# Direction of torque vector tilting a spinning top?

1. Dec 18, 2013

1. The problem statement, all variables and given/known data

http://puu.sh/5Qrxt.jpg [Broken]

2. Relevant equations

τ = dL/dt

3. 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 in to 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

Last edited by a moderator: May 6, 2017
2. Dec 18, 2013

### tiny-tim

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

3. Dec 18, 2013

Right, I understand that solution, but I don't understand what the problem is with my reasoning that gives the answer in to 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.

Last edited: Dec 18, 2013
4. Dec 18, 2013

### tiny-tim

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​

5. Dec 18, 2013