A hoop of mass M = 100 grams and radius R = 50 cm rolls without slipping down a

ID: 2116047 • Letter: A

Question

A hoop of mass M = 100 grams and radius R = 50 cm rolls without slipping down a hill of

height 2 meters. The fact that it does not slip means that, when the center of mass moves at a

speed vCM, the point of the hoop in contact with the floor at that instant does not move. The

angular speed of the hoop is, thus, %uF077 = vCM / R. Determine: a) The speed of the hoop at the

bottom of the hill; b) If we replace the hoop by a cylindrical full garbage can with the same mass

and radius, what would be its speed at the bottom of the hill?. Note: You may find useful to

know that Ihoop = M R2 and Icylinder = 1/2 M R2

Part A)

The Hoop

Conservation of Energy applies and we must consider translational and rotational KE, so

PE = KEr + KEt

mgh = .5Iw^2 + .5mv^2

I for the hoop = mr^2 and w = v/r, so...

mgh = .5mr^2v^2/r^2 +_.5mv^2

mgh = .5mv^2 + .5mv^2 (mass cancels)

gh = v^2

(9.8)(2) = v^2

v = 4.43 m/s

Part B)

The disk (garbage can)

mgh = .5Iw^2 + .5mv^2

I for the dsik = .5mr^2 and w = v/r, so...

mgh = (.5).5mr^2v^2/r^2 +_.5mv^2

mgh = .25mv^2 + .5mv^2 (mass cancels)

gh = .75v^2

(9.8)(2) = v^2

v = 5.11 m/s

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.