Question 21.9: The motion of a football is observed using a slow-motion pro......
The motion of a football is observed using a slow-motion projector. From the film, the spin of the football is seen to be directed 30° from the horizontal, as shown in Fig. 21-24a. Also, the football is precessing about the vertical axis at a rate \dot{\phi} = 3 rad/s. If the ratio of the axial to transverse moments of inertia of the football is \frac{1}{3} , measured with respect to the center of mass, determine the magnitude of the football’s spin and its angular velocity. Neglect the effect of air resistance
Since the weight of the football is the only force acting, the motion is torque-free. In the conventional sense, if the z axis is established along the axis of spin and the Z axis along the precession axis, as shown in Fig. 21-24b, then the angle θ = 60°. Applying Eq. 21-37, the spin is
\dot{ψ} = \frac{I – I_{z}}{I_{z}} \dot{\phi} \cos θ = \frac{I – \frac{1}{3} I}{\frac{1}{3} I}(3) \cos 60°
= 3 rad/s
Using Eqs. 21-34, where He = \dot{\phi} (Eq. 21-36), we have
\theta = \text{constant }\\ \dot{\phi} =\frac{H_G}{I}\\\dot{\psi}=\frac{I-I_z}{II_z}H_G\cos\theta (21-36)
ω_{x} = 0
ω_{y} = \frac{H_{G}\sin θ}{I} = \frac{3I \sin 60°}{I} = 2.60 rad/s
ω_{z} = \frac{H_{G} \cos θ}{I_{z}} = \frac{3I \cos 60°}{\frac{1}{3} I} =4.50 rad/s
Thus,
ω= \sqrt{(ω_{x})² + (ω_{y})² + (ω_{z})² }
= \sqrt{(0)² + (2.60)² + (4.50)² }
= 5.20 rad/s