Question 8.13: WARMING UP GOAL Find a moment of inertia and apply the rotat...
WARMING UP
GOAL Find a moment of inertia and apply the rotational analog of Newton’s second law.
PROBLEM A baseball player loosening up his arm before a game tosses a 0.150-\mathrm{kg} baseball, using only the rotation of his forearm to accelerate the ball (Fig. 8.27). The forearm has a mass of 1.50 \mathrm{~kg} and the length from the elbow to the ball’s center is 0.350 \mathrm{~m}. The ball starts at rest and is released with a speed of 30.0 \mathrm{~m} / \mathrm{s} in 0.300 \mathrm{~s}. (a) Find the constant angular acceleration of the arm and ball. (b) Calculate the moment of inertia of the system consisting of the forearm and ball. (c) Find the torque exerted on the system that results in the angular acceleration found in part (a).
STRATEGY The angular acceleration can be found with rotational kinematic equations, while the moment of inertia of the system can be obtained by summing the separate moments of inertia of the ball and forearm. The ball is treated as a point particle. Multiplying these two results together gives the torque.
(a) Find the angular acceleration of the ball.
The angular acceleration is constant, so use the angular velocity kinematic equation with \omega_{i}=0 :
\omega=\omega_{i}+\alpha t \quad \rightarrow \quad \alpha=\frac{\omega}{t}
The ball accelerates along a circular arc with radius given by the length of the forearm. Solve v=r \omega for \omega and substitute:
\alpha=\frac{\omega}{t}=\frac{v}{r t}=\frac{30.0 \mathrm{~m} / \mathrm{s}}{(0.350 \mathrm{~m})(0.300 \mathrm{~s})}=286 \mathrm{rad} / \mathrm{s}^{2}
(b) Find the moment of inertia of the system (forearm plus ball).
Find the moment of inertia of the ball about an axis that passes through the elbow, perpendicular to the arm:
I_{\text {ball }}=m r^{2}=(0.150 \mathrm{~kg})(0.350 \mathrm{~m})^{2}=1.84 \times 10^{2} \mathrm{~kg} \cdot \mathrm{m}^{2}
Obtain the moment of inertia of the forearm, modeled as a rod rotating about an axis through one end, by consulting Table 8.1:
\begin{aligned}I_{\text {forearm }} &=\frac{1}{3} M L^{2}=\frac{1}{3}(1.50 \mathrm{~kg})(0.350 \mathrm{~m})^{2} \\&=6.13 \times 10^{2} \mathrm{~kg} \cdot \mathrm{m}^{2}\end{aligned}
Sum the individual moments of inertia to obtain the moment of inertia of the system (ball plus forearm):
I_{\text {system }}=I_{\text {ball }}+I_{\text {forearm }}=7.97 \times 10^{2} \mathrm{~kg} \cdot \mathrm{m}^{2}
(c) Find the torque exerted on the system.
Apply Equation 8.13,
\sum \tau = I \alpha [8.13]
using the results of parts (a) and (b):
\begin{aligned}\tau &=I_{\text {system }} \alpha=\left(7.97 \times 10^{2} \mathrm{~kg} \cdot \mathrm{m}^{2}\right)\left(286 \mathrm{rad} / \mathrm{s}^{2}\right) \\&=22.8 \mathrm{~N} \cdot \mathrm{m}\end{aligned}
REMARKS Notice that having a long forearm can greatly increase the torque and hence the acceleration of the ball. This is one reason it’s advantageous for a pitcher to be tall: the pitching arm is proportionately longer. A similar advantage holds in tennis, where taller players can usually deliver faster serves.