Torque exerted on rolling cylinder

In summary, the cylinder will roll rightward if the angle between the string and the cylinder is equal to 2 pi radians.
  • #1
maxhell
3
1
Homework Statement
Torque and rolling
Relevant Equations
Torque=Ia and F=ma
Hello,
i have tried to calculate the acceleration (COM) of the cylinder (even though in the question they asked about the angular acceleration) and the answer is:
𝑎(𝑐𝑜𝑚)=𝐹(𝑟/𝑅−𝑐𝑜𝑠(𝑡ℎ𝑒𝑡𝑎))/(𝐼/𝑅2+𝑀)

and my answer is with minus (I/R^2 - M) . in their solution they wrote in the torque equation--> f(static)R-Fr=Ia(Com)/R but i don't understand why , because the magnitude of a(com) is positive ( chose the right direction as positive) and f(static)-Fr is <0 so why take f(s)R-Fr and not like i did Fr-f(s)R=I(a(com)/R
(at b))

why isn't obvious that the cylinder will roll rightward?
thanks for the help
https://ibb.co/QPYKG3p https://ibb.co/MMTzRtS
 
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  • #2
Hi @maxhell and welcome to PF.

It would help us to have the original, accurate, complete question, word-for word and with any supplied diagram(s). Otherwise we are just guessing and/or relying on you having interpreted the question correctly.

Some general points though…

Is the object actually a rolling yo-yo, pulled by a string? If not, what is it?

Is there any sliding? If it’s a yo-yo, is the string being wound onto (or off) the reel as the yo-yo moves? Is the yo-yo moving left or right (you imply right).?

(Note, if it's a yo-yo it can be made to mve left or right depending on the angle (0 - 90º) of the string. If this isn't clear, imagine what happens when string is vertical (angle = 90º).

In your solution, note you are dividing by ##\frac I {r^2} – M##. That should sound alarm bells. If you take the case where ##M = \frac I {r^2}## then you would have infinite acceleration. You don’t want that!

Last (but not least), check use of sign conventions carefully when mixing rotational and linear motions. For example, for something rolling to the right (+x direction), if using a = αR you want to ensure that a positive angular acceleration (α) gives a positive linear acceleration (a). You would need to use the convention ‘clockwise is positive’ for α (angular acceleration in this case.

Edited - for typo's and clarity only.
 
  • #3
Steve4Physics said:
Hi @maxhell and welcome to PF.

It would help us to have the original, accurate, complete question, word-for word and with any supplied diagram(s). Otherwise we are just guessing and/or relying on you having interpreted the question correctly.

Some general points though…

Is the object actually a rolling yo-yo, pulled by a string? If not, what is it?

Is there any sliding? If it’s a yo-yo, is the string being wound onto (or off) the reel as the yo-yo moves? Is the yo-yo moving left or right (you imply right).?

(Note, if it's a yo-yo it can be made to mve left or right depending on the angle (0 - 90º) of the string. If this isn't clear, imagine what happens when string is vertical (angle = 90º).

In your solution, note you are dividing by ##\frac I {r^2} – M##. That should sound alarm bells. If you take the case where ##M = \frac I {r^2}## then you would have infinite acceleration. You don’t want that!

Last (but not least), check use of sign conventions carefully when mixing rotational and linear motions. For example, for something rolling to the right (+x direction), if using a = αR you want to ensure that a positive angular acceleration (α) gives a positive linear acceleration (a). You would need to use the convention ‘clockwise is positive’ for α (angular acceleration in this case.

Edited - for typo's and clarity only.
hi thanks for your help,this is the question:
the system in the picture is placed on a floor with friction. the cord that wrapped around the inner cylinder is pulled by force F with angle theta
the system is rolling without slipping on the plane.
the system moment of inertia is given by I and is of mass M
1.what is the acceleration of the system
2.what is the direction of the movement

here the cylinder is rotate counterclockwise and the rolls to right
so that is why i took Fr-f(s)R=Ia(com)
but the answer is when you do the opposite f(s)R-Fr=Ia(com) and i don't understand why.
 

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  • #4
maxhell said:
Homework Statement:: Torque and rolling
Relevant Equations:: Torque=Ia and F=ma

why isn't obvious that the cylinder will roll rightward?
Even if ##\theta=\pi/2##?
maxhell said:
here the cylinder is rotate counterclockwise and the rolls to right
Do you mean that those are the directions you are taking as positive? Clearly it cannot actually rotate counterclockwise but move right.
Which direction are you taking as positive for the frictional force?
 
Last edited:
  • #5
maxhell said:
here the cylinder is rotate counterclockwise and the rolls to right
Disagree! (Also, I'd call it a yo-yo or a spool, to distinguish it from a simple cylinder.)

If the yo-yo moves to the right wihtout slipping (rolls), then it must rotate clockwise, not counterclockwise. (In this case the string will get wound onto the inner part of the yo-yo, which is a bit counterintuitive.)

Watch this demo':
 
  • #6
Thnks for the video i understand now that i was wrong and the torque of friction is bigger than the torque of the force.
And when the angle is 90 degree than the friction is the only force in the x-axis and it will roll to left and than the troque of the tenstion is bigger than the torque of friction. right ?
 
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  • #7
maxhell said:
Thnks for the video i understand now that i was wrong and the torque of friction is bigger than the torque of the force.
And when the angle is 90 degree than the friction is the only force in the x-axis and it will roll to left and than the troque of the tenstion is bigger than the torque of friction. right ?
Yes
 

Related to Torque exerted on rolling cylinder

1. What is torque exerted on a rolling cylinder?

Torque exerted on a rolling cylinder is the measure of the force that causes the cylinder to rotate about its axis. It is the product of the force applied to the cylinder and the distance from the axis of rotation to the point where the force is applied.

2. How is torque calculated for a rolling cylinder?

To calculate torque exerted on a rolling cylinder, you need to know the force applied to the cylinder and the distance from the axis of rotation to the point where the force is applied. The formula for torque is T = F x r, where T is torque, F is force, and r is the distance from the axis of rotation.

3. What factors affect the torque exerted on a rolling cylinder?

The torque exerted on a rolling cylinder is affected by several factors, including the force applied, the distance from the axis of rotation, the mass and shape of the cylinder, and the surface it is rolling on. Friction and air resistance can also impact the torque exerted on a rolling cylinder.

4. How does torque affect the motion of a rolling cylinder?

The torque exerted on a rolling cylinder determines its rotational motion. If the torque is greater than the opposing forces, the cylinder will accelerate in the direction of the torque. If the torque is equal to the opposing forces, the cylinder will maintain a constant rotational speed. And if the torque is less than the opposing forces, the cylinder will decelerate or stop rotating.

5. How is torque exerted on a rolling cylinder useful in real life?

Understanding torque exerted on a rolling cylinder is important in many practical applications, such as designing and controlling vehicles and machinery. It is also essential in sports, where torque is used to generate power and control movements, such as in throwing a ball or swinging a golf club.

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