Change of the angular velocity vector

7 replies · 2K views · last reply
athrun200
Messages
275
Reaction score
0

Homework Statement


I don't know how to solve part(b).
See the photo Q3 for the question.


Homework Equations





The Attempt at a Solution


See the photo 2010Q3

I found the torque and hence the new angular momentum vector. But I wonder how could this help me to find the angle between the new and old angular velocity vector.
 
Attachments
  • Problem statement with text, followed by a diagram of a solid sphere of mass M and radius R with a point mass M on its surface, a z-axis through the sphere's center and the point mass, a y-axis, and the angular velocity vector omega at angle theta from the y-axis.
    Q3.jpg
    47.9 KB · Views: 536
  • Handwritten working on lined paper: matrices for the inertia tensors of a sphere and a point mass, their sum, the angular momentum L = I omega in components, and a free-body style sketch with vectors omega, r, mg and axes x, y, z with the torque cross product F x F.
    2010Q3.jpg
    14.9 KB · Views: 530
Your result for the moment of inertia tensor looks good to me.

In your expression for the torque, I don't see where the factor of 1/2 comes from when writing R/2. [EDIT: Otherwise, I think your expression for the final angular momentum is correct].

Are you familiar with Euler's equations for rigid body rotation? I think they will lead to the answer fairly quickly. If not, then you will need to relate the final angular momentum to the final angular velocity components using the (new) final components of the moment of inertia tensor in the fixed inertial frame.
 
Last edited:
Ha...That was a stupid mistake. That should be R not R/2.
 
I have learned Euler equation. But I am not sure what values to be put for those ω.
See my photo attached below
 
Attachments
  • Handwritten set of equations: three Euler equations for torque components Tx, Ty, Tz with moments of inertia Ix, Iy, Iz and angular velocities, followed by three substituted equations involving -Rmg sin θ, 7/5 MR² and 2/5 MR² terms.
    Euler.jpg
    20.6 KB · Views: 527
Maybe a sign error for the last term of your first equation.

Use these equations at time t = 0 to find the rate of change of each component of ω at time t = 0.
 
I used both method to solve this problem.
But I discovered that the new angular velocitys (only the x component) obtained from 2 methods are not the same.
Which version is wrong?
 
Attachments
  • Handwritten derivation on lined paper: torque equations numbered 1 to 3 with terms for R m g sin θ, I_x and ω_x, ω_y, ω_z, followed by initial conditions at t=0 and solved expressions for ω̇_z, ω̇_y and ω̇_x.
    method1.jpg
    25.2 KB · Views: 527
  • Handwritten physics derivation on lined paper: L' = I w', components of angular velocity w_x = -Rmg sinθ Δt / I_x, w_y = -ω sinθ, w_z = ω cosθ, then w·w' = |w||w'| cosα and cosα expressed in terms of ω, R, m, g, θ, Δt.
    method2.jpg
    19.5 KB · Views: 497
First version looks good.

For the second version, note that ##\tau = \dot L## holds only in an inertial frame. At the end of the small time interval the point mass will have rotated out of the yz plane in the fixed inertial frame. So, the moment of inertia tensor will no longer be diagonal in the inertial frame.