A thin hoop has a radius R = 0.060 m and a mass m= 0.20 kg. The hoop starts from

Question

A thin hoop has a radius R = 0.060 m and a mass m= 0.20 kg.  The hoop starts from a height of 2.4 m, from rest, and rolls down on an inclined place.  There is no frictional energy loss during the motion.  The moment of inertia of a thin hoop is MR^2.

a) The moment of inertia of the hoop is

b) The total mechanical energy of the hoop at any point of its motion is

c) the velocity of the center of mass of the hoop at the bottom of the inclined plane is

d) The angular speed of the rotations of the hoop at the bottom of the place is

e) the angular momentum of the hoop at the bottom of the inclined plane is

Please include details of how you came up with the answer! Thank you.

Explanation / Answer

a)

MOI = MR^2 = 0.2*0.06^2 = 0.00072 kg-m^2

b)

Total energy = PE at starting point = mgh = 0.2*9.81*2.4 = 4.7088 J

c)

At bottom of plane, total energy = KE due to rolling + KE due to translation

= 1/2*I*w^2 + 1/2*mv^2

But no-slip condition says, v = Rw

Thus, Total energy = 1/2*I*(v/R)^2 + 1/2*mv^2

= 1/2*(mR^2)*(v/R)^2 + 1/2*mv^2

= mv^2

Thus, Energy conservation gives, mgh = mv^2

v = sqrt (gh)

= sqrt (9.81*2.4)

= 4.85 m/s

d)

w = v/R

= 4.85/ 0.06

= 80.87 rad/s

e)

Angular momentum = Iw

= 0.00072*80.87

= 0.0582 kg-m/s

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