Question

A string is wrapped around a disk of mass 1.7 kg and radius 0.11 m. Starting from rest, you pull the string with a constant force 7 N along a nearly frictionless surface. At the instant when the center of the disk has moved a distance 0.15 m, your hand has moved a distance of 0.22 m.
a. At this instant, what is the speed of the center of mass of the disk?
b. At this instant, how much rotational kinetic energy does the disk have relative to its center of mass?

Answer 1

Answer:

Explanation:

a) work done in moving the disk by 0.15 m = its kinetic energy = 0.5 mv²

Fd = 0.5 mv²

2 fd = mv²

√( $\frac{2fd}{m}$) = v

v = 1.11 m/s

b) rotational kinetic energy = F ( 0.22 m - 0.15 m) = 0.49 J

Answer 2

Answer:

(a) Vcm = 1.43m/s

(b) KEcm = 2.59J

Explanation:

Given

m = mass of the solid disk = 1.7kg

Radius R = 0.11m

F = 7N

Let the distance moved by the the center of mass be x = 0.15m

And the distance moved by in unwinding the rope be d = 0.22m

The total workdone on the disk is causing it to rotate and also move its center of mass is F×(d+x)

W = 7×(0.15+0.22)

= 7× 0.37 = 2.59J

By work-energy theorem,

W = ΔKE = 1/2mVcm² + 1/2×I×ω²....(1)

I = moment of inertia = 1/2MR² and

ω = angular speed = Vcm/R

So substituting these expressions into the equation above we have

W = 1/2MVcm² + 1/2×1/2MR²×(Vcm/R)²

W = 1/2MVcm² + 1/4MVcm²

W = 3/4MVcm²

Vcm² = 4/3×W/M

Vcm = √(4/3×W/M)

Vcm = √(4/3×2.59/1.7)

Vcm = 1.43m/s

KE = workdone = 2.59J.