Chapter 10, Problem 50P

The rotor (flywheel) of a toy gyroscope has a mass of 0.140 kg. Its moment of inertia about its axis is 1.20 × 10⁻⁴ kg·m². The mass of the frame is 0.0250 kg. The gyroscope is supported on a single pivot (see Figure 10.67) with its center of mass a horizontal distance of 4.00 cm from the pivot. The gyroscope is precessing in a horizontal plane at the rate of 1 revolution in 2.20 s. (a) Find the upward force exerted by the pivot. (b) Find the angular speed with which the rotor is spinning about its axis, expressed in rev/min. (c) Copy the diagram and draw vectors to show the angular momentum of the rotor and the torque acting on it.

Figure 10.67

Problem 50.

10.50. Set Up: The precession angular velocity is $\Omega -\frac{wr}{I\omega }$. The total mass of the gyroscope is m_r + m_f = 0.140 kg + 0.0250 kg = 0.165 kg.

$\Omega =\frac{2\pi \text{rad}}{T}=\frac{2\pi \text{rad}}{2.20\text{s}}=2.856\text{rad/s}$

Solve: (a) F_p = w_tot = (0.165 kg)(9.80 m/s²) = 1.62 N

(b) $w=\frac{wr}{I\Omega }=\frac{(0.165\text{kg})(9.80{\text{m/s}}^2)(0.0400\text{m})}{(1.20\times {10}^{-4}\text{kg}\cdot {\text{m}}^{\text{2}})(2.856\text{rad/s})}=189\text{rad/s}=\text{1}\text{.80}\times {\text{10}}^{\text{3}}\text{rev/min}$

(c) If the figure in the problem is viewed from above, $\stackrel{\to }{\tau }$ is in the direction of the precession and $\stackrel{\to }{L}$ is along the axis of the rotor, away from the pivot.