A solid uniform 3.33 kg disk has string of negligible mass wrapped around its ri

Question

A solid uniform 3.33 kg disk has string of negligible mass wrapped around its rim, with one end of the string tied to the ceiling, as shown in Figure 1. The disk is released from rest and turns as it falls as the string unwraps. At the instant its center has fallen 2.25 m, (a) how fast is it moving, and (b) how much rotational kinetic energy does it have? The radius was not given. I know a is v =gh/t = 3.68 m/s or at least I think. Can some help me solve part b. I have KE = 1/2 (IW^2) I know W= v/r , but r is not given. I found this equation :mr2/2 , but I do not know if that means (mr^2)/(2) , but still do not know r.

Figure 1: A solid uniform 3.33 kg disk has string of negligible mass wrapped around its rim, with one end of the string tied to the ceiling, as shown in Figure 1. The disk is released from rest and turns as it falls as the string unwraps. At the instant its center has fallen 2.25 m, (a) how fast is it moving, and (b) how much rotational kinetic energy does it have? The radius was not given. I know a is v =gh/t = 3.68 m/s or at least I think. Can some help me solve part b. I have KE = 1/2 (IW^2) I know W= v/r , but r is not given. I found this equation :mr2/2 , but I do not know if that means (mr^2)/(2) , but still do not know r.

Explanation / Answer

As the disk falls, it loses potential energy and gains translational and rotational kinetic energy.

PE lost = mgh (h = 2.25m)

KE gained = KEtranslational + KErotational

= 0.5mv2 + 0.5Iw2

(I = moment of inertia, w = Angular velocity of disk)

Now, since the rope is not slipping on the disc, therefore the linear and angular valocity are related as :

w = v/R (R = radius of disk)

Also, I = 0.5mR2, about an axis which is perpendicular to the disk and passes through its centre

Rotational KE = 0.5Iw2 = 0.5(0.5mR2)(v/R)2 = 0.25mv2

KE gained = 0.5mv2 + 0.25mv2

= 0.75mv2

Conserving energy,

PE lost = KE gained

mgh = 0.75mv2 .......... (i)

Rotational KE = 0.25mv2 = mgh/3 (From eq (i) )

= (3.33 x 9.8 x 2.25)/3

= 24.4755 J

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