The angle of tilt of a gyroscope

[Leo Liu]

Homework Statement

Below

Relevant Equations

$\tau=\dot{\vec L}$

Problem

Solution

My question
For the torque equation, I calculated the torque of the wheel at the point at the upper end of the string L' and wrote $\dot{\vec L_x}=\tau_x=-Mg(l+L\sin\beta) \text{[the direction of x is out of paper]}$ rather than the equation highlighted by green colour, which is measured at CM of the wheel. Since the wheel is rotating about the vertical axis (the auxiliary line) and is not spinning about the axis which passes through CM and is parallel to the auxiliary line, I don't understand why the solution's equation is valid. Could someone explain it and tell me whether my method works?

Thanks in advance :)

 

[haruspex]

The tension T acts on the wheel's axle at the point where it is attached. Your method effectively makes the axle, hence the lever arm, longer.

 

[Steve4Physics]

... I don't understand why the solution's equation is valid. Could someone explain it and tell me whether my method works?

EDITed

Problems with the question are:
The title is incorrect: the required angle ($\beta$) is not the angle of tilt of the gyroscope.
The symbols A and l are both used for the length of the axle.
The symbols B, L and L’ are all used for the length of the string (choosing ‘L’ is a poor choice anyway as it is the standard symbol for angular momentum).

I’ve changed my answer as I think I now realize what’s happening.

There are two torques acting on the axle/wheel causing precession around the central axis:
a) the torque produced by the wheel’s weight (acts clockwise on diagram);
b) the torque produced by the tension in the string (acts anticlockwise on diagram).

Their (vector) sum gives the net torque. You should be able to show that (with the allowed approximations) this is the same as $Mgl$. The author has simply used $Tl$ which is effectively the same because $Mg≈T$. This is then equated to the rate of change of angular momentum.

 

[Leo Liu]

Their (vector) sum gives the net torque. You should be able to show that (with the allowed approximations) this is the same as $Mgl$. The author has simply used $Tl$ which is effectively the same because $Mg≈T$. This is then equated to the rate of change of angular momentum.

Hi Steve, thank you for your reply.

User haruspex has pointed out that the length of my torque arm was incorrect, as the pivot point should be the tip of the axle as opposed to the point at which the cord is attached to the ceiling. However, I've just noticed that if I calculate the torque about the point on the ceiling, the length of the arm has to be x, and the torque, therefore, is -Mgx. So I don't think my approach is wrong. May I know what your view is?

 

[Steve4Physics]

User haruspex has pointed out that the length of my torque arm was incorrect, as the pivot point should be the tip of the axle as opposed to the point at which the cord is attached to the ceiling. However, I've just noticed that if I calculate the torque about the point on the ceiling, the length of the arm has to be x, and the torque, therefore, is -Mgx. So I don't think my approach is wrong. May I know what your view is?

I don't think @haruspex actually said that the length of your torque arm was incorrect, just that it was longer than the one in the model answer.

Also, no one has said "the pivot point should be the tip of the axle as opposed to the point at which the cord is attached to the ceiling". I guess you have (incorrectly) inferred this. (And I don't really like the term 'pivot point' in this context.)

In ths problem, I prefer to think about the perpendicular distance from the axis of rotation. So when calculating torque, I would measure distance from the centre of rotation and use the appropriate force component. Choose the 'pivot point' to be the centre of rotation.

The torque produced by $Mg$ is then indeed $Mgx$ (where $x = l+L\sin\beta$) as you say. I fully agree.

But I believe you need to also consider the torque produced by the tension. Tension does not act through the centre of rotation and has a vertical component. Therefore it produces an additional torque which contributes to the precession, See post #3.

 

[Leo Liu]

But I believe you need to also consider the torque produced by the tension. Tension does not act through the centre of rotation and has a vertical component.

But shouldn't the tension be a central force which produces no torque if I set the frame on that point?

 

[haruspex]

I don't think @haruspex actually said that the length of your torque arm was incorrect, just that it was longer than the one in the model answer.

Seems to me there are several ways to think about the problem.

For clarity, I'll label the centre of the gyroscope G, the point of attachment of the rod to the wire A, the point of attachment of wire to ceiling C, and the point below that in the line of the rod O.

We need to find the torque $\tau$ the rod exerts on the gyroscope. This determines the rate of precession of the gyroscope's spin axis. Given the torque, the horizontal distance GO is irrelevant. This is a key point to grasp, and I feel this is at the root of @Leo Liu's confusion.

The model answer considers the torque about G exerted on the rod by the tension T.
This is clearly $Tl\cos(\beta)=Mgl$. Since there must be no net torque on the rod, $\tau=Mgl$.

More simply, we could consider the balance about A, leading directly to $\tau=Mgl$.

Your approach considers the torque balance on the rod about O. This involves $\tau$, $Mg(l+L'\sin(\beta))$ and $T\cos(\beta)L'\sin(\beta))$, and results in the same equation as above.

 

[Steve4Physics]

But shouldn't the tension be a central force which produces no torque if I set the frame on that point?

This is not a simple system with a point mass on the end of a string. The tension from the string acts through the end of the rod, not through the rod & wheel's centre of mass.

EDIT: And of course the string is moving - tracing out a cone (just in case you hadn't realized this). END EDIT

The tension acting at the end of the rod has 2 components. The radial component $(Tsin\beta)$ is indeed a central force and does not contribute to the precessional torque. But the component of T parallel the axis of rotation $(Tcos\beta)$ does contribute to the precessional torque (in the same way that Mg contributes).

If it helps understanding, simply imagine the tension force acting on the end of the rod replaced by two completely separate forces: a horizontal force and a vertical force.

 

[Leo Liu]

For clarity, I'll label the centre of the gyroscope G, the point of attachment of the rod to the wire A, the point of attachment of wire to ceiling C, and the point below that in the line of the rod O.

the horizontal distance GO is irrelevant

I am sorry that I still don't quite understand what you meant because of my stupidity. Since the gyro is precessing about O instead of A, it makes more sense to me that distance GO is the arm.

The model answer considers the torque about G exerted on the rod by the tension T.

Doesn't this method only give information about the precession about CM instead of O?

Thank you.

 

[haruspex]

Since the gyro is precessing about O instead of A, it makes more sense to me that distance GO is the arm.

The gyroscope does not "know" where O is. It is a very common error in mechanics to treat components as experiencing forces that do not act directly on them.
All the gyroscope feels is the force of gravity, Mg, and the torque τ, lift Mg, and pull (centripetal force) from the rod; all the rod feels is the equal and opposite forces and torque from the gyroscope, and the force from the string.
Since we are ignoring the mass of the rod, the forces and torques on that must balance. Taking moments about the point where the string attaches to the rod:
$\tau=Mgl$.
We can also write:
$T\cos(\beta)=Mg$
$T\sin(\beta)=M(l+L'\sin(\beta))\Omega^2$.