A string is wrapped around a disk of mass m = 2.5 kg and radius R = 0.08 m. Star

Question

A string is wrapped around a disk of mass m = 2.5 kg and radius R = 0.08 m. Starting from rest, you pull the string with a constant force F = 7 N along a nearly frictionless surface. At the instant when the center of the disk has moved a distance x = 0.12 m, your hand has moved a distance of d = 0.22 m.

(a) At this instant, what is the speed of the center of mass of the disk?
vcm =  m/s

(b) At this instant, how much rotational kinetic energy does the disk have relative to its center of mass?
Krot =  J

Explanation / Answer

a)

KE = W

which is,

(1/2)mv^2 = Fd (in this case, d is x which is .10m)

v^2 = (2Fx)/m

v = sqrt[(2 x 7 x 0.12) / 2.5]

v = 0.82 m/s

b)

rotational KE = F(d - x) = 7(0.22 - 0.12) = 0.7 J

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