Question

Assume that a 1.00-kg ball is thrown solely by the action of the forearm, which rotates about the elbow joint under the action of the triceps muscle, Fig. 8–46. The ball is accelerated uniformly from rest to 8.5 m/s in 0.38 s, at which point is released. Calculate (a) the angular acceleration of the arm, and (b) the force required of the triceps muscle. Assume that the forearm has a mass of 3.7 kg and rotates like a uni- form rod about an axis at its end.

Answer 1

(a) The angular acceleration of the arm is approximately
`\( 22.37 \, \text{rad/s}^2 \).`

(b) The force required of the triceps muscle is approximately
`\( 76.74 \, \text{N} \).`

Step-by-step explanation:

(a) The angular acceleration
`(\( \alpha \))` can be calculated using the formula
`\( \alpha = (\Delta \omega)/(\Delta t) \)`, where
`\( \Delta \omega \)` is the change in angular velocity and
`\( \Delta t \)` is the change in time. In this case, the ball is released, so the angular velocity
`(\( \omega \))` becomes
`\( 0 \)` at the end. The initial angular velocity is
`\( 0 \)` because the ball starts from rest. Therefore,
`\( \Delta \omega = \omega_{\text{final}} - \omega_{\text{initial}} = 8.5 \, \text{rad/s} \).` Plugging in the values,
`\( \alpha = \frac{8.5 \, \text{rad/s}}{0.38 \, \text{s}} \)`, which gives
`\( \alpha \approx 22.37 \, \text{rad/s}^2 \)`.

(b) The force required
`(\( F \))` can be found using the equation
`\( \tau = I \alpha \)` , where
`\( \tau \)` is the torque,
`\( I \)` is the moment of inertia, and
`\( \alpha \)` is the angular acceleration. The moment of inertia for a uniform rod rotating about its end is
`\( I = (1)/(3)mL^2 \),` where
`\( m \)` is the mass and
`\( L \)` is the length. The torque can be expressed as
`\( \tau = rF \)`, where
`\( r \)` is the length from the axis to the point where the force is applied. In this case,
`\( r = L \)`. Substituting these into the torque equation, we get
`\( F = (I \alpha)/(r) \).` Plugging in the values,
`\( F = ((1)/(3) * 3.7 * (L)^2 * 22.37)/(L) \)`, which simplifies to
`\( F \approx 76.74 \, \text{N} \).`

In summary, (a) the angular acceleration of the arm is approximately
`\( 22.37 \, \text{rad/s}^2 \)`, and (b) the force required of the triceps muscle is approximately
`\( 76.74 \, \text{N} \).`