# Velocity and Torques Problem

Tags:
1. Mar 31, 2015

### Okazaki

1. The problem statement, all variables and given/known data
A uniform cylindrical spool of mass M and radius R unwinds an essentially
massless rope under the weight of a mass m. If R = 12 cm, M = 400 gm and m = 50 gm, find the speed of m after it has descended 50 cm starting from rest.

Solve the problem twice: once using Newton's laws for torques, and once by application of energy conservation principles.

2. Relevant equations
I(spool) = (1/2)MR^2
τ = R * Fsin(90) (*it will be 90 degrees here due to the way the Gravitational Force pulls down)
τ = Iά
at = ά * R (tangential acceleration)
vf = sqrt(vi2 + 2at*Δx)

3. The attempt at a solution

So I used τ = R * Fsin(90) to come up with the torque (since I converted everything into meters and kilograms, my answer ended up being -5.88 x 10^-2 N-m)

Then, I basically plugged in values for I:
I = 0.5 * 0.4 kg * (0.12 m)^2 = 0.00288 kg m^2

After that, I thought about solving for ά. But ά is the angular acceleration, which is kind of useless in this case, since we're trying to find the velocity of mass m after is has fallen 50 cm. So I set
ά = τ/I = at / R

at = 0.12 m * -5.88 x 10^-2 N-m/0.00288 kg m^2 = -20.4 m/s^2,

and from here I used the equation: vf = sqrt(vi2 + 2at*Δx) and got -4.52 m/s

So, I honestly don't even know if I solved this problem right (more or less, where to even start if I was going to use conservation of energy principles to solve it again.) Any help would be greatly appreciated.

2. Mar 31, 2015

### TSny

Hello, Okazaki. Welcome to PF.

What force produces the torque on the cylinder? Is it the tension in the string or is it the weight of mass m?

3. Mar 31, 2015

### Okazaki

Thanks. :)

I'm assuming it's the gravitational force, since the tension forces cancel out (I attached the related picture.)

#### Attached Files:

• ###### physicsquestion.PNG
File size:
4.6 KB
Views:
157
4. Mar 31, 2015

### Aceix

I don't think information on the masses and radius is very important since there is nothing like friction whatsoever. I attempted the question using Newton's laws of linear motion and got a final velocity of ~3.162m/s.

5. Mar 31, 2015

### TSny

When considering the forces that act on the cylinder alone, the tension force does not cancel. From the figure you can see that it's the tension force that acts on the rim of the cylinder that causes the cylinder to spin.

The tension is an unknown force. So, you will need to set up a second equation that contains the tension. Try applying Newton's second law to the mass on the end of the string.

6. Mar 31, 2015

### Okazaki

But if you apply that, how are you supposed to find the acceleration of the object? Then, not only do you not know tension force, but you don't know acceleration either....

7. Mar 31, 2015

### TSny

But you indicated that you know a relation between $a$ and $\alpha$. That gives you a third equation for the three unknowns: $T$, $a$, and $\alpha$.

8. Mar 31, 2015

### Okazaki

So...then:

F = ma
a = (-mg - T)/m

and:

at = a = rά = (R * RTsin(90))/(0.5MR^2) = 2T/m (or is this M?)

So:

(-mg - T)/m = 2T/m
T = (-m^2*g)/(2m + m)
= -0.163 N

Then:
a = (-T - mg)/m
= (-0.163N - 0.49N)/0.05 kg ?

9. Mar 31, 2015

### TSny

Are the signs correct? What direction are you choosing for the positive motion of m? Does the tension force on m act in the same direction as the force of gravity?

OK. You're analyzing the cylinder. So, the mass is M, right?

Looks like the correct approach. Note that you get a linear acceleration greater than g, so something's wrong.
You will need to fix the signs in your F = ma equation for m.

10. Mar 31, 2015

### Okazaki

Yeah...It was supposed to be M (I don't know why I thought it was m), meaning
T = (Mmg)/(2m + M) = -0.392, so -T = 0.392.
The acceleration then works out to be -1.96 m/s^2, and from there I found a velocity of 1.4 m/s

And I did have a few sign errors (which I worked out.) Now, both answers work (since I got help and figured out how to analyze the problem with the conservation of energy approach.)

Thanks for the help!

11. Mar 31, 2015

### TSny

OK, Good work!