# Finding out the rotational speed of a mass

Homework Statement:
If a ##1.5## kg mass is rotating at uniform speed on a table and another ##2## kg mass is connected with a string which goes through the hole on the table at the center of rotation of the ##1.5## kg mass making the ##2## kg mass hanging then what should be the speed of rotation of the ##1.5## kg mass to keep the ##2## kg mass stationary?

Coefficient of sliding friction of the ##1.5## kg mass is ##μ_k=0.2## and the string is inextensible.
Relevant Equations:
Newtonian Mechanics **My Attempt:** Here, I considered for 2 kg mass the resultant motion is zero, which means it's acceleration (a) would be zero. So, if we consider the Tension force from the 1.5 kg mass to keep 2 kg mass from falling is L, then 2×9.8-L=ma=2× 0 ⇒ L=19.6 N. But where does the tension force from 1.5 kg comes from? I thought it's from the centrifugal (or centripetal) force. But should I take into account friction now? If the friction is fk then can I write this ##\frac{mv^2}{r}-f_k-L=0## in the next step to find out speed of the 1.5 kg mass?

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• Delta2

Delta2
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From what I know sliding friction acts in direction opposite to velocity, and since the velocity has tangential direction, sliding friction has tangential direction too. So the correct equation for me is $$\frac{mv^2}{r}-L=0$$, sliding friction has only tangential component, not radial component.

• nazmulhasanshipon
Thank you so much. Another question of mine is: If the speed of 1.5kg mass decrease due to that friction, then how can I draw the effective Velocity vs. Time graph of the 2kg mass?
Thank you so much. Another question of mine is: If the speed of 1.5kg mass decrease due to that friction, then how can I draw the effective Velocity vs. Time graph of the 2kg mass?

Delta2
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Thank you so much. Another question of mine is: If the speed of 1.5kg mass decrease due to that friction, then how can I draw the effective Velocity vs. Time graph of the 2kg mass?
I am afraid then the problem becomes quite more complex, the 2kg mass will do linear motion with non constant acceleration, and I am not sure at all what sort of motion the 1.5kg mass will do (I think its velocity and hence friction will have a normal as well as tangential component). I 'll invite some other users to this thread maybe they will be able to offer us their lights, @ergospherical ,@haruspex ,@TSny .

• nazmulhasanshipon
I am afraid then the problem becomes quite more complex, the 2kg mass will do linear motion with non constant acceleration, and I am not sure at all what sort of motion the 1.5kg mass will do (I think its velocity and hence friction will have a normal as well as tangential component). I 'll invite some other users to this thread maybe they will be able to offer us their lights, @ergospherical ,@haruspex ,@TSny .
Thanks, remember the Coefficient of sliding friction of the 1.5 kg mass is ##μ_k=0.2## and the string is inflexible.

jbriggs444
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Thanks, remember the Coefficient of sliding friction of the 1.5 kg mass is ##μ_k=0.2## and the string is inflexible.
I think we should interpret the string as being "inextensible" rather than inflexible. If the string is inflexible then the mass down below is irrelevant and you have a coat hangar through the table, not a string.

• nazmulhasanshipon
I think we should interpret the string as being "inextensible" rather than inflexible. If the string is inflexible then the mass down below is irrelevant and you have a coat hangar through the table, not a string.
Sorry, I am not native english speaker. I translated it, maybe it was a mistranslation.

kuruman
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For the 2-kg mass to be stationary, the centripetal force ##F_c## on the 1.5 kg mass must match the hanging weight. Now ##F_c=\dfrac{mv^2}{r}##. Because friction changes the kinetic energy and hence the centripetal force, the value of the speed at which the centripetal force matches the hanging weight is only instantaneous. Thus the mass cannot be stationary if friction is taken into account.
To @nazmulhasanshipon: Are you sure you have translated the statement of the problem faithfully?

For the 2-kg mass to be stationary, the centripetal force ##F_c## on the 1.5 kg mass must match the hanging weight. Now ##F_c=\dfrac{mv^2}{r}##. Because friction changes the kinetic energy and hence the centripetal force, the value of the speed at which the centripetal force matches the hanging weight is only instantaneous. Thus the mass cannot be stationary if friction is taken into account.
To @nazmulhasanshipon: Are you sure you have translated the statement of the problem faithfully?
Yes, that was an assignment problem. My english wasn't very good but I mentioned all the information presented there. By the way, 2 kg mass must have Tension force on the 1.5 kg mass, which provides the centripetal force. Right? But does those two force act in opposite direction?

kuruman
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There is one tension in the string connecting the two masses. The tension force acting on the 2-kg mass is up in the opposite direction from gravity. The tension force acting on the 1.5-kg mass is directed towards the center.

• nazmulhasanshipon
haruspex
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a 1.5 kg mass is rotating at uniform speed on a table
...
what should be the speed of rotation of the ##1.5## kg mass to keep the ##2## kg mass stationary?

Coefficient of sliding friction of the ##1.5## kg mass is ##μ_k=0.2## and the string is inextensible.
The question strikes me as ambiguous. Is the table stationary or rotating?
If stationary then, as @kuruman notes, the hanging mass cannot remain stationary. It will only be instantaneously so.

But if the table is rotating and the mass stationary wrt it then the relevant coefficient is static friction, not kinetic, and the answer will be a range, not a single value.

@nazmulhasanshipon , please confirm that the original question a) does not make it clear whether the table is rotating and b) does specify only kinetic friction.
From what I know sliding friction acts in direction opposite to velocity, and since the velocity has tangential direction, sliding friction has tangential direction too. So the correct equation for me is $$\frac{mv^2}{r}-L=0$$, sliding friction has only tangential component, not radial component.
Again, it depends how we interpret the question.
Friction acts to oppose relative motion of the surfaces in contact. If the platform is rotating at steady speed then the static friction is purely radial.

But if the table is static then, as you say, at the instant that the hanging block is stationary, friction will be kinetic and entirely tangential. That makes its value irrelevant, suggesting the other interpretation.

@haruspex , the table is static. If the table is static and we take the friction into account, then it is apparent that the rotating body will slow down, as it slows down the radius will decrease. Will there be any radial friction then?
The question strikes me as ambiguous. Is the table stationary or rotating?
If stationary then, as @kuruman notes, the hanging mass cannot remain stationary. It will only be instantaneously so.

But if the table is rotating and the mass stationary wrt it then the relevant coefficient is static friction, not kinetic, and the answer will be a range, not a single value.

@nazmulhasanshipon , please confirm that the original question a) does not make it clear whether the table is rotating and b) does specify only kinetic friction.

Again, it depends how we interpret the question.
Friction acts to oppose relative motion of the surfaces in contact. If the platform is rotating at steady speed then the static friction is purely radial.

But if the table is static then, as you say, at the instant that the hanging block is stationary, friction will be kinetic and entirely tangential. That makes its value irrelevant, suggesting the other interpretation.

kuruman
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@haruspex , the table is static. If the table is static and we take the friction into account, then it is apparent that the rotating body will slow down, as it slows down the radius will decrease. Will there be any radial friction then?
If there is a radial component to the velocity, there will be a radial component to the kinetic friction. This problem is fine as long as the table contact with the 1.5-kg mass is frictionless. Once you introduce friction, the 1.5-kg mass does not execute uniform circular motion and there is no answer to the question "What should be the speed of rotation of the 1.5-kg mass to keep the 2-kg mass stationary?" because the 2-kg mass cannot be stationary. So what's your pleasure?

If there is a radial component to the velocity, there will be a radial component to the kinetic friction. This problem is fine as long as the table contact with the 1.5-kg mass is frictionless. Once you introduce friction, the 1.5-kg mass does not execute uniform circular motion and there is no answer to the question "What should be the speed of rotation of the 1.5-kg mass to keep the 2-kg mass stationary?" because the 2-kg mass cannot be stationary. So what's your pleasure?
If we take into account friction and forget about keeping it stationary then, due to friction the radius decrease and the mass also comes near the center gradually, does this mean that it will experience radial friction when at the same time it is rotating and it's radius is decreasing?
I know because of circular motion it will feel tangential friction. But if the radius is decreasing, will it experience radial friction along the radius?

kuruman
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At any point on the trajectory of the mass, draw an arrow in the direction of the velocity. The force of kinetic friction will be directed at 180o from that. If the velocity has a radial component, the kinetic friction will have a radial component. It's as simple as that.

• Delta2
jbriggs444
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At any point on the trajectory of the mass, draw an arrow in the direction of the velocity. The force of kinetic friction will be directed at 180o from that. If the velocity has a radial component, the kinetic friction will have a radial component. It's as simple as that.
There is a somewhat interesting result of this fact.

Consider a moving object. Its velocity has an x component. Its velocity has a y component. If the object is subject to kinetic friction, the magnitude of the kinetic friction is fixed. It does not depend on speed.

If you make the object move faster in the x direction then the force component in the -x direction increases.
If you make the object move faster in the y direction then the force component in the -y direction increases.

But if magnitude of the vector sum is constant, the interesting thing is that if you increase the velocity in the x direction, the force in the y direction must decrease. This is why you can do doughnuts in the parking lot. Spinning your wheels fast in the x direction lets them slip quite easily when they move slowly in the y direction.

• • nazmulhasanshipon and kuruman
kuruman
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There is a somewhat interesting result of this fact.

Consider a moving object. Its velocity has an x component. Its velocity has a y component. If the object is subject to kinetic friction, the magnitude of the kinetic friction is fixed. It does not depend on speed.

If you make the object move faster in the x direction then the force component in the -x direction increases.
If you make the object move faster in the y direction then the force component in the -y direction increases.

But if magnitude of the vector sum is constant, the interesting thing is that if you increase the velocity in the x direction, the force in the y direction must decrease. This is why you can do doughnuts in the parking lot. Spinning your wheels fast in the x direction lets them slip quite easily when they move slowly in the y direction.
I never thought of the doughnut as an application. I was always annoyed by that fact because it stands in the way of getting closed-form solutions to projectile motion with air resistance.

There is a somewhat interesting result of this fact.

Consider a moving object. Its velocity has an x component. Its velocity has a y component. If the object is subject to kinetic friction, the magnitude of the kinetic friction is fixed. It does not depend on speed.

If you make the object move faster in the x direction then the force component in the -x direction increases.
If you make the object move faster in the y direction then the force component in the -y direction increases.

But if magnitude of the vector sum is constant, the interesting thing is that if you increase the velocity in the x direction, the force in the y direction must decrease. This is why you can do doughnuts in the parking lot. Spinning your wheels fast in the x direction lets them slip quite easily when they move slowly in the y direction.
Okay, but if the mass is subject to kinetic friction, the radius of it's rotation will decrease and the object will be spinning simultaneously. I understand, due to spin there would be tangential friction. But the object is moving to the center simultaneously too, does that mean it will experience a radial friction i.e friction along the radius because as you can see the hanging mass will make it's radius decrease if the spinning body experience friction?

kuruman
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Okay, but if the mass is subject to kinetic friction, the radius of it's rotation will decrease and the object will be spinning simultaneously. I understand, due to spin there would be tangential friction. But the object is moving to the center simultaneously too, does that mean it will experience a radial friction i.e friction along the radius because as you can see the hanging mass will make it's radius decrease if the spinning body experience friction?
You can answer that question yourself. See #15.

You can answer that question yourself. See #15.
I understood that, but for the problem I mentioned in the last comment, does my velocity has radial component? (I have given all the information I was given. This question was set by Bangladesh ministry of education as an assignment. Everyone knows the problem has no numerical solution but I have to point out the concept behind it and bring some new formulas out of it. Some of the teacher said a simplified version of this problem is available on [this book](https://salmanisaleh.files.wordpress.com/2019/02/fundamentals-of-physics-textbook.pdf) )

haruspex
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Okay, but if the mass is subject to kinetic friction, the radius of it's rotation will decrease and the object will be spinning simultaneously. I understand, due to spin there would be tangential friction. But the object is moving to the center simultaneously too, does that mean it will experience a radial friction i.e friction along the radius because as you can see the hanging mass will make it's radius decrease if the spinning body experience friction?
The radius is not necessarily decreasing. Maybe the initial condition includes a velocity component away from the hole, so the radius increases at first and decreases later. While the radius increases the friction will have a radial component towards the hole, at maximum radius only tangential friction, then radially outward friction as the radius decreases.

• nazmulhasanshipon
kuruman
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Here is a screenshot from the reference you gave. This version is fine because there is no friction and has an answer to the question being asked. Whoever added the complication of friction, did not modify the question being asked to take friction into account.

Edit: I remember puzzling over this problem as an undergraduate more than 50 years ago. It was in the 1964 edition of "Physics" by Halliday and Resnick. Last edited:
• nazmulhasanshipon and Delta2
Here is a screenshot from the reference you gave. This version is fine because there is no friction and has an answer to the question being asked. Whoever added the complication of friction, did not modify the question being asked to take friction into account.

Edit: I remember puzzling over this problem as an undergraduate more than 50 years ago. It was in the 1964 edition of "Physics" by Halliday and Resnick.

View attachment 286780
@kuruman where can I find it's solution, sir?

haruspex
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@kuruman where can I find it's solution, sir?
The solution to the frictionless version? Consider the tension and the balance of forces on each mass.

kuruman