Rotation of a Spherical Top

In summary: However, if you take the perspective that the sphere is free to rotate about its own center of mass, then there would be no external torque and the e3 axis would precess. I'm not sure why this axis would not precess in this case, but it seems to be the case.In summary, the e3 axis of the sphere does not precess while there is torque about e2 from the movement of the particle.
  • #1

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?
 
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  • #2
Angular momentum is the sum of what for this problem?
 
  • #3
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} ##
 
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  • #4
Write the angular moments in terms of their masses and R.
 
  • #5
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.
 
  • #6
... and θ (t)?
 
  • #7
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?
 
  • #8
Ben Johnson said:
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.
 
  • #9
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.
 
  • #10
Ben Johnson said:
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.
 

What is a spherical top?

A spherical top is a type of molecule or object that has a spherical shape and can rotate freely around its center of mass.

How does a spherical top rotate?

A spherical top rotates around its center of mass in three-dimensional space, similar to how a spinning top rotates around its tip.

Why is the rotation of a spherical top important in scientific research?

The rotation of a spherical top is important in scientific research because it can provide information about the structure and properties of the molecule or object. It can also help determine the physical constants and fundamental properties of the system.

What factors affect the rotation of a spherical top?

Factors that affect the rotation of a spherical top include its moment of inertia, angular velocity, and any external forces or torques acting on the object.

How is the rotation of a spherical top studied in the laboratory?

The rotation of a spherical top can be studied in the laboratory using techniques such as nuclear magnetic resonance (NMR) spectroscopy or microwave spectroscopy. These techniques involve applying external electromagnetic fields to the sample and measuring the resulting changes in its rotational behavior.

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