A thin light string is wrapped around the outer rim of a uniform hollow cylinder

Question

A thin light string is wrapped around the outer rim of a uniform hollow cylinder of mass 4.10 kg having inner and outer radii as shown in the figure (Figure 1) . The cylinder is then released from rest. You may want to review ( pages 309 - 315) . Part A How far must the cylinder fall before its center is moving at 6.47 m/s ? d = m SubmitMy AnswersGive Up Part B If you just dropped this cylinder without any string, how fast would its center be moving when it had fallen the distance in part A? v = m/s SubmitMy AnswersGive Up Part C Why do you get two different answers? The cylinder falls the same distance in both cases.

Explanation / Answer

a)

ro = outer radius = 35 cm = 0.35 m

ri = inner radius = 20 cm = 0.20 m

m = mass = 4.10 kg

I = moment of inertia of the cylinder = (0.5) m (ro2 + ri2) = (0.5) (4.10) (0.352 + 0.202) = 0.33 kgm2

T = tension force

a = acceleration

force equation is given as

mg - T = ma

T = mg - ma eq-1

Torque is given as

T ro = I a/ro

using eq-1

(mg - ma) = Ia/ro2

(4.10 x 9.8) - 4.10 a = (0.33) a/(0.35)2

a = 5.91 m/s2

Vo = initial speed = 0 m/s

Vf = final speed = 6.47 m/s

d = distance dropped

using the equation

Vf2 = Vo2 + 2 a d

(6.47)2 = 02 + 2 (5.91) d

d = 3.5 m

when the rope is not there

a = 9.8 m/s2

Vo = initial speed = 0 m/s

Vf = final speed = 6.47 m/s

d = distance dropped

using the equation

Vf2 = Vo2 + 2 a d

(6.47)2 = 02 + 2 (9.8) d

d = 2.13 m

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