Precession of a Gyroscope — Question 9

Question 9

Holding the axis fixed requires a compensating torque

A horizontal gyroscope has spin angular momentum $5\hat x\,\mathrm{kg\,m^2/s}$ and gravity torque $2\hat y\,\mathrm{N\,m}$. An external holder keeps its axis fixed while preserving constant spin speed.

1. Find the torque exerted by the holder.

2. After removing the holding torque, find the initial rate of change of angular momentum and the slow-precession rate if a steady regime is established.

Question 9 - Solution

Holding the axis fixed requires a compensating torque

(a) Derivation and calculation

A fixed axis with constant spin has fixed angular momentum. Hence $$\frac{d\vec L}{dt}=\vec0=\vec\tau_g+\vec\tau_{\mathrm{holder}}.$$ Therefore $$\boxed{\vec\tau_{\mathrm{holder}}=-2\hat y\ \mathrm{N\,m}}.$$ The holder cancels the gravitational torque, not the nonzero spin momentum.

(b) Result and interpretation

Immediately after removal, the remaining torque gives $$\boxed{\frac{d\vec L}{dt}=2\hat y\ \mathrm{N\,m}}.$$ If a spin-dominated steady regime is established, $$\Omega\hat z\times5\hat x=2\hat y\Rightarrow \boxed{\vec\Omega=.400\hat z\ \mathrm{rad/s}}.$$ This steady estimate does not prescribe the immediate axis motion after arbitrary release; transient nutation can occur.

Physical check

The exact initial angular-momentum derivative follows from torque even when steady precession has not yet developed.