Chapter 8, Problem 39P

(a)

To determine

To find: The angular acceleration of the arm.

Answer to Problem 39P

The angular acceleration of the arm is $72\text{rad/s}$ .

Explanation of Solution

Given information:

Assume that a $1.00\text{kg}$ ball is thrown solely by the action of the forearm, which rotates about the elbow joint under the action of the triceps muscle. The ball is accelerated uniformly from rest to $8.5\text{m/s}$ in $0.38\text{s}$ , at which point it is released.

Formula used:

  $\alpha =\frac{\omega -{\omega }_0}{\Delta t}$

Here

  $\alpha$ is the angular acceleration,

  $\omega$ is the final angular velocity,

  ${\omega }_0$ is the initial angular velocity,

  $\Delta t$ is the time.

Calculation:

Use kinematic equations in rotational dynamics to final the angular acceleration and then apply torque condition to find the required force of the triceps muscle.

The angular acceleration of the atom is given as follows

  $\alpha =\frac{\omega -{\omega }_0}{\Delta t}$

  $\alpha =\frac{v}{\left(\Delta t\right)r}$

Substituting the values as shown below,

  $\begin{array}{l}\alpha =\frac{\text{8}\text{.5}\text{m/s}}{\left(\text{0}\text{.38}\text{s}\right)\left(\text{0}\text{.31}\text{m}\right)}\\ \alpha =72{\text{rad/s}}^{\text{2}}\end{array}$

Therefore, the angular acceleration of the arm is $72\text{rad/s}$ .

Conclusion:

The angular acceleration of the arm is $72\text{rad/s}$ .

(b)

To determine

To find: The value of the force.

Answer to Problem 39P

The required force of the triceps muscle is $620\text{N}$ .

Explanation of Solution

Given information:

Assume that a ball is thrown solely by the action of the forearm, which rotates about the elbow joint under the action of the triceps muscle. The ball is accelerated uniformly from rest to $8.5\text{m/s}$ in $0.38\text{s}$ , at which $1.00\text{kg}$ high point it is released.

Formula used:

  $I=m_B{d}_B^2+\frac{1}{3}{m}_{A}d$

Here,

  $I$ is the moment of inertia,

  $m_A,m_B$ is the mass ,

  $d_A,d_B$ is the distance.

Calculation:

The moment of inertia of the ball-arm system to find the required force. The moment of inertia of the ball-arm system is given as follows:

  $I=m_B{d}_B^2+\frac{1}{3}{m}_{A}d$

Substituting the values as shown below,

  $\begin{array}{l}I=\left(1.0\text{kg}\right){\left(0.31\text{m}\right)}^2+\frac{1}{3}\left(3.7\text{kg}\right){\left(0.31\text{m}\right)}^2\\ I=0.215\text{kg}{\text{.m}}^{\text{2}}\end{array}$

Thus, the moment of inertia of the ball-arm system is $0.215\text{kg}\cdot {\text{m}}^{\text{2}}$ .

The torque acting on the system is given as follows:

  $\begin{array}{l}\tau =Fr\text{'}\sin 90°\\ I\alpha =Fr\text{'}\end{array}$

Rearrange the equation $I\alpha =Fr\text{'}\text{for}F$

  $F=\frac{I\alpha }{r\text{'}}$

Rearrange the equation for $F$

  $F=\frac{I\alpha }{r\text{'}}$

Substituting the values as shown below,

  $\begin{array}{l}F=\frac{\left(\text{0}\text{.215}\text{kg}\cdot {\text{m}}^{\text{2}}\right)\left(\text{72}{\text{rad/s}}^{\text{2}}\right)}{\text{0}\text{.025}\text{m}}\\ F=620\text{N}\end{array}$

Therefore, the required force of the triceps muscle is $620\text{N}$ .

Conclusion:

The required force of the triceps muscle is $620\text{N}$ .