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:

(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.

Answer 2

The speed of the center of mass of the disk will be equal to 1.43m/s, while the rotational kinetic energy will be 2.59J.

We can arrive at this answer as follows:

Let us assume that the distance moved by the center of mass will be represented by the letter "d" in the formulas. As long as the distance traveled when unwinding the rope is represented by the letter "D".

• Let's start by calculating the total work that allows the disk to move its center of mass. For this we will use the formula:

$W=F*(D+d)\\W=7*(0.15+0.22)\\W= 2.59J$

Thus, we can say that we found rotational kinetic energy.

• To find the speed of the center of mass of the disk, we will use the following formula:

$W=\frac{1}{2} MVcm^2+\frac{1}{2}*I*\omega^2\\W=\frac{1}{2} MVc^2+\frac{1}{2}*\frac{1}{2}*[\frac{Vcm}{R} ]^2\\W= \frac{1}{2} MVc^2+\frac{1}{4}MVcm^2\\W= \frac{3}{4}MVcm^2\\Vcm^2= \frac{4}{3} *\frac{W}{M} \\Vcm=\sqrt{\frac{4}{3}*\frac{W}{M} }\\Vcm= \sqrt{\frac{4}{3}*\frac{2.59}{1.7} }\\Vcm= 1.43m/s$

That way we can confirm that we found the velocity of the center of mass of the disk.

More information:

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