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A spool of mass M sits upright on a table...

by bhimberg
Tags: mass, sits, spool, table, upright
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bhimberg
#1
Jan25-08, 10:26 AM
P: 19
1. The problem statement, all variables and given/known data
A spool with radius r and mass M sits upright on a table, a distance l from the edge. The string hangs off the edge of the table with a mass m attached. The table is frictionless: what will the velocity of the weight be when the spools center of mass reaches the edge of the table (l is measured from the spools center of mass to the edge).


2. Relevant equations
(Torque)=I(alpha)
I(alpha)=.5(M^2)(theta'')=F x r = F*r*sin(angle between F and r)
r(theta'')+x''=y'', where x is at the com of the spool and y is the com of the weight

what I'm having trouble with is the tension. The F in the torque equation should be the tension on M at r due to m. I've written down the equation:
my''=mg-T, where T is Mx''.

The thing is, the tension should be at its maximum when M (or Mx'') is stationary. The above equation has tension at 0 at that point. Perhaps I'm confusing total tension with the the force applied due to tension?

3. The attempt at a solution
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Doc Al
#2
Jan26-08, 08:52 AM
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For this problem you are assuming, correctly, that the tension is the only force on the spool. Thus T = Mx''. The only way that x'' can equal zero is if another force acts on the spool.
bhimberg
#3
Jan28-08, 07:13 AM
P: 19
I ended up using T=mg-Mx'', and crossing this with torque to find the rotational acceleration of M. That is, the total tension instead of the tension on M due to m.

(torque)=|(mg-Mx'') x r|= (mg-Mx'')r. This made sense to me since, once Mx''=mg, the tension disappears. Is this correct? More important to me atm is this relation:

my''=-mg+mx''. This says the acceleration of the weight, m, is zero when mx''=g. This doesn't sound right to me, but was gotten from a free body diagram. Any suggestions?

Doc Al
#4
Jan28-08, 03:28 PM
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A spool of mass M sits upright on a table...

Quote Quote by bhimberg View Post
I ended up using T=mg-Mx'', and crossing this with torque to find the rotational acceleration of M. That is, the total tension instead of the tension on M due to m.
I have no idea what you mean by "total" tension. There's only one tension. Since it's the only force acting on M, it happens that T = Mx''.

(torque)=|(mg-Mx'') x r|= (mg-Mx'')r. This made sense to me since, once Mx''=mg, the tension disappears. Is this correct?
No.

More important to me atm is this relation:

my''=-mg+mx''. This says the acceleration of the weight, m, is zero when mx''=g. This doesn't sound right to me, but was gotten from a free body diagram. Any suggestions?
Show me how you got that from a free body diagram.
bhimberg
#5
Jan28-08, 04:29 PM
P: 19
Thanks, you've helped a ton. In response to your request:

The weight has the force of gravity acting on it, -mg. It, and heres the confusing part, also feels the tension in the string caused by the spool (mass M), which is moving in the -x direction, and is thereby translated to the positive y (along the string). So my''=Mx''-mg.

Another way to put it is The net force on the weight is the tension on the weight due to the spool minus the force of gravity on the weight.

My understanding of tension is that it acts in the opposite direction of the net force on an object, and acts along the line connecting said object to another object. So if the spool, mass M, were moving along positive x with an acceleration x'', the tension due to the spool would be Mx'' along negative x: -Mx''.
Doc Al
#6
Jan28-08, 05:28 PM
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Quote Quote by bhimberg View Post
Thanks, you've helped a ton. In response to your request:

The weight has the force of gravity acting on it, -mg. It, and heres the confusing part, also feels the tension in the string caused by the spool (mass M), which is moving in the -x direction, and is thereby translated to the positive y (along the string). So my''=Mx''-mg.

Another way to put it is The net force on the weight is the tension on the weight due to the spool minus the force of gravity on the weight.
Yes, the net force on the weight (mass m) = +T - mg. The signs depend on your coordinate system.

The net force on the spool is just T.

My understanding of tension is that it acts in the opposite direction of the net force on an object, and acts along the line connecting said object to another object. So if the spool, mass M, were moving along positive x with an acceleration x'', the tension due to the spool would be Mx'' along negative x: -Mx''.
I don't know where you got the idea that tension acts opposite to the net force on an object. It's simple: String tension can only pull. The string pulls on the spool with tension T; since that's the only force on the spool, the net force on the spool is T.

To solve this problem, you need to apply Newton's 2nd law to the spool (to both translational and rotational motion) and the weight. And you need to connect them via the constraint equation relating the accelerations (which you already have in your first post).
bhimberg
#7
Jan29-08, 05:46 AM
P: 19
But aren't there two Ts? One T for the tension due to the weight, another due to M?


1) d(theta)'' = -x'' -y''
2) Mx''=T
3) my''=T-mg
4) (Torque)=T x r = Tr (since r and T are always orthogonal)

Four equations, 4 unknowns. If the T in 2 is the T in 3. Yet if they are the same, the net force on the weight equals the net force in the spool - the force of gravity on the weight. A force due to the weight can't act on the weight. I'm just a little confused here.

Thanks again. I have to hand this in today, in 2 hours, and will be reworking this problem before I do because of this thread.
Doc Al
#8
Jan29-08, 07:51 AM
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Quote Quote by bhimberg View Post
But aren't there two Ts? One T for the tension due to the weight, another due to M?
No. There's only one piece of string with a single tension from spool to weight.


1) d(theta)'' = -x'' -y''
2) Mx''=T
3) my''=T-mg
4) (Torque)=T x r = Tr (since r and T are always orthogonal)

Four equations, 4 unknowns. If the T in 2 is the T in 3. Yet if they are the same, the net force on the weight equals the net force in the spool - the force of gravity on the weight.
I don't understand what you are saying here. The net force on the weight is T-mg. The net force on the spool is T. Those aren't the same.

(Edit: Describe the orientation of the spool with respect to the edge and define your variables. This way I can better understand the signs used in your equations.)
FedEx
#9
Jan29-08, 09:40 AM
P: 336
I am sorry to say but i cant understand any of your explanations, bhimberg.

(Tension)(Radius) = (Moment Of Inertia)(Angular Acceleration)

mg - T = ma

And lets consider that the spool is like a hollow cylinder.And lets assume that the spool moves with out sliding. From this you will get that a = g/2

Apply the equations of rolling without sliding and then find out the velocity.

We can also apply law of conservation of energy but that will complicate things.
Doc Al
#10
Jan29-08, 09:46 AM
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Quote Quote by FedEx View Post
Apply the equations of rolling without sliding and then find out the velocity.
The way I read the problem, "a spool... sits upright on a table" means that it's on its end, not rolling at all. And since the table is frictionless, it couldn't roll without slipping anyway.
FedEx
#11
Jan29-08, 10:13 AM
P: 336
Indeed, You are totally correct Doc,as you are always are.Rolling is not possible without friction.

Coming to the prob if the spool is upright than how does it move on the table?I am not able to figure out the type of motion which the spool will follow.and of course there can never be two tensions in the same string.
Doc Al
#12
Jan29-08, 10:19 AM
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Quote Quote by FedEx View Post
Coming to the prob if the spool is upright than how does it move on the table?I am not able to figure out the type of motion which the spool will follow.and of course there can never be two tensions in the same string.
The tension does two things to the spool: (1) it produces an acceleration of the center of mass, and (2) it produces an angular acceleration about the center of mass.

The motion of the spool is the combination of both accelerations.
FedEx
#13
Jan29-08, 10:36 AM
P: 336
So if we look from above it will look like a ring. But will it have angular acc about the centrer? seeing that there is no friction.
Doc Al
#14
Jan29-08, 11:49 AM
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Sure it will have an angular acceleration. The string tension exerts a torque on it. (What does the absence of friction have to do with it?)


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