# I have a diffcult question about torque -- me

• garylau
In summary: If you think R/r should be negative then you have to think about what that means in the problem. Is it really negative, or is the force in the wrong direction?easier to just think about it for the simple cases like this. If you think R/r should be negative then you have to think about what that means in the problem. Is it really negative, or is the force in the wrong direction?In summary, the conversation discusses a pulley system with two cylinders and two cords used to raise a mass. The tension in the cord attached to the mass, the angular acceleration of the cylinders, and the acceleration of the mass are all
garylau

## Homework Statement

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.

## Homework Equations

F=mg
Torque=maR=I*alpha

## 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

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

haruspex said:
Your signs are inconsistent. You have taken α as positive anticlockwise, then written a=αr. That makes a positive down.
oh i see
thank
But how to make the sign of the torque and the sign of the net force consistent in general?

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

haruspex said:
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.
but in convention
the anti-clockwise direction is the positive direction of the torque,right?

garylau said:
but in convention
the anti-clockwise direction is the positive direction of the torque,right?
So write that the torque is -Fx.

haruspex said:
So write that the torque is -Fx.
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)

garylau said:
then horizontal force F to the right should be negative??
)
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

garylau
haruspex said:
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
So in my original problem

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

haruspex said:
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
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?

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garylau said:
So in my original problem

the sign of R/r should be negative so that i can get the correct answer
Yes.
garylau said:
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?
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.

garylau

## 1. What is torque and how is it measured?

Torque is a measure of the turning force of an object. It is calculated by multiplying the force applied to an object by the distance from the point of rotation to the point where the force is applied. The unit of measurement for torque is Newton-meters (N*m) in the metric system and foot-pounds (ft*lbs) in the imperial system.

## 2. How does torque affect the motion of an object?

Torque is responsible for causing rotational motion in an object. When a torque is applied to an object, it causes the object to rotate around its axis. The amount of torque applied determines the speed and direction of the rotation.

## 3. What factors affect the torque of an object?

The torque of an object is affected by three main factors: the magnitude of the force applied, the distance from the point of rotation to where the force is applied, and the angle at which the force is applied. Increasing any of these factors will result in an increase in torque.

## 4. How is torque used in everyday life?

Torque is used in many everyday objects and activities, such as opening a door, riding a bike, and driving a car. It is also essential in machines and tools, such as wrenches and drills, which use torque to rotate and tighten or loosen bolts and screws.

## 5. What is the difference between torque and force?

While both torque and force are measures of the application of physical energy, they are different concepts. Force is a push or pull exerted on an object, while torque is a twisting or turning force. Torque is also dependent on the distance from the point of rotation, whereas force is not necessarily affected by distance.

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