Rotation of a Spherical Top

[Ben Johnson]

Homework Statement

A solid sphere of mass M and radius R rotates freely in space with an angular velocity ω about a fixed diameter. A particle of mass m, initially at one pole, moves with constant velocity v along a great circle of the sphere. Show that, when the particle has reached the other pole, the rotation of the sphere will have been retarded by an angle

α=ωT(1-√[2M/(2M+5m)])

Homework Equations

The Attempt at a Solution

Apologies for the ugly formatting, can someone please link me an article on how to type equations?
For the problem, I can find the solution if I set the sphere's rotation about the z axis and I assume that angular momentum about the z axis is constant. My question lies with the assumption that angular momentum about the z axis is constant. Angular momentum is constant if there is no external torque. However, the particle's movement from the top pole to the bottom pole implies that there is a torque.

What am I missing?

 

[Bystander]

Angular momentum is the sum of what for this problem?

 

[Ben Johnson]

Angular momentum of the sphere as it rotates about the z axis in the space frame and and a particle of mass m moves along the circumference of the sphere in the xz plane as this plane rotates in the body frame.

$L_z = I_z \omega_z$
$L_{e2} = I_{e2} \omega_{e2}$

 

[Bystander]

Write the angular moments in terms of their masses and R.

 

[Ben Johnson]

The moment of inertia about the e3 axis at t=0 is the moment of inertia of the sphere (since the particle lies on the e3 axis).

$I = \frac{2R^2M}{5}$

At time t the moment of inertia about the e3 axis is the moment of the sphere plus a contribution from the particle, a distance d from the z axis.

$I = \frac{2R^2M}{5} + d^2$
​

It can be shown that the distance d is given by
$d = mR^2 sin^2 \theta$
$\theta = \frac{vt}{R}$

Substituting into equation for I
$I = \frac{2R^2M}{5} + mR^2 sin^2 \theta$The e3 axis should precess, should it not? I can't see how the e3 axis remains parallel to the z axis after t=0.

 

[Bystander]

... and θ (t)?

 

[Ben Johnson]

By trigonometry,
$\theta (t) = \frac{vt}{R}$

I can solve the problem correctly if I assume that e3 does not precess, my question is why does this axis not precess while there is torque about e2 from the movement of the particle?

 

[Bystander]

mass m, initially at one pole, moves with constant velocity v

No torque. And, no, I'm not exactly happy with the appeal to the original problem statement myself.

 

[Ben Johnson]

Constant linear velocity v, constant angular velocity
$\omega_{e2} = \dot{\theta}(t)$
$\omega_{e2} = \frac{v}{R}$

In the body frame, the angular velocity introduces a centrifugal force which must be balanced by a centripetal force. The centrifugal force is given by
$F_{cf} = m ( \omega_{e2} \times r) \times \omega_{e2}$
$F_{cf} = m R \omega_{e2}^2 \hat{r}$

The balancing centripetal force is in the negative $\hat{r}$ direction
$F_{cp} = - m R \omega_{e2}^2 \hat{r}$

Taking torque to be
$\Gamma = r \times F$
$\Gamma_{e2} = 0$
since r and F both lie in the $\hat{r}$ direction.

Is this the reason torque is zero?
The reason I thought torque was not zero is somewhere in my notes I have written
$\Gamma = r \times \omega$
Using this logic,
$\Gamma_{e2} = R \hat{r} \times \omega \hat{e_2}$
which is a nonzero quanitity.

 

[TSny]

The e3 axis should precess, should it not? I can't see how the e3 axis remains parallel to the z axis after t=0.

Hello, Ben.

The problem states that "A solid sphere of mass M and radius R rotates freely in space with an angular velocity ω about a fixed diameter."

It seems to me that this could be interpreted to mean that the sphere is mounted on a fixed axle like a globe but is otherwise free to rotate about this axle. If so, then there would be an external torque on the axle to keep it always aligned parallel to the z axis. As you noted, in this case you will get the answer stated in the problem.