A string is wrapped around a disk of mass m = 2.5 kg and radius R = 0.11 m. Star

Question

A string is wrapped around a disk of mass m = 2.5 kg and radius R = 0.11 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 = 0.11 m, your hand has moved a distance of d = 0.30 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:

For the first one, you just set the Kinetic Energy equal to the work done on the disk.

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

So v = m/s

b) rotational KE = F(d-x) J

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