Direction of torque vector tilting a spinning top?

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Homework Help Overview

The discussion revolves around understanding the direction of the torque vector in relation to the angular momentum of a spinning top. The original poster is grappling with the implications of the right-hand rule and the relationship between torque and angular momentum, particularly in the context of a rotating system.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to reconcile their understanding of torque direction with the right-hand rule, questioning why their reasoning leads to a different conclusion. They explore the implications of applying torque to a system with existing angular momentum.

Discussion Status

Participants are actively engaging with the original poster's reasoning, providing insights into the nature of torque and angular momentum. Some participants clarify that torque adds to the angular momentum vector rather than causing a rotation of the axis itself. There is an ongoing exploration of the concepts without a clear consensus.

Contextual Notes

The original poster expresses uncertainty about the formula τ = dL/dt, indicating a potential gap in their understanding of its application. The discussion includes references to specific diagrams and examples, such as the spinning basketball, which may not be fully detailed in the thread.

Esoremada
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Homework Statement



http://puu.sh/5Qrxt.jpg

Homework Equations



τ = dL/dt

The Attempt at a Solution



I understand the initial and final angular momentum vectors in the diagram, and the answer seems to make sense.

But when I initially solved it my logic was that since the axle itself is rotating clockwise, then the torque vector causing that would point into the page due to the right hand rule - why is this reasoning wrong?

To be honest I've never used τ = dL/dt before, it's on our formula sheet but I don't remember learning it
 
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Hi Esoremada! :smile:
Esoremada said:
… when I initially solved it my logic was that since the axle itself is rotating clockwise, then the torque vector causing that would point into the page due to the right hand rule …

the torque to cause a constant rotation is zero

(same as the force to cause a constant velocity is zero)

in the diagram, the vector L (the angular momentum) for the initial rotation is pointing into the page

the new L is the vector for the new angular momentum

the change in angular momentum, ∆L, is the difference, ie the short vector to the right

all the above is on the RHS of τ = dL/dt

the torque that the question asks for is on the LHS … obviously, it equals lim ∆L/∆t on the RHS :wink:
 
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tiny-tim said:
Hi Esoremada! :smile:the torque to cause a constant rotation is zero

(same as the force to cause a constant velocity is zero)

in the diagram, the vector L (the angular momentum) for the initial rotation is pointing into the page

the new L is the vector for the new angular momentum

the change in angular momentum, ∆L, is the difference, ie the short vector to the right

all the above is on the RHS of τ = dL/dt

the torque that the question asks for is on the LHS … obviously, it equals lim ∆L/∆t on the RHS :wink:

Right, I understand that solution, but I don't understand what the problem is with my reasoning that gives the answer into the page:

The entire system has no torque to begin with, just angular momentum
The axle itself is then tilted to the right (from the point of view of the diagram)
This tilt is a clockwise rotation about the centre of mass (again from the point of view of the diagram)
A clockwise rotation requires a clockwise torque, which points IN TO the page

What exactly is wrong with this reasoning?
Let me give another example. Here's a basketball: O

The basketball is in space spinning clockwise if you look down from above (clockwise about the y axis)

So looking straight on from my 'diagram' you see the centre moving to the left.

Now a clockwise torque is applied from this point of view to cause the basketball to accelerate in the clockwise direction (about the x axis). It seems like the increasing rotational speed from this torque would have no connection to the initial rotation about the y axis. Furthermore; applying the solution given to my initial question this basketball after a single full rotation would have had ZERO torque applied since the initial angular momentum vector is pointing straight down again.
 
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Esoremada said:
The entire system has no torque to begin with, just angular momentum
The axle itself is then tilted to the right (from the point of view of the diagram)
This tilt is a clockwise rotation about the centre of mass (again from the point of view of the diagram)
A clockwise rotation requires a clockwise torque, which points IN TO the page

… basketball …

that simply isn't the way angular momentum works

if you have a stationary horizontal disc, and you apply a vertical torque to the disc, it makes the disc rotate

but that isn't because it moves the axis (it doesn't), it's because it adds a vertical vector to the original angular momentum vector (which was zero) (and more torque just adds more to the vector, without changing its direction)

a torque does not rotate, it adds

a torque adds a vector to the angular momentum vector​
 
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tiny-tim said:
that simply isn't the way angular momentum works

if you have a stationary horizontal disc, and you apply a vertical torque to the disc, it makes the disc rotate

but that isn't because it moves the axis (it doesn't), it's because it adds a vertical vector to the original angular momentum vector (which was zero) (and more torque just adds more to the vector, without changing its direction)

a torque does not rotate, it adds

a torque adds a vector to the angular momentum vector​

Alright thanks! I've been thinking of torque too much in terms of physical movement rather than the vectors themselves, I think I get it now.
 

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