1. Jan 25, 2017

### garylau

1. The problem statement, all variables and given/known data
The sketch shows, from two views, a simple pulley system for raising large masses. A light, inextensible cord attached to a mass m is wound, without slipping, on a cylinder of radius r. Rigidly joined to this cylinder and mounted on the same central shaft is a cylinder of radius R. The friction between the cylinders and the shaft is negligible. The two cylinders together have mass M, moment of inertia, I and radius of gyration k. Around the larger cylinder is wound, without slipping, anther light, inextensible cord, as shown. An operator pulls with tension T1 on this cord to raise the mass m.The mass m accelerates upwards with vertical acceleration a. (i) Using Newton's laws or otherwise, write an expression for T2, the tension in the cord attached to mass m, in terms of m, g and a. (ii) Using Newton's laws for rotation or otherwise, derive an expression for the angular acceleration, α, of the cylinders, in terms of variables given in the diagram. (iii) Showing all working, derive an expression for the acceleration a of mass m in terms of some or all of the parameters T1, m, M, I, r, R and the gravitational acceleration g.

2. Relevant equations
F=mg
Torque=maR=I*alpha

3. The attempt at a solution

i have done my work in the paper but i found the answer get wrong in(iii)and i don't know whether it is correct or not

can i write the answer like this?

the correct answer is a =( RT1 – rmg)/( I/r + rm)

but my answer is something like...(I/r-rm)...

Do i do it wrong?thank you

#### Attached Files:

File size:
28.5 KB
Views:
29
• ###### 16344105_361846214200159_1900791846_n.jpg
File size:
31.8 KB
Views:
32
Last edited: Jan 25, 2017
2. Jan 25, 2017

### haruspex

Your signs are inconsistent. You have taken α as positive anticlockwise, then written a=αr. That makes a positive down.

3. Jan 25, 2017

### garylau

oh i see
thank
But how to make the sign of the torque and the sign of the net force consistent in general?

4. Jan 25, 2017

### haruspex

If not using vectors then you just have to stop and think. If there is a horizontal force F you are taking as positive to the right and it acts a distance x above the axis then the moment Fx is positive clockwise, etc.

5. Jan 25, 2017

### garylau

but in convention
the anti-clockwise direction is the positive direction of the torque,right?

6. Jan 25, 2017

### haruspex

So write that the torque is -Fx.

7. Jan 25, 2017

### garylau

then horizontal force F to the right should be negative??

it seems it is contradicted with the "other sign of convention"(which usually we take right side to be positive and left side to be negative)

8. Jan 25, 2017

### haruspex

No, you can keep right as positive for forces, displacements, velocities and accelerations. The problem is the torque arm, x.
If the force is displacement x below the axis then Fx will be positive anticlockwise, as desired. But with the force x above the axis it becomes -Fx. So, for these cases, you could define x as the displacement from the force to the axis, up being positive.
Unfortunately, this breaks down when you consider vertical forces with horizontal displacements. Now you have to make the displacement from the force to the axis positive to the left to get the right result.
So the simplest thing, in scalars, is just to figure it out each time: if a positive right force acts above the axis it gives a clockwise moment, so use -Fx.

For a more systematic solution you need to use vectors and cross products. A set of conventions arranges that the sign comes out right. Specifically:
• $\vec \tau=\vec r\times\vec F$, not the other way around
• the right hand rule

9. Jan 25, 2017

### garylau

So in my original problem

the sign of R/r should be negative so that i can get the correct answer?

10. Jan 25, 2017

### garylau

for these cases, you could define x as the displacement from the force to the axis, up being positive.

how about the vector x pointing in random direction?

#### Attached Files:

• ###### 22222.png
File size:
51.7 KB
Views:
14
11. Jan 25, 2017

### haruspex

Yes.
i was not proposing you do that. I was just illustrating there is no simple way to determine it.

Using the vector approach, you would represent the first argument to the cross product (displacement from axis to force line) with your right index finger, and the second, the force, with your right middle finger. Your thumb indicates the direction of the result. This has to be combined with the convention that torques and angular motions also use a right hand thread rule. E.g. if the vector points away from you then the motion will appear clockwise.