# Frictionless Pulley with two weights

Firstly, nice forum!
Secondly, I am Swedish but I will try to use the correct scientific terms in explaining the problem and my proposed solution.

## Homework Statement

There is a string of negligible mass which "hangs" over a pulley attached to a frictionless horizontal axis. On each end of the string there is a attached a weight. One of the weights (i've chosen the one on the right hand side if drawn the "normal way") is twice the mass of the weight of the other.
The system starts out at rest.

[pulley] = Radius R, and Moment of Inertia I = (1/5)*m*R².
[weights] = A = m and B = 2m.
[string tension] = left side (A): T ; right side (B): S.
[gravity] = As always g denoted the pull of the earth (9.82 m/s² here in Sweden).

When the weights are released from rest (and left to fall freely), what will be the tension on the string on the two sides of the pulley?

## Homework Equations

In my attempted solution below I used:
Newtons II equation where the sum of the forces are equal to the mass times the acceleration (F = m*a).
I also used Eulers II law for momentum (torque) which in this case would be that the momentum is equal to the moment of inertia times the angular acceleration (M = I*alpha)
I also used that alpha = R * a (along the x-axis (which I set to be positive upward))

## The Attempt at a Solution

So, as listed above I will call the momentum about the z-axis of the pulley M, the angular acceleration alpha. I called the "left" pulley A and the "right" pulley B.
I set the x-axis to the positive in the upward direction. And I assumed that the string was not elastic in any way.

First I used newtons second law to describe the forces acting on pulley A to be
F = T - mg = m * a
and respectively for B:
F = S - 2mg = m * a

Then I used Eulers II with the moments origo in the center of the pulley to be
M = RT - RS = I * alpha (and since alpha = R * a) = T-S = I * a.

I then simply solved the equations using both NII and EII to get my expressions for what the tensions on S and T are.

I got that T =m(a((R²/5)-1)+2g)
and S = m(a(1-(R²/5))+g)

It should be added I am at the moment of typing a little tired. Also I should probably have made a picture, but I hope my explanation of the problem will be enough.

Usually I would be satisfied with my answer but the "hint" to the question was that I could use that acceleration (dv/dt) could be written as v*dv/ds (using the chain rule)! Why and how could that be used? I am now worried that I have made some very basic assumption or something wrong. Anyways help much appreciated!

Related Introductory Physics Homework Help News on Phys.org
Doc Al
Mentor
First I used newtons second law to describe the forces acting on pulley A to be
F = T - mg = m * a
OK.
and respectively for B:
F = S - 2mg = m * a
Two errors on the right hand side:
(1) You have m, but should have 2m.
(2) Careful with signs: If the acceleration of A is +a, what must be the acceleration of B?

Then I used Eulers II with the moments origo in the center of the pulley to be
M = RT - RS = I * alpha (and since alpha = R * a) = T-S = I * a.
Two problems:
(1) Careful with signs. S > T, so use RS - RT = I*alpha.
(2) alpha = a/R, not a*R.

Usually I would be satisfied with my answer but the "hint" to the question was that I could use that acceleration (dv/dt) could be written as v*dv/ds (using the chain rule)! Why and how could that be used?
That hint doesn't seem to apply to this problem. No calculus is needed here.

Doc Al said:
OK.
Two errors on the right hand side:
(1) You have m, but should have 2m.
(2) Careful with signs: If the acceleration of A is +a, what must be the acceleration of B?
then the acceleration of B must of course be negative! thanks!

So the new NII for B is: S - 2mg = -2ma

Doc Al said:
Two problems:
(1) Careful with signs. S > T, so use RS - RT = I*alpha.
(2) alpha = a/R, not a*R.
Gah! stupid mistake! I misinterpreted some old notes of mine. Of course alpha = a/R!

I am heading out to a wedding now (my girlfriends sister is getting married) I will be sure to post the completed solution when I get back tomorrow! Thank you so much for the help!

ideasrule
Homework Helper
I am heading out to a wedding now (my girlfriends sister is getting married) I will be sure to post the completed solution when I get back tomorrow! Thank you so much for the help!
Best of luck for your gf's sister and her husband!

[pulley] = Radius R, and moment of inertia I = (1/5)mR².
[weights] = A = m and B = 2m.
[string tension] = left side (A): T ; right side (B): S.
[gravity] = As always g denotes the pull of the earth (9.82 m/s² here in Sweden).

The (all new!) attempt at a solution
So, as listed above I will call the momentum about the z-axis of the pulley M, the angular acceleration alpha. I called the "left" pulley A and the "right" pulley B.
I set the x-axis to the positive in the upward direction. And I assumed that the string was not elastic in any way. Also I have come to realise that the main reason to use Euler II in this question is to solve so we get an expression that doesn't assume that the acceleration is known, so that the tension of the strings will only depend on the weight (m) (and g) of the pulley and weights on the string.

First I used Newtons second law to describe the forces acting on pulley A to be:
F = T - mg = ma
and respectively for B:
F = S - 2mg = -2ma

Then I used Eulers II law with the moments origo in the center of the pulley (my y-axis is positive going upwards, and the pulley is rotating clockwise!) to be:
M = RT - RS = I * alpha (and since alpha = a/R) = R(T - S) = mR²a/(5R) = T-S=ma/5.

I now solved the NII equations for T and S which made them:
T = ma + mg and S = 2mg - 2ma
In combination with EII:
ma+mg - 2mg+2ma = ma/5
Solving for a:
15a-5g = a -> 14a = 5g -> a = 5g/14

Thus I inserted the acceleration a into my equations for T and S:
T = 5mg/14 + 14mg/14 = 19mg/14
S = 28mg/14 - 10mg/14 = 9mg/7

Could this be the correct answer?

Doc Al
Mentor
First I used Newtons second law to describe the forces acting on pulley A to be:
F = T - mg = ma
and respectively for B:
F = S - 2mg = -2ma
OK.
Then I used Eulers II law with the moments origo in the center of the pulley (my y-axis is positive going upwards, and the pulley is rotating clockwise!) to be:
M = RT - RS = I * alpha (and since alpha = a/R) = R(T - S) = mR²a/(5R) = T-S=ma/5.
You still have a sign problem; see my previous post.