A string is wrapped around a disk of mass m = 2.2 kg and radius R = 0.10 m. Star

Question

A string is wrapped around a disk of mass m = 2.2 kg and radius R = 0.10 m. Starting from rest, you pull the string with a constant force F = 6 N along a nearly frictionless surface. At the instant when the center of the disk has moved a distance x = 0.10 m, your hand has moved a distance of d = 0.23 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

I got part b) the rotational energy and I tried the steps on this same question for part a but it was incorrect we tried doing 1/2mv^2=Fd and it was also incorrect.

Explanation / Answer

a) work done by force = change in KE

F.d = mv^2 /2 + Iw^2 /2

I = m r^2 / 2 and w = v/r so Iw^2 /2 = (m r^2 /2 ) (v/r) = m v^2 / 4

6 x 0.23 = 2.2 ( v^2 / 2 + v^2 / 4)

v = 0.914 m/s

b) rotational KE = Iw^2 /2 = mv^2 /4 = 2.2 x 0.914^2 / 4 = 0.46 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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