Determining angular velocity of weight attached to a cord and cylinder

[I_Try_Math]

Homework Statement

A cord is wrapped around the rim of a solid cylinder of radius 0.25 m, and a constant force of 40 N is exerted on the cord shown, as shown in the following figure. The cylinder is mounted on frictionless bearings, and its moment of inertia is $6 kg \cdot m^2$. (a) Use the work energy theorem to calculate the angular velocity of the cylinder after 5.0 m of cord have been removed. (b) If the 40-N force is replaced by a 40-N weight, what is the angular velocity of the cylinder after 5.0 m of cord have unwound?

Relevant Equations

$W_{AB} = K_B - K_A$
$a_t = r\alpha$
$\omega_f^2 = \omega_0^2 + 2\alpha(\Delta\theta)$

$a_t = r\alpha$
$9.8 \: m/s^2 = (.25 \: m)\alpha$
$\alpha = 39.2 \: rad/s^2$

Since the cylinder rotates through 20.01 radians after 5 m of cord has been removed:
$\omega_f^2 = \omega_0^2 + 2\alpha(\Delta\theta)$
$\omega_f^2 = 2(39.2 \: rad/s^2)(20.01 \: rad)$
$\omega_f = 39.6 \: rad/s$

Textbook answer key says $\omega_f = 8 \: rad/s$. Can't see where my calculation is wrong. Any help is appreciated.

 

[haruspex]

Why do you think the tangential acceleration is g?

 

[kuruman]

Also, the problem is asking you explicitly to use the work-energy theorem. Doing that might work better for you than going the acceleration route.

 

[I_Try_Math]

Why do you think the tangential acceleration is g?

Oh for some reason I keep getting confused with problems like this. I didn't take into account that some of the work done by gravity is going to rotate the cylinder not just accelerate the weight down.

 

[haruspex]

Oh for some reason I keep getting confused with problems like this. I didn't take into account that some of the work done by gravity is going to rotate the cylinder not just accelerate the weight down.

So it is part b you are interested in?

 

[I_Try_Math]

So it is part b you are interested in?

Yes, apologies, I meant to indicate that in my post. I'm still trying to understand it. I can't wrap my head around why a constant force of 40 N is different than a weight of 40 N. If weight = mg = 40 N then it is by definition equivalent to a constant force of 40 N (assuming g is constant)? Which would mean the answer for part (b) would be the same as part (a)?

 

[jbriggs444]

Yes, apologies, I meant to indicate that in my post. I'm still trying to understand it. I can't wrap my head around why a constant force of 40 N is different than a weight of 40 N. If weight = mg = 40 N then it is by definition equivalent to a constant force of 40 N (assuming g is constant)? Which would mean the answer for part (b) would be the same as part (a)?

What forces act on the weight? Is it in free fall?
What forces act on the cylinder? Is it subject to 40 N?

 

[I_Try_Math]

What forces act on the weight? Is it in free fall?
What forces act on the cylinder? Is it subject to 40 N?

On the weight, there are only the forces of weight and tension acting on it. I suppose it can't be in free fall because of tension which I believe makes my earlier reasoning flawed.

For the cylinder, I'm less certain about it. The tension is where I start to get tripped up. The force that acts on it would mg - T?

 

[jbriggs444]

On the weight, there are only the forces of weight and tension acting on it. I suppose it can't be in free fall because of tension which I believe makes my earlier reasoning flawed.

For the cylinder, I'm less certain about it. The tension is where I start to get tripped up. The force that acts on it would mg - T?

What is $m$ there?

Does gravity act directly on the cylinder? What force(s) act directly on the cylinder?

 

[kuruman]

Hats off to the author of this question for it gets to the root of a common misconception in cases of masses hanging from pulleys. Its figure of the juxtaposed pulleys leads one down the garden path to reach an "obvious" but incorrect conclusion, thus bringing forth the misconception.

One looks at the static figure on the left and imagines that the pulleys are at rest perhaps blocked from moving by a brake. They are not seen moving, are they? Well, in that case there is no acceleration, and the tension in each of the two strings is 40 N. No problem here.

Now the pulleys are unblocked and allowed to accelerate. What's the difference between the two pictures? Isn't it obvious that the 40 N weight on the pulley on the right can be replaced by a 40 N force of gravity on a 4-kg mass so that the two free body diagrams of the pulleys are the same?

Spoiler

No because the tensions are not the same. On the left, the external force of 40 N accelerates only the pulley and the tension is 40 N. On the right, the external 40 N force of gravity accelerates both the pulley and the 4-kg hanging mass and the tension is less than 40 N.

 

[I_Try_Math]

What is $m$ there?

Does gravity act directly on the cylinder? What force(s) act directly on the cylinder?

Oh sorry I could've been more clear but $m$ would refer to to the mass of the weight. So gravity acts on the cylinder as well as a force equal to $m_{weight}g -T$? I think the net force on the cylinder would be $m_{weight}g -T$?

 

[kuruman]

The center of mass of the cylinder does not accelerate. Therefore, the net force on the cylinder is zero. However, the cylinder has angular acceleration about the center of mass. What does that mean?

 

[I_Try_Math]

The center of mass of the cylinder does not accelerate. Therefore, the net force on the cylinder is zero. However, the cylinder has angular acceleration about the center of mass. What does that mean?

So the net force is zero but the net torque is nonzero. After reading your post I think I finally understand why the two situations in the problem are different. For some reason I'm still having trouble coming up with correct answer.

 

[jbriggs444]

Oh sorry I could've been more clear but $m$ would refer to to the mass of the weight.

The force of gravity on the weight acts directly on the weight.
The force of gravity on the weight does not act directly on the cylinder.

Try to think in terms of what forces act directly on the object of interest.

So gravity acts on the cylinder as well as a force equal to $m_{weight}g -T$? I think the net force on the cylinder would be $m_{weight}g -T$?

What forces act directly on the cylinder?

 

[kuruman]

For some reason I'm still having trouble coming up with correct answer.

Then you have to show your work. We cannot help you without knowing what you did and what troubles you. Remember that this problem requires you to use the work-energy theorem.

 

[I_Try_Math]

The force of gravity on the weight acts directly on the weight.
The force of gravity on the weight does not act directly on the cylinder.

Try to think in terms of what forces act directly on the object of interest.


What forces act directly on the cylinder?

Would it be just gravity and tension?

 

[I_Try_Math]

Then you have to show your work. We cannot help you without knowing what you did and what troubles you. Remember that this problem requires you to use the work-energy theorem.

I figured out how to use the work-energy theorem to solve it. Turns out I was making a different mistake unrelated to my original misconception about the two situations in the problem being the same. Thank you for the hints.

 

[jbriggs444]

Would it be just gravity and tension?

No.

 

[kuruman]

I figured out how to use the work-energy theorem to solve it. Turns out I was making a different mistake unrelated to my original misconception about the two situations in the problem being the same. Thank you for the hints.

You are welcome. Finding your mistake is important but does not guarantee that you answer is correct. How did you ascertain that the answer you got is correct?

 

[I_Try_Math]

You are welcome. Finding your mistake is important but does not guarantee that you answer is correct. How did you ascertain that the answer you got is correct?

Oh I checked it against the answer key which unfortunately isn't always the most reliable. My answer was $7.99 \ rad/s$ whereas the key has it as $8 \ rad/s$. My calculation looked like:

$mgh= \frac{1}{2}I\omega^2 + \frac{1}{2}m(r\omega)^2$
With $m=4.08 \ kg$ and $h=5 \ m$ you get $\omega=7.99 \ rad/s$.

This is the correct problem solving strategy and answer, correct?

 

[I_Try_Math]

No.

I must be misunderstanding the question. I mean at least gravity acts on the cylinder due to its own mass. It's just that it doesn't result in a net force on the cylinder, no?

 

[kuruman]

Oh I checked it against the answer key which unfortunately isn't always the most reliable. My answer was $7.99 \ rad/s$ whereas the key has it as $8 \ rad/s$. My calculation looked like:

$mgh= \frac{1}{2}I\omega^2 + \frac{1}{2}m(r\omega)^2$
With $m=4.08 \ kg$ and $h=5 \ m$ you get $\omega=7.99 \ rad/s$.

This is the correct problem solving strategy and answer, correct?

If your calculation took account of the significant figures implied by the given variables, it will be in ine with the solution manual. Read about that here.

 

[haruspex]

I must be misunderstanding the question. I mean at least gravity acts on the cylinder due to its own mass. It's just that it doesn't result in a net force on the cylinder, no?

… because it is balanced by?

 

[jbriggs444]

I must be misunderstanding the question. I mean at least gravity acts on the cylinder due to its own mass. It's just that it doesn't result in a net force on the cylinder, no?

I asked for a list of forces acting directly on the cylinder. Your response was that gravity and tension are the only two such forces.

Yes, gravity acting directly on the cylinder is such a force. With magnitude $Mg$ (where $M$ is the mass of the cylinder). Since we are not given $M$, this one cannot be calculated from the information at hand. Fortunately, we do not need it.

Yes, tension is another such force. With a magnitude that could be calculated but need not be.

Both of those forces are downward. Yet the cylinder remains in place. There must be another force acting: The support force provided by the axle that holds the cylinder in place.

We do not need to know the magnitudes of these forces to solve the problem. But we should list them so that we can do a proper job of energy accounting, totting up the work done by each. Let us do that.

Gravity on the cylinder: Does the center of mass of the cylinder move? No. So zero work done by that force.

Gravity on the weight: Does the center of mass of the weight move? Yes. Multiply distance moved by force to get work performed.

Axle supporting the cylinder: Does the cylinder move in the direction of that support force? No. So zero work done by the axle.

Friction from the axle: We are told that the bearings are frictionless. So zero work there.

Tension: There are a few ways to argue that the work done by the cord on the system consisting of the cylinder plus weight totals to zero. One way is to note that the attachment forces at either end are equal and opposite (assuming an ideal cord) and that the motions at either end are equal. So net work done by the cord is zero.