String wrapped around a Disk with 4 Rods

[Oribe Yasuna]

A string is wrapped around a uniform disk of mass M = 1.3 kg and radius R = 0.05 m. (Recall that the moment of inertia of a uniform disk is (1/2)MR2.) Attached to the disk are four low-mass rods of radius b = 0.08 m, each with a small mass m = 0.3 kg at the end. The device is initially at rest on a nearly frictionless surface. Then you pull the string with a constant force F = 33 N. At the instant when the center of the disk has moved a distance d = 0.034 m, a length w = 0.023 m of string has unwound off the disk.

(a) At this instant, what is the speed of the center of the apparatus?
v =

(b) At this instant, what is the angular speed of the apparatus?
omega₁ =

(c) You keep pulling with constant force 33 N for an additional 0.039 s. Now what is the angular speed of the apparatus?
omega₂ =

Homework Equations

I_disk = 1/2 MR^2
I = m₁r₁^2 + m₂r₂^2 ...

torque = r x F

The Attempt at a Solution

I = 1/2 MR^2 + m₁r₁^2 + ... m₄r₄^2
I = 1/2 (1.3)(0.05)^2 + 4*(0.3)(0.08)^2
I = 0.009305 kg*m^2

torque = r x f
torque = 1.65 N*m

I have no idea where to go from this.

 

[Mister T]

Was there no figure accompanying the problem?

 

[Oribe Yasuna]

Was there no figure accompanying the problem?

There was.

My bad.

 

[Mister T]

At the instant when the center of the disk has moved a distance d = 0.034 m, a length w = 0.023 m of string has unwound off the disk.

The fact that these two numbers are different tells you what?

 

[Oribe Yasuna]

The fact that these two numbers are different tells you what?

That the disk is rotating and the system is moving at different rates?

 

[Mister T]

Right. So from those two numbers you can tell how far it's moved (translated) and how many rotations it's made.

 

[Oribe Yasuna]

Right. So from those two numbers you can tell how far it's moved (translated) and how many rotations it's made.

I don't understand how to get velocity from that. Is there a formula?

 

[Mister T]

Have you tried using $F=ma$?

 

[Oribe Yasuna]

Have you tried using $F=ma$?

F = ma
a = m/F
a = 33 N / (1.3 kg + 0.3 kg + 0.3 kg + 0.3 kg + 0.3 kg)
a = 13.2 m/s^2

Maybe...
0.034 * 13.2 = 0.4488 m^2 / s^2
0.023 * 13.2 = 0.3036 m^2 / s^2
0.4488 + 0.3036 = 0.7524 m^2 / s^2
sqrt 0.7524 = 0.8674... m/s

I'm just guessing through units, but is this it?

 

[Mister T]

Guessing is not a good idea. Don't you remember the formula that relates distance, velocity and acceleration? Distance times acceleration does give you something with units of $v^2$ but that doesn't mean it's equal to $v^2$.

 

[Oribe Yasuna]

Guessing is not a good idea. Don't you remember the formula that relates distance, velocity and acceleration? Distance times acceleration does give you something with units of $v^2$ but that doesn't mean it's equal to $v^2$.

I don't remember a formula like that at all.

All I could find was:
d = v₁ + 1/2 at^2
d = 1/2 at^2
0.034 + 0.023 = 1/2 13.2 t^2
0.057 = 6.6t^2
0.008636364 = t^2
0.092932038 = t

t = (v_f - v_i) / a
0.092932038 = v_f / 13.2
1.2267... = v_f

That doesn't look right though.

 

[Mister T]

Why did you add the two distances d and w together?

 

[Oribe Yasuna]

Why did you add the two distances d and w together?

Well, I tried calculating them separately using the same formula and got different times.

But there can only be one time, so I thought it was wrong.

 

[Mister T]

Well, I tried calculating them separately using the same formula and got different times.

Well, of course if you use two different distances d and w you get two different times.

But there can only be one time, so I thought it was wrong.

And if you use a third distance d+w you get a third time. If there's only one right answer then at least two of those three have to be wrong.

You need to find a way to justify using one of them. Getting the right answer is not a justification. The justification needs to be based on sense-making, not on answer-making.

 

[Oribe Yasuna]

You need to find a way to justify using one of them. Getting the right answer is not a justification. The justification needs to be based on sense-making, not on answer-making.

Alright. So the first question is asking for the velocity of the system.

The rotation of the disk is related to angular velocity, so that's not it.

So the center of the system (center of the disk) has moved a distance of 0.034 m, which is d in the equation.
d = v₁ + 1/2 at^2
d = 1/2 at^2
0.034 = 1/2 13.2 t^2
0.034 = 6.6t^2
0.005151515 = t^2
0.071774056 = t

t = (v_f - v_i) / a
0.071774056 = v_f / 13.2
0.947417543 = v_f

Does this make sense?

 

[Mister T]

Yes! If you remember to attach the units: 0.947 m/s.

Next, repeat that same calculation, but this time you use the rotational motion versions of the linear motion equations.

 

[Oribe Yasuna]

Yes! If you remember to attach the units: 0.947 m/s.

Next, repeat that same calculation, but this time you use the rotational motion versions of the linear motion equations.

d = 1/2 at^2
0.023 = 1/2 (13.2)t^2
0.059032605 = t

omega_f = omega_i + RFt / I
omega_f = (0.05)(33)(0.059032605) / 0.009305
omega_f = 10.46789879 rad./s

Okay, does that look correct?

 

[Mister T]

d = 1/2 at^2
0.023 = 1/2 (13.2)t^2
0.059032605 = t

Is 13.2 m/s² the linear acceleration of the string? The reason I ask is because 0.023 m is the linear displacement of the string relative to the center of rotation.

 

[Oribe Yasuna]

Is 13.2 m/s² the linear acceleration of the string? The reason I ask is because 0.023 m is the linear displacement of the string relative to the center of rotation.

No, it's the acceleration of the system of rods and disk.

Would the string's movement be measured in torques?

 

[Mister T]

Would the string's movement be measured in torques?

The string moves a distance of 0.023 m relative to the center of rotation of the disk. What does this tell you about the disk's rotational motion?