Amit & Deepali Sinha Foundation

Two Perpendicular Spins Compose Like Vectors

Irodov 1.55DifficultyUniqueness

Problem Irodov 1.55

Two rigid bodies rotate about two axes that are fixed in space, mutually perpendicular, and intersecting at one point. Body 1 rotates about its axis with constant angular velocity ω1=3.0\omega_1=3.0 rad/s, body 2 about its axis with constant angular velocity ω2=4.0\omega_2=4.0 rad/s. Find the modulus of the angular velocity of body 2 relative to body 1, and the modulus of the angular acceleration of body 2 relative to body 1.

The two given spins drawn at the point where the axes cross: ω_1 along the first axis and ω_2 along the second, perpendicular to it. Each arrow's length is in proportion to its stated modulus.

The two given spins drawn at the point where the axes cross: ω1\vect\omega_1 along the first axis and ω2\vect\omega_2 along the second, perpendicular to it. Each arrow’s length is in proportion to its stated modulus.

Given
  • ω1=3.0\omega_1=3.0 rad/s
  • ω2=4.0\omega_2=4.0 rad/s
  • axes fixed, perpendicular, intersecting
Find
∣ω2/1∣|\vect\omega_{2/1}|; ∣β2/1∣|\vect\beta_{2/1}|

Solution

Step 1Relative angular velocity, as a vector sum

Taking i\vect i along axis 1 and j\vect j along axis 2,

ω1=ω1i,ω2=ω2j,ω2/1=ω2−ω1=−ω1i+ω2j.\vect\omega_1 = \omega_1\vect i, \qquad \vect\omega_2 = \omega_2\vect j, \qquad \vect\omega_{2/1} = \vect\omega_2-\vect\omega_1 = -\omega_1\vect i+\omega_2\vect j.

Since i⊥j\vect i\perp\vect j, the modulus follows from the two perpendicular legs.

∣ω2/1∣=ω12+ω22|\vect\omega_{2/1}| = \sqrt{\omega_1^2+\omega_2^2}
ω=ω12+ω22=5.0 rad/s\omega = \sqrt{\omega_1^2+\omega_2^2} = 5.0\ \text{rad/s}
(1)
Step 2Relative angular acceleration, from the rate of turning of ω2\vect\omega_2

Body 1’s own axis direction does not change in body 1’s frame, so all of the relative angular acceleration comes from ω2\vect\omega_2 turning in that frame.

β2/1=(dω2dt)frame of body 1=−ω1×ω2=−ω1ω2(i×j)=−ω1ω2k.\vect\beta_{2/1} = \left(\frac{d\vect\omega_2}{dt}\right)_{\text{frame of body 1}} = -\vect\omega_1\times\vect\omega_2 = -\omega_1\omega_2(\vect i\times\vect j) = -\omega_1\omega_2\vect k.
β=ω1ω2=12 rad/s2\beta = \omega_1\omega_2 = 12\ \text{rad/s}^2
(2)

directed perpendicular to the plane containing the two axes.

The same two spins carrying both results. The heavier diagonal, from the tip of ω_1 to the tip of ω_2, is the difference ω_2/1. The crossed circle at right is β_2/1, pointing into the plane the axes span.

The same two spins carrying both results. The heavier diagonal, from the tip of ω1\vect\omega_1 to the tip of ω2\vect\omega_2, is the difference ω2/1\vect\omega_{2/1}. The crossed circle at right is β2/1\vect\beta_{2/1}, pointing into the plane the axes span.

Check

Dimensionally ω12+ω22\sqrt{\omega_1^2+\omega_2^2} is a rad/s and ω1ω2\omega_1\omega_2 a rad/s2^2, as required. And since ω1⊥ω2\vect\omega_1\perp\vect\omega_2 throughout the motion (both fixed in direction), ω2/1\vect\omega_{2/1} changes only through ω2\vect\omega_2’s apparent turning, never through a changing modulus, consistent with the vector β2/1\vect\beta_{2/1} coming out perpendicular to ω2/1\vect\omega_{2/1} itself: β2/1⋅ω2/1=−ω1ω2k⋅(−ω1i+ω2j)=0\vect\beta_{2/1}\cdot\vect\omega_{2/1} = -\omega_1\omega_2\vect k\cdot(-\omega_1\vect i+\omega_2\vect j) = 0.

Remark

The relative angular acceleration here has nothing to do with either body speeding up: both spins are constant. It exists purely because the two fixed axes are not parallel, so the relative vector ω2−ω1\vect\omega_2-\vect\omega_1 rotates in direction even though each piece is constant, the same reason a compass needle held fixed in the lab appears to swing if you rotate yourself around it.