of hoop at bottom = (1/2)*m*V^2

given, m = mass of hoop = 2.95 kg

So, RKE = 0.5*2.95*4.32^2

RKE = 27.5 J

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