Yo-yo on an accelerating conveyor belt

In summary: Oh, I see. So if the centre of the yo-yo were at the origin and it were to roll without slipping, the linear acceleration would be ##R \alpha##?Yes, that's correct.
  • #1
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Homework Statement
A yo-yo is placed on a conveyor belt accelerating ##a_C = 1 m/s^2## to the left. The end of the rope of the yo-yo is fixed to a wall on the right. The moment of inertia is ##I = 200 kg \cdot m^2##. Its mass is ##m = 100kg##. The radius of the outer circle is ##R = 2m## and the radius of the inner circle is ##r = 1m##. The coefficient of static friction is ##0.4## and the coefficient of kinetic friction is ##0.3##.

Find the initial tension in the rope and the angular acceleration of the yo-yo.
Relevant Equations
##T - f = ma##
##\tau_P = -fr##
##\tau_G = Tr##
##I_P = I + mr^2##
##I_G = I + mR^2##
##a = \alpha R##
First off, I was wondering if the acceleration of the conveyor belt can be considered a force. And I'm not exactly sure how to use Newton's second law if the object of the forces is itself on an accelerating surface.
Also, I don't know whether it rolls with or without slipping.
I thought I could use ##a_C = \alpha R## for the angular acceleration, but the acceleration of the conveyor belt is not the only source of acceleration, since the friction and the tension also play a role.
I can't find a way to combine these equations to get the
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  • #2
sudera said:
if the acceleration of the conveyor belt can be considered a force
It certainly results in a frictional force, but it is not immediately clear how great that force is.
sudera said:
whether it rolls with or without slipping.
Again, cannot immediately say. Assume it does not slip, calculate the frictional force necessary to keep pace with the belt, then check whether the static frictional force can be that large.
sudera said:
could use aC=αR ... but the acceleration of the conveyor belt is not the only source of acceleration
Quite so. For that equation to represent rolling contact the linear acceleration term needs to be the relative acceleration of the two bodies.
If α is the rotational acceleration, what is the yo-yo's acceleration relative to the wall?
 
  • #3
haruspex said:
For that equation to represent rolling contact the linear acceleration term needs to be the relative acceleration of the two bodies.
If α is the rotational acceleration, what is the yo-yo's acceleration relative to the wall?
##T-f = ma_F## where ##a_F## denotes the acceleration (to the right) due to the forces. ##a_{total} = a_F - a_c##.
If I'm correct, this would give ##\alpha R = a_{total} = \frac{T-f}{m} - 1 m/s^2##.
However, I still don't know ##T## or ##f##.

Thanks for the help!
 
  • #4
sudera said:
##T-f = ma_F## where ##a_F## denotes the acceleration (to the right) due to the forces. ##a_{total} = a_F - a_c##.
If I'm correct, this would give ##\alpha R = a_{total} = \frac{T-f}{m} - 1 m/s^2##.
However, I still don't know ##T## or ##f##.

Thanks for the help!
You do not need to consider forces to answer my question - it's a matter of kinematics.
 
  • #5
haruspex said:
You do not need to consider forces to answer my question - it's a matter of kinematics.
I believe it would be ##a = \alpha R - a_C## then (with ##\alpha## in the clockwise direction). But what can I do with this, exactly?
 
  • #6
sudera said:
I believe it would be ##a = \alpha R - a_C## then (with ##\alpha## in the clockwise direction). But what can I do with this, exactly?
Forget the belt for a moment. Concentrate on the wall, the string and the yo-yo. If the yo-yo turns through some angle θ, what happens to its distance from the wall?
 
  • #7
haruspex said:
Forget the belt for a moment. Concentrate on the wall, the string and the yo-yo. If the yo-yo turns through some angle θ, what happens to its distance from the wall?

Its distance would increase/decrease by ##\theta R## if it rolls without slipping.
 
  • #8
sudera said:
Its distance would increase/decrease by ##\theta R## if it rolls without slipping.
I said to ignore the belt, so whether it slips on that is irrelevant. Focus on the string. Assume that stays taut.
 
  • #9
haruspex said:
I said to ignore the belt, so whether it slips on that is irrelevant. Focus on the string. Assume that stays taut.
Oh, I think I see what you mean. If it turns an angle ##\theta##, a distance of ##r \theta## of string would wind around the yo-yo, so the yo-yo would move closer to the wall with this distance.
 
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  • #10
sudera said:
Oh, I think I see what you mean. If it turns an angle ##\theta##, a distance of ##r \theta## of string would wind around the yo-yo, so the yo-yo would move closer to the wall with this distance.
Right. What can you conclude from that regarding acceleration relative to the wall?
 
  • #11
haruspex said:
Right. What can you conclude from that regarding acceleration relative to the wall?
The acceleration relative to the wall would be ##r \alpha##.
 
  • #12
sudera said:
The acceleration relative to the wall would be ##r \alpha##.
Right.
In terms of r, R and α, what is the linear acceleration of the point of the yo-yo that is in contact with the belt?
 
  • #13
haruspex said:
Right.
In terms of r, R and α, what is the linear acceleration of the point of the yo-yo that is in contact with the belt?

Still ignoring the acceleration of the belt, the linear acceleration of that point would be ##R \alpha##. If you mean without ignoring the belt's acceleration, I think it would be the same if it rolls without slipping.
 
  • #14
sudera said:
Still ignoring the acceleration of the belt, the linear acceleration of that point would be Rα.
No. Remember that the centre of the yo-yo is accelerating towards the wall.
 
  • #15
haruspex said:
No. Remember that the centre of the yo-yo is accelerating towards the wall.
Is it ##-R \alpha## then?
 
  • #16
@sudera :
sudera said:
Is it ##-R \alpha## then?
Be sure to distinguish between the acceleration relative to the center of the yo-yo and the acceleration relative to the wall.

In general, consider any three points A, B, and C that can move relative to each other. Let ##\vec a_{A/B}## be the acceleration of A relative to B, ##\vec a_{B/C}## the acceleration of B relative to C, and ##\vec a_{A/C}## the acceleration of A relative to C. How are these three vectors related?
 
  • #17
TSny said:
@sudera :

Be sure to distinguish between the acceleration relative to the center of the yo-yo and the acceleration relative to the wall.

In general, consider any three points A, B, and C that can move relative to each other. Let ##\vec a_{A/B}## be the acceleration of A relative to B, ##\vec a_{B/C}## the acceleration of B relative to C, and ##\vec a_{A/C}## the acceleration of A relative to C. How are these three vectors related?

Oh, I see. ##\vec a_{A/C} = \vec a_{A/B} + \vec a_{B/C}## so this means that for the point touching the belt, its acceleration is the sum of the acceleration of the center relative to the wall and the acceleration of the point relative to the center. So its linear acceleration is ##r \alpha - R \alpha##?
 
  • #18
sudera said:
Oh, I see. ##\vec a_{A/C} = \vec a_{A/B} + \vec a_{B/C}## so this means that for the point touching the belt, its acceleration is the sum of the acceleration of the center relative to the wall and the acceleration of the point relative to the center. So its linear acceleration is ##r \alpha - R \alpha##?
Yes.
So if it is not slipping, what equation can you write?
 
  • #19
haruspex said:
Yes.
So if it is not slipping, what equation can you write?

Its linear acceleration is also the acceleration of the belt I think, so ##-a_C = r \alpha - R \alpha##?
 
  • #20
sudera said:
Its linear acceleration is also the acceleration of the belt I think, so ##-a_C = r \alpha - R \alpha##?
Right. So what torque is needed for α and what frictional force does that imply?
 
  • #21
haruspex said:
Right. So what torque is needed for α and what frictional force does that imply?
##\tau_C = I_C \alpha##, where ##I = 200 kg \cdot m^2## is the moment of inertia around the center and ##\tau_C## the torque on the center.

##\alpha = \dfrac{a_C}{R-r}## so the magnitude of the torque needed is ##200 \dfrac{a_C}{R-r}##.

I'm not sure what this implies for the frictional force because both the rope and the friction cause the torque.
 
  • #22
sudera said:
##\tau_C = I_C \alpha##, where ##I = 200 kg \cdot m^2## is the moment of inertia around the center and ##\tau_C## the torque on the center.

##\alpha = \dfrac{a_C}{R-r}## so the magnitude of the torque needed is ##200 \dfrac{a_C}{R-r}##.

I'm not sure what this implies for the frictional force because both the rope and the friction cause the torque.
Yes, both the frictional force and the tension in the rope contribute to the torque. They also both contribute to the linear acceleration, so you have two equations available.
 
  • #23
haruspex said:
Yes, both the frictional force and the tension in the rope contribute to the torque. They also both contribute to the linear acceleration, so you have two equations available.
##\tau_c = -T r + f R = \dfrac{a_c}{R-r}##
##m a = T - f = m \alpha r##
I don't know if I should include the belt's acceleration (in the form of ## + m a_C## on the left-hand side) as a force. Does an accelerating surface count as a force? I feel like it should, but on the other hand, if you put something on a slippery conveyor belt, it would stay in place so it would not receive a force.
 
  • #24
sudera said:
Does an accelerating surface count as a force?
No.
So combine your equations to get an expression for f.
What is the maximum static frictional force available?
 
  • #25
The equations give ##f = 300 N## and ##T = 400 N##. The maximum static frictional force is ##\mu_s n = \mu_s m g = 392.4 N## so it rolls without slipping.

I'm just wondering now what I'd have to do if the maximum was less than 300 N. The acceleration of the contact point of the yo-yo on the belt, relative to the wall, wouldn't equal the acceleration of the conveyor belt anymore. Can you still find the unknowns in that case?
 
  • #26
sudera said:
The equations give ##f = 300 N## and ##T = 400 N##. The maximum static frictional force is ##\mu_s n = \mu_s m g = 392.4 N## so it rolls without slipping.

I'm just wondering now what I'd have to do if the maximum was less than 300 N. The acceleration of the contact point of the yo-yo on the belt, relative to the wall, wouldn't equal the acceleration of the conveyor belt anymore. Can you still find the unknowns in that case?
Once you have shown that it slips, you know the frictional force from the kinetic coefficient.
 
  • #27
haruspex said:
Once you have shown that it slips, you know the frictional force from the kinetic coefficient.
Oh, I see. Thank you so much for your elaborate help and your patience! I really appreciate you walking me through the problem step by step.
 
  • #28
sudera said:
Oh, I see. Thank you so much for your elaborate help and your patience! I really appreciate you walking me through the problem step by step.
You are welcome. It was an instructive problem.
 

What is a yo-yo on an accelerating conveyor belt?

A yo-yo on an accelerating conveyor belt is a thought experiment that explores the relationship between acceleration and motion. It involves a yo-yo placed on a moving conveyor belt that is accelerating, causing the yo-yo to move in a circular motion.

What forces are acting on the yo-yo on an accelerating conveyor belt?

The yo-yo on an accelerating conveyor belt experiences two main forces: the tension force from the string and the frictional force from the conveyor belt. The tension force pulls the yo-yo towards the center of the circular motion, while the frictional force acts in the opposite direction of the conveyor belt's motion.

How does the acceleration of the conveyor belt affect the motion of the yo-yo?

The acceleration of the conveyor belt has a direct effect on the motion of the yo-yo. As the conveyor belt accelerates, the yo-yo's circular motion will also accelerate, causing it to move faster and with a larger radius. The direction of the acceleration also affects the direction of the yo-yo's motion.

What happens if the acceleration of the conveyor belt is greater than the acceleration of gravity?

If the acceleration of the conveyor belt is greater than the acceleration of gravity, the yo-yo will experience a net upward force and will eventually rise off the conveyor belt. This is because the tension force from the string will be greater than the weight of the yo-yo, causing it to move in an upward trajectory.

What real-world applications can be related to the yo-yo on an accelerating conveyor belt?

The concept of a yo-yo on an accelerating conveyor belt can be related to various real-world scenarios, such as a car turning on a curved road or a satellite orbiting around a planet. It also helps explain the relationship between acceleration and circular motion, which is important in fields such as physics and engineering.

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