Force exerted by rod on a mass moving in vertical circle

In summary, (A) and (B) are obviously wrong but I think both (C) and (D) are correct. At the top, the forces acting on the mass are tension and weight, both directed downwards so the equation of motion will be: $$\text{Tension}+\text{Weight}=m.a$$ $$\text{Tension}=m.a-\text{Weight}$$ Based on that equation, tension will be minimum at the top position. The direction of tension is always towards the center as the mass moves in a circle. At the horizontal position, tension will always be towards the center to provide the centripetal force. At the bottommost position, it is also directed towards the center since the net force of
  • #1
songoku
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Homework Statement
A mass connected to one end of a rigid rod rotates at constant speed in a vertical plane about the
other end of the rod (please see below). The force exerted by the rod on the mass is
A. zero everywhere.
B. constant in magnitude.
C. always directed towards the centre.
D. a minimum at the top of the circular path.
Relevant Equations
centripetal force
1632490273634.png


(A) and (B) are obviously wrong but I think both (C) and (D) are correct.

At the top, the forces acting on the mass are tension and weight, both directed downwards so the equation of motion will be:
$$\text{Tension}+\text{Weight}=m.a$$
$$\text{Tension}=m.a-\text{Weight}$$

Based on that equation, tension will be minimum at the top position

But the direction of tension is always towards the center as the mass moves in circle.

Am I missing something?

Thanks
 
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  • #2
To help decide if (C) is correct, draw a free body diagram for the ball when the rod is horizontal.
 
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  • #3
TSny said:
To help decide if (C) is correct, draw a free body diagram for the ball when the rod is horizontal.

Doesn't a rod always push the connected object as compared to a string which will always pull a connected object? Or is it that tension in a rod can be either push or pull depending on the scenario?

At horizontal position, tension will always be towards center to supply the centripetal force. At bottom most position also, it's directed towards the center since net force of centripetal force must point towards the center. But at the topmost position, tension could point towards or away from center since mg force is already pointing towards the center.
 
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  • #4
vcsharp2003 said:
Doesn't a rod always push the connected object as compared to a string which will always pull a connected object? Or is it that tension in a rod can be either push or pull depending on the scenario?
Consider the instant when the rod is vertical and the ball is at the top of the circle. Compare the case where the rod is rotating extremely slowly to the case where the rod is rotating extremely rapidly.

Would it be possible for the rod to rotate at a speed such that the rod is neither pushing or pulling on the ball when the ball is at the top?

What can you say about the direction of the net force that the rod exerts on the ball when the rod is horizontal?
 
  • #5
TSny said:
Would it be possible for the rod to rotate at a speed such that the rod is neither pushing or pulling on the ball when the ball is at the top?
Yes, if the value of mg equals the value of centripetal force, so no tension force is needed i.e mg force supplies the full centripetal force.

TSny said:
What can you say about the direction of the net force that the rod exerts on the ball when the rod is horizontal?
The net force would be the vector sum of vertically downward mg force and the horizontal radially inward tension force. So the direction of net force is going to be pointing downward and inward but at some acute angle to the vertical.

The free body diagram is easy for positions A,B and C, but at position D there are two possibilities since tension could point up or down as shown in diagram.
163249617786042953069448618665.jpg
 
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  • #6
vcsharp2003 said:
Yes, if the value of mg equals the value of centripetal force, so no tension force is needed i.e mg force supplies the full centripetal force.The net force would be the vector sum of vertically downward mg force and the horizontal radially inward tension force. So the direction of net force is going to be pointing downward and inward but at some acute angle to the vertical.
Are you sure that the rod's force on the ball would have a downward component rather than an upward component?

Anyway, I think we need to hold off until @songoku has a chance to answer. It's his homework problem.
 
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  • #7
songoku said:
Homework Statement:: A mass connected to one end of a rigid rod rotates at constant speed in a vertical plane about the
other end of the rod (please see below). The force exerted by the rod on the mass is
A. zero everywhere.
B. constant in magnitude.
C. always directed towards the centre.
D. a minimum at the top of the circular path.
Relevant Equations:: centripetal force

View attachment 289635

(A) and (B) are obviously wrong but I think both (C) and (D) are correct.

At the top, the forces acting on the mass are tension and weight, both directed downwards so the equation of motion will be:
$$\text{Tension}+\text{Weight}=m.a$$
$$\text{Tension}=m.a-\text{Weight}$$

Based on that equation, tension will be minimum at the top position

But the direction of tension is always towards the center as the mass moves in circle.

Am I missing something?

Thanks
Are you sure that the mass is rotating at a constant speed? If it were true then the law of conservation of energy would be violated for bottom and top positions. According to me, it should be rotating with a varying speed, and as mass rises its speed must decrease in order to satisfy the law of conservation of energy. So, speed at bottom position should be higher than speed at top position.
 
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  • #8
vcsharp2003 said:
Are you sure that the mass is rotating at a constant speed? If it were true then the law of conservation of energy would be violated for bottom and top positions.
We can assume that an outside agent forces the rod to rotate at constant angular speed.
 
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  • #9
TSny said:
We can assume that an outside agent forces the rod to rotate at constant speed.
Ok. Then the +ve work done by external agent on the mass rod system, as the mass rises will provide for extra energy at top, i.e. ##W + KE_b = mg(2R) + KE_t## where KE at bottom and top are equal.
 
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  • #10
TSny said:
To help decide if (C) is correct, draw a free body diagram for the ball when the rod is horizontal.
Let say the mass is at position of 3 o'clock. The force acting on the mass will be tension to the left (center of circle) and weight downwards. Is this correct?

Thanks
 
  • #11
songoku said:
Let say the mass is at position of 3 o'clock. The force acting on the mass will be tension to the left (center of circle) and weight downwards. Is this correct?

Thanks
It is not quite correct. The rod does not exert only a tension force to the left.

The mass is moving at constant speed in a circle. So, when the mass is at 3 o'clock, what must be the direction of the net force on the mass?

Use a free-body diagram of the mass when it is at 3 o'clock to deduce the directions of the vertical and horizontal components of the force that the rod exerts on the mass. (The question is concerned with the force exerted by the rod on the mass.)
 
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  • #12
songoku said:
Let say the mass is at position of 3 o'clock. The force acting on the mass will be tension to the left (center of circle) and weight downwards. Is this correct?

Thanks
Yes.
And generally, you can think of the rod as having to provide two forces, one associated with gravity and one associated with the rotation.
Consider how the magnitudes and directions of these change over time, and what that tells you about the magnitude and direction of the sum.
 
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  • #13
TSny said:
It is not quite correct. The rod does not exert only a tension force to the left.

The mass is moving at constant speed in a circle. So, when the mass is at 3 o'clock, what must be the direction of the net force on the mass?
haruspex said:
Yes.
And generally, you can think of the rod as having to provide two forces, one associated with gravity and one associated with the rotation.
Consider how the magnitudes and directions of these change over time, and what that tells you about the magnitude and direction of the sum.

Ah I see. Actually I have another question that I want to ask after I understand about this question. My next question will be related to weight of the mass, what effect it would have on the motion. It turns out that both hints also cover that question.

There should be upward force acting on the mass to balance weight so the net vertical force on the mass is zero to assure the mass moves with constant speed. This is provided by the rod so the force on the mass by the rod will be to the left and upwards and the resultant force by the rod will be slanted to the top left (option (C) is wrong)

Would it be the same if the rod is changed to string? I always think that if the object moving in vertical circle, the tension will always only be directed to center of circle. If the object moves with constant speed, it means that the string must also provide upwards force to counter the weight, yes?

The possibility for string to only provides force directed to center if the object moves with varying speed. Is this correct?

Thanks
 
  • #14
songoku said:
If the object moves with constant speed, it means that the string must also provide upwards force to counter the weight, yes?
... and since string cannot provide a force other than along itself, the speed cannot be constant.
 
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  • #15
haruspex said:
... and since string cannot provide a force other than along itself, the speed cannot be constant.
So treatment for string and rod are is not the same. Sorry but I don't understand the reason. Why rod can provide upwards force while string can not?

Thanks
 
  • #16
songoku said:
So treatment for string and rod are is not the same. Sorry but I don't understand the reason. Why rod can provide upwards force while string can not?

Thanks
The string can provide vertical force while it is itself vertical, but it cannot provide a tangential force.
Can you push a ball along the ground with a vertical stick?
Can you push a ball along the ground with a vertical string?
 
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  • #17
Thank you very much for the help and explanation TSny, vcsharp2003, haruspex
 
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  • #18
haruspex said:
The string can provide vertical force while it is itself vertical, but it cannot provide a tangential force.
Can you push a ball along the ground with a vertical stick?
Can you push a ball along the ground with a vertical string?

In topmost position, the rod would exert an upwards push force on the ball; it couldn't pull the ball downwards at this position. Does that sound right?
 
  • #19
vcsharp2003 said:
In topmost position, the rod would exert an upwards push force on the ball; it couldn't pull the ball downwards at this position. Does that sound right?
It depends how fast the rotation is. Which is larger, g or ##r\omega^2##?
 
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  • #20
haruspex said:
It depends how fast the rotation is. Which is larger, g or ##r\omega^2##?
That's difficult to say. I know g is acceleration if an object is in free fall. My guess argument is as follows: here the object is restrained so it cannot be in free fall and hence its centripetal acceleration is less than g.
 
  • #21
haruspex said:
It depends how fast the rotation is. Which is larger, g or ##r\omega^2##?
If centripetal acceleration > g then T will be downwards. ##(mg + T = mr\omega^2)##
If centripetal acceleration < g then T will be upwards. ##(mg - T = mr\omega^2)##
If centripetal acceleration = g then T will be zero and ball is in free fall. ##(mg = mr\omega^2)##
 
  • #22
vcsharp2003 said:
If centripetal acceleration > g then T will be downwards. ##(mg + T = mr\omega^2)##
If centripetal acceleration < g then T will be upwards. ##(mg - T = mr\omega^2)##
If centripetal acceleration = g then T will be zero and ball is in free fall. ##(mg = mr\omega^2)##
Right. Though my question in post #19 was rhetorical.
 
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  • #23
haruspex said:
Right. Though my question in post #19 was rhetorical.

I am trying to reason why in the horizontal position the net force needs to be radially inward.

If the ##mg## force acting downwards did not get balanced then there would a downward vertical component of net force on the ball, which would then produce a a tangential acceleration causing it's speed to increase. We know the speed of ball is constant at all positions and therefore there should be zero tangential acceleration at all positions including the horizontal position. In other words, the net force should always be in a radial inward direction at all positions of the ball.
 
  • #24
vcsharp2003 said:
I am trying to reason why in the horizontal position the net force needs to be radially inward.
... We know the speed of ball is constant at all positions and therefore there should be zero tangential acceleration at all positions including the horizontal position. In other words, the net force should always be in a radial inward direction at all positions of the ball.
Either you analyze the forces from the point of reference of the mass or the bar, you need to treat the problem as a vectors addition.
One vector is vertical and constant.
The other vector has inward radial direction and its magnitude depends on the angular velocity.

Your post #5 shows the free body diagram of the forces acting on the mass.
The magnitude of T could be very small for the case of a Ferris whell, the main force acting on the hanging basket would be the persistently mg.
The magnitude of T could be very high for the case of an airplane propeller.

Still, for both cases, the resultant net force at any position of the circular trajectory, except top and bottom points, could never be purely vertical or radial.
 
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  • #25
vcsharp2003 said:
I am trying to reason why in the horizontal position the net force needs to be radially inward.

If the ##mg## force acting downwards did not get balanced then there would a downward vertical component of net force on the ball, which would then produce a a tangential acceleration causing it's speed to increase.
Well, imo, it is impossible for the tangent velocity to be constant unless a motor is applied to the system.that being said, we can ignore the tangent acceleration since the motor will keep the velocity constant.
 
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  • #26
Lnewqban said:
Either you analyze the forces from the point of reference of the mass or the bar, you need to treat the problem as a vectors addition.
One vector is vertical and constant.
The other vector has inward radial direction and its magnitude depends on the angular velocity.

Your post #5 shows the free body diagram of the forces acting on the mass.
The magnitude of T could be very small for the case of a Ferris whell, the main force acting on the hanging basket would be the persistently mg.
The magnitude of T could be very high for the case of an airplane propeller.

Still, for both cases, the resultant net force at any position of the circular trajectory, except top and bottom points, could never be purely vertical or radial.
But speed of ball in vertical motion is constant, so tangential acceleration needs to be zero at every point of its path. The rod, unlike a string, can exert an upward force on ball when in a horizontal position in addition to a radial force from rod on the ball.
 
  • #27
Father_Ing said:
Well, imo, it is impossible for the tangent velocity to be constant unless a motor is applied to the system.that being said, we can ignore the tangent acceleration since the motor will keep the velocity constant.
Shouldn't tangential acceleration be zero as explained in my previous post?
 
  • #28
The problem is asking about the force exerted by the rod on the mass.
That total or net force changes as the position of the mass along the circular trajectory changes.
That force is the only thing preventing the mass from following its natural trajectory: a vertical one which would be accelerated downwards by the action of gravity.

Sorry to say I see no contradiction respect to a rotation that may be or not angularly accelerated.
Let's add certain angular acceleration to the rod, in such a way that the magnitude of the tangential acceleration becomes 2g: how would that change things?

If, in a different assumption, we make the value of the circle's radius infinite, we end up with a classic elevator problem and we could compare tension of the cable and weight during accelerated and non-accelerated vertical movements.

If, again, we tilt the plane of rotation until making it perfectly horizontal, the rod simultaneously will force the mass to rotate and to not fall or to keep the same height: two forces, one rotating horizontally and another moving vertically, that become one net force (not purely vertical or horizontal).
 
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  • #29
vcsharp2003 said:
I am trying to reason why in the horizontal position the net force needs to be radially inward.
Not sure what you mean by "need" there.
We know it is radially inward because we are told the mass moves in a circle at uniform speed. Presumably this is arranged by a varying torque applied at the axle on which the rod rotates.
We do not have to worry about what provides that torque, but an easy way would be that the rod extends the same distance the other side of the axle and carries an identical counterbalancing mass.
 
  • #30
vcsharp2003 said:
But speed of ball in vertical motion is constant, so tangential acceleration needs to be zero at every point of its path. The rod, unlike a string, can exert an upward force on ball.
Is this an action-reaction force? Which means there will be a force downward (with the same in magnitude as the force you have stated) that applies on the rod?
 
  • #31
Father_Ing said:
Is this an action-reaction force? Which means there will be a force downward (with the same in magnitude as the force you have stated) that applies on the rod?
Yes. All real forces are "action-reaction" forces. If the rod pushes up on the mass, the mass pushes down on the rod. That is Newton's third law.
 
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  • #32
haruspex said:
Not sure what you mean by "need" there.
We know it is radially inward because we are told the mass moves in a circle at uniform speed. Presumably this is arranged by a varying torque applied at the axle on which the rod rotates.
Is it not possible for the end of the rod to be the one which keep the speed constant? As what @vcsharp2003 is trying to say.
 
  • #33
Rikudo said:
Is it not possible for the end of the rod to be the one which keep the speed constant? As what @vcsharp2003 is trying to say.
I understand that question even less.
Some mechanism acts to keep the speed constant. We do not know and do not care what it is. We can deduce the force it must be exerting on the mass to achieve that, because we know that the net result of that force and mg is a radially inward force providing the centripetal acceleration.
 
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  • #34
haruspex said:
Not sure what you mean by "need" there
I was just trying to decide whether ##a_t>0## or ##a_t=0##, where ##a_t## stands for tangential acceleration in horizontal position. I wasn't trying to understand why speed is constant, which you've explained through a varying torque mechanism.
 
  • #35
vcsharp2003 said:
I was just trying to decide whether ##a_t>0## or ##a_t=0##, where ##a_t## stands for tangential acceleration in horizontal position. I wasn't trying to understand why speed is constant, which you've explained through a varying torque mechanism.
Tangential acceleration is the component of acceleration parallel to the velocity. If speed is constant, that must be zero.
(In vectors, ##0=\frac{d}{dt}v^2=\frac{d}{dt}\vec v^2=2\vec v.\dot{\vec v}=2||\vec v|| a_t##.)
 
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