Which way will the conical object move?

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Homework Statement


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Homework Equations



v = ωr

The Attempt at a Solution




Honestly speaking I have very little idea about the problem . I am not understanding the setup clearly . What role does the rails play while the cones move on them .Are the cones fixed to the rails ? Does the tuning of the object -right or left , means rotation of the object clockwise –anticlockwise or it means shifting of the object as a whole towards left or right as it moves forward ?

I would really appreciate if somebody could help me with the problem .

Many Thanks
 

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  • #2
haruspex
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I agree the question is far from clear. I shall assume the rails are horizontal, that the picture is the plan view ( looking down from above), that the rails are not parallel, that the dashed line is perpendicular to CD, and that the roller rolls without slipping.
Suppose the roller goes straight. After rotating through some angle, consider how far its points of contact have moved.
 
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  • #3
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Hello ,

Thanks for replying.

I shall assume the rails are horizontal, that the picture is the plan view ( looking down from above), that the rails are not parallel, that the dashed line is perpendicular to CD, and that the roller rolls without slipping.
I agree .

Suppose the roller goes straight. After rotating through some angle, consider how far its points of contact have moved.
If 'r' is the radius of the cones and angle turned is θ , then the points of contacts have moved by ##rθ## .
 
  • #4
haruspex
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If 'r' is the radius of the cones and angle turned is θ , then the points of contacts have moved by ##rθ## .
But the radius where the cone touches rail AB is reducing, no?
 
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  • #5
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But the radius where the cone touches rail AB is reducing, no?
This is exactly where I am having trouble understanding ( the motion of the cone ) . So you are suggesting that the cones do not touch the floor ,instead they rest on the rails such that the the distance of the point , where the left cone touches the rail AB from mid point O , decreases as the object moves forward ??

But the radius where the cone touches rail AB is reducing, no?
So , now how should I use this fact to proceed further ?
 
  • #6
haruspex
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So , now how should I use this fact to proceed further ?
How does that affect the forward movement of that cone compared with the other (assuming both cones turn through the same angle, without slipping)?
 
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  • #7
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How does that affect the forward movement of that cone compared with the other (assuming both cones turn through the same angle, without slipping)?
The right cone moves faster as compared to left ??
 
  • #8
haruspex
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The right cone moves faster as compared to left ??
Yes.
 
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  • #9
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Is my understanding in first part of post#5 correct ?
 
  • #10
haruspex
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Is my understanding in first part of post#5 correct ?
Yes.
 
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  • #11
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Does that mean the object as a whole translates forward and also rotates anticlockwise (as seen from the top) ??
 
  • #12
haruspex
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Does that mean the object as a whole translates forward and also rotates anticlockwise ??
As viewed from above, yes.
 
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  • #13
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As viewed from above, yes.
and this is equivalent to object turning left i.e option (4) is correct ??
 
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  • #14
haruspex
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and this is equivalent to object turning left i.e option (4) is correct ??
Yes.
 
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  • #15
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Fantastic analysis of the problem :bow: .

Thanks a lot haruspex .
 
  • #16
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Hi ,

I still have some niggling doubts .

What role does friction play in this problem ?
 
  • #17
haruspex
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Hi ,

I still have some niggling doubts .

What role does friction play in this problem ?
You have to assume some friction or the system would be unstable right from the start. Do you see why? Draw a diagram in the vertical plane. If the double cone were to slide slightly to one side, what would happen to its mass centre... up or down or the same?
 
  • #18
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Please check this reasoning -

As the left cone moves forward , the radius decreases . As a result vL(speed of CM of left cone) > ωr . There is a tendency to slip forward and static friction acts backward . This increases ω .

Now , in case of right cone vR(speed of CM of left cone) < ωr . There is a tendency to slip backward and static friction acts forward.

The result of the two friction forces is to rotate the system anti - clockwise .

Does it make sense ?
 
  • #19
haruspex
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Please check this reasoning -

As the left cone moves forward , the radius decreases . As a result vL(speed of CM of left cone) > ωr . There is a tendency to slip forward and static friction acts backward . This increases ω .

Now , in case of right cone vR(speed of CM of left cone) < ωr . There is a tendency to slip backward and static friction acts forward.

The result of the two friction forces is to rotate the system anti - clockwise .

Does it make sense ?
Yes, I believe that works. An equivalent model is two wheels of different sizes fixed to the same shaft (and locked to it) on a frictional floor.
 
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  • #20
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Yes, I believe that works.
Ok .

But would you agree that this is essentially the same reasoning we had discussed earlier in the thread ( your proposed way of looking at the problem :smile:) . I mean in the earlier reasoning also , static friction was also involved although we didn't mention it ??
 
  • #21
haruspex
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Ok .

But would you agree that this is essentially the same reasoning we had discussed earlier in the thread ( your proposed way of looking at the problem :smile:) . I mean in the earlier reasoning also , static friction was also involved although we didn't mention it ??
Oh yes, it's the same reasoning in effect. I thought you were just looking for a simpler way to convey it.
 
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  • #22
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An equivalent model is two wheels of different sizes fixed to the same shaft (and locked to it) on a frictional floor.
But we need to assume that the system rolls without slipping ??
 
  • #23
haruspex
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But we need to assume that the system rolls without slipping ??
Well, not exactly. The trouble is that there has to be some slipping of a sort. Consider my asymmetric wheel model. Both wheels start off pointing N, say, but as they roll the larger wheel goes ahead, dragging the system around to point in a different direction. This means the wheels have slipped enough to change direction. It's like trying to turn a corner in a car without a differential fitted to the axle of the non-steering wheels. It wears out the tyres.

Edit: sorry, I messed up the analogy. I should have written:
It's like trying to turn a corner in a car without toe-in fitted to the steering mechanism. It wears out the tyres.
 
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  • #24
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Ok . But anyways a terrific analogy :smile:

Do you think there can be another way to look at this problem ? Can there be a reasoning based on energy concepts (specifically potential energy considerations ) ?
 
  • #25
haruspex
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Ok . But anyways a terrific analogy :smile:

Do you think there can be another way to look at this problem ? Can there be a reasoning based on energy concepts (specifically potential energy considerations ) ?
Perhaps, but that looks tough to me. Wouldn't you have to consider the work done against friction under the hypothesis of an arbitrary trajectory?
 

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