Normal acceleration of the Belt on the Pulley

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  • #1
MPavsic
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I have a series of pulleys where the belt is running around them in a way to describe a sine curve. The pulleys are stationary and the belt is running from left to right. For every particle of the belt I can use standard formula to calculate their normal acceleration, when in contact with the pulley.

Now suppose that we have very long belt which is stationary and pulleys are now moving from left to right. Can I use the same formula to calculate normal acceleration of the points on the belt when in contact with the pulley?

Now the points of the belt are in simple harmonic motion and are not circling, looking from this perspective.
BeltPulleys.png
 

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  • #2
tnich
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Summary: rotating reference frame, normal acceleration
You list "rotating reference frame" here. You can solve your problem using the concept of reference frames, but not rotating ones. Consider two reference frames, one in which the linear velocity of the pulleys is zero, and the other in which the (horizontal) velocity of the belt is zero. Are these frames inertial? What is the effect of a change of inertial reference frame on acceleration of an object?
By the way, the motion of the belt over the pulleys is not truly sinusoidal.
 
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  • #3
jbriggs444
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You have mentioned a sine curve and simple harmonic motion. But you have also mentioned pulleys and drawn circular shapes. Which is it? Are the curves circular or sinusoidal? Or is it a third possibility -- since simple harmonic motion is incompatible with a sinusoidal profile for the belt.
 
  • #4
256bits
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Summary: rotating reference frame, normal acceleration

I can use standard formula to calculate their normal acceleration
What would have been the standard formula you did use to calculate the NORMAL force?
 
  • #5
MPavsic
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Hi, thank you all for your reply.
I am self-learner, brainstorming how to solve specific problem. In my study I am mainly using internet and my favorite book titled Vector Mechanic for Engineers Static and Dynamics. Please take my question as simplification of my depicted problem.
My thinking is; every mass particle on the moving belt, when in contact with the pulley will have Normal acceleration which is pointing from the belt mass particle toward the center of instant rotation of the pulley, in case if the belt is moving across pulleys.
My simplified question is; will the mass particle have the same normal acceleration if the pulleys are moving and the belt is standing still. As I can anticipate from your answers there is no difference, or can I use the same calculation procedure?
Subtract position of point P from instant center of rotation to get r
Calculate relative velocity of point P relative to instant center of rotation V rel= omega x r
Calculate normal acceleration An=omega x V rel
Where x = vector cross product
Thanks.
 
  • #6
tnich
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The two situations you mention - moving belt and moving pulley's - are the same. Imagine that you are on a train moving at the same constant velocity as the belt while the pulleys remain motionless. Even though the belt does not appear to move from your point of view, the acceleration of the belt doesn't change.
This is true as long as one reference frame is not accelerating relative to the other.
In the moving belt reference frame with belt speed v, and pulley radius R, the equations of motion of a point on the belt as it passes over a pulley are $$x_B = R-R\cos(\frac {vt}R)$$
$$y_B = R\sin(\frac {vt}R)$$
Differentiating twice with respect to ##t## gives
$$\ddot x = \frac{v^2}R \cos(\frac {vt}R)$$
$$\ddot y = -\frac{v^2}R \sin(\frac {vt}R)$$
and the magnitude of the acceleration is ##(\ddot x^2 + \ddot y^2)^{\frac12}##
For moving pulleys, the equations of motion of a point on the belt are
$$x_B = R-vt-R\cos(\frac {vt}R)$$
$$y_B = R\sin(\frac {vt}R)$$
and the second derivatives don't change.
 
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  • #7
sophiecentaur
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Is there another way of looking a this? It strikes me that the points on the wheels will be moving on a Cycloid (or a Trochoid- if the belt has finite thickness). (See Wiki for a starter). Alternate wheels will be rotating in opposite directions, as if one set were rolling along a floor and the alternate set would be rolling along the 'ceiling'. The motion of the belt would also follow a cycloid when it is in contact with a wheel.
 
  • #8
MPavsic
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Is there another way of looking a this? It strikes me that the points on the wheels will be moving on a Cycloid (or a Trochoid- if the belt has finite thickness). (See Wiki for a starter). Alternate wheels will be rotating in opposite directions, as if one set were rolling along a floor and the alternate set would be rolling along the 'ceiling'. The motion of the belt would also follow a cycloid when it is in contact with a wheel.
Hi, the wheels are counter rotating and the belt si not sliding on the pulleys. I am a little bit confused, because I was using wrong equations in my model.
Is the following thinking correct?
When the belt touches the pulley and follows the rotation path of the pulley; I can use the same equations to calculate forces on the belt as for the mass particles of the pulley?
 
  • #9
jbriggs444
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When the belt touches the pulley and follows the rotation path of the pulley; I can use the same equations to calculate forces on the belt as for the mass particles of the pulley?
Yes. The acceleration of a small piece of belt that is in contact with the surface of the pulley is the same as that of the small piece of pulley under that bit of belt.
 
  • #10
sophiecentaur
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Thinking aloud here and some of the following has already been written - just trying to tidy things up in my own head as much as anything:
My cycloid idea may be interesting but not particularly significant, perhaps,
Definitely not SHM; it's all motion in (part of) a circle.
If the wheels are moving at constant rotational speed and if there is no load then there need be no (?) mean tension on the belt; there's a lot of symmetry. (Shown by the equations of @tnich above) Tangential forces on each part of the circumference must be the same as the belt tension at that point and they will integrate to zero.
All this seems to indicate that the net force on the belt is radial and equal to ω2/R Tension on the belt is normal to the radius so why would it affect that radial force - except perhaps where the belt crosses the line of centres?
Someone tell me if / where this is wrong please.
 
  • #11
MPavsic
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Yes. The acceleration of a small piece of belt that is in contact with the surface of the pulley is the same as that of the small piece of pulley under that bit of belt.
OK thanks, I thought that I am dead wrong.
So I could use rotating reference frame equations to describe forces on the belt when touching the pulley?
The belt particle is having curved trajectory around imaginary center of rotation, can I measure its angular velocity and angular momentum as the pulley wasn’t even there? I think I can or?

Regarding SHM, from Wikipedia "Uniform circular motion describes the movement of an object traveling a circular path with constant speed. The one-dimensional projection of this motion can be described as simple harmonic motion."
The belt particle is very close to SHM looking in Y axis.
 
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  • #12
sophiecentaur
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can I measure its angular velocity and angular momentum as the pulley wasn’t even there? I think I can or?
I think you can safely do that. If the force were greater or less than that, the belt would leave the surface or sink into it.
The one-dimensional projection of this motion can be described as simple harmonic motion."
Yes that's true BUT the sinusoidal motion is normal to the line of centres and not radial. (It's the projection of uniform circular motion along one cartesian axis as time varies.
 
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  • #13
tnich
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So I could use rotating reference frame equations to describe forces on the belt when touching the pulley?
I am uncomfortable with your use of the term "reference frame" here. In a rotating reference frame (the one the would be applicable in the case) the pulley and the part of the belt in contact with it would be at rest, and the belt and pulley would be subject to an apparent centrifugal acceleration outward from the center of the pulley. This is a rather convoluted way to think of the normal acceleration. Given that do this you have to first find the centripetal acceleration in an inertial reference frame, there is no point in using a rotating reference frame.

You may be confusing the concept of a rotating reference frame with polar coordinates. It would make sense to consider the centripetal acceleration in polar coordinates in an inertial frame, since it would be constant and normal to the belt.
The belt particle is very close to SHM looking in Y axis.
To add to @sophiecentaur's comments, the motion of the belt around the pulley differs significantly from SHM. In your belt/pulley system, the belt is moving vertically at the point where it is in contact with two pulleys. In SHM, the absolute value of the slope of the curve is never greater than v for ##y = R\sin(\frac{vt} R)##. That is a big difference.
 
  • #14
sophiecentaur
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There is a difference between the slope of the belt in the spatial domain (vertical at the crossover from one wheel to the other) and the time derivative of the vertical position, ( the 45 degrees of a sinusoid when crossing the x axis.
 
  • #15
tnich
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There is a difference between the slope of the belt in the spatial domain (vertical at the crossover from one wheel to the other) and the time derivative of the vertical position, ( the 45 degrees of a sinusoid when crossing the x axis.
Agreed. That is the point I was trying to make with my example.
 
  • #16
MPavsic
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Thank you all for your replies. As I mentioned I am self-learner, approaching to my problem directly, following my intuition regarding equations and calculation procedures. I might publish my complete problem on this forum in a while.
 

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