Chapter 8, Problem 42P

(a)

To determine

To Find: The value of angular acceleration.

Answer to Problem 42P

  $10.8{\text{rad/s}}^{\text{2}}$

Explanation of Solution

Given information:

A hammer thrower accelerates the hammer (mass $m=7.30\text{kg}$ ) from rest within four full turns (revolutions) and releases it at a speed of $28.0\text{m/s}$ .

Formula used:

  $\omega =\frac{v}{r}$

Here,

  $\omega$ is the angular velocity,

  $v$ is the linear velocity,

  $r$ is the radius.

Calculation:

Mass of the hammer, $m=7.30\text{kg}$

Number of rotations, $n=4$

Total angular displacement, $\theta =2\pi n$$=8\pi$

The hammer starts from rest, hence ${\omega }_0=0$

Let $\alpha$ be the angular acceleration.

Radius of the circular motion, $r=1.2\text{m}$

Speed with which hammer is released is $v=28\text{m/s}$ .

Angular speed with which hammer is released,

  $\omega =\frac{v}{r}$

  $\omega =\frac{\text{28}}{\text{1}\text{.2}}\text{rad/s}$

From third kinematic equation:

  ${\omega }^2={\omega }_0^2+2\alpha \theta$

  ${\omega }^2=0+2\alpha \theta$

  $\begin{array}{l}\alpha =\frac{{\left(\frac{28}{1.2}\right)}^2}{2\times 8\pi }{\text{rad/s}}^{\text{2}}\\ \alpha =10.8{\text{rad/s}}^{\text{2}}\end{array}$

Conclusion:

The value is $10.8{\text{rad/s}}^{\text{2}}$

(b)

To determine

To Find: The tangential acceleration

Answer to Problem 42P

The value is $a_T=12.96{\text{m/s}}^{\text{2}}$

Explanation of Solution

Given information:

A hammer thrower accelerates the hammer (mass $m=7.30\text{kg}$ ) from rest within four full turns (revolutions) and releases it at a speed of $28.0\text{m/s}$ .

Formula used:

  $a_T=\alpha r$

Here,

  $\alpha$ is the angular acceleration,

  $a_T$ is the tangential acceleration,

  $r$ is the radius.

Calculation:

The tangential acceleration $a_T=\alpha r$

  $\begin{array}{l}{a}_T=\left(10.8\times 1.2\right){\text{m/s}}^{\text{2}}\\ a_T=12.96{\text{m/s}}^{\text{2}}\end{array}$

Conclusion:

The value is $a_T=12.96{\text{m/s}}^{\text{2}}$ .

(c)

To determine

To Find: The centripetal acceleration

Answer to Problem 42P

The centripetal acceleration is

  $653.3{\text{m/s}}^{\text{2}}$ ..

Explanation of Solution

Given information:

A hammer thrower accelerates the hammer (mass $m=7.30\text{kg}$ ) from rest within four full turns (revolutions) and releases it at a speed of $28.0\text{m/s}$ .

Formula used:

  $a_C=\frac{v^2}{r}$

Here,

  $a_C$ is the centripetal acceleration,

  $v$ is the linear velocity,

  $r$ is the radius.

Calculation:

The centripetal acceleration $a_C=\frac{v^2}{r}$

  $\begin{array}{l}=\frac{{28}^2}{1.2}{\text{m/s}}^{\text{2}}\\ =653.3{\text{m/s}}^{\text{2}}\end{array}$

Conclusion:

The centripetal acceleration is $653.3{\text{m/s}}^{\text{2}}$

(d)

To determine

To Find: Net acceleration of the hammer

Answer to Problem 42P

The value is $4769.82\text{N}$ .

Explanation of Solution

Given information:

A hammer thrower accelerates the hammer (mass $m=7.30\text{kg}$ ) from rest within four full turns (revolutions) and releases it at a speed of $28.0\text{m/s}$ .

Formula used:

  $a=\sqrt{a_T^2+a_C^2}$

Here,

  $a_C$ is the centripetal acceleration,

  $a_T$ is the tangential acceleration,

  $a$ is the total acceleration.

Calculation:

The net acceleration of the hammer $a=\sqrt{a_T^2+a_C^2}$

  $\begin{array}{l}a=\sqrt{{12.96}^2+{653.3}^2}{\text{m/s}}^{\text{2}}\\ a=653.4{\text{m/s}}^{\text{2}}\end{array}$

Hence, net force exerted on hammer: $F=ma$

  $\begin{array}{l}=7.3\times 653.4\text{N}\\ =4769.82\text{N}\end{array}$

Conclusion:

The value is $4769.82\text{N}$ .

(e)

To determine

To find: The angle between the force and the radius.

Answer to Problem 42P

The angle between the force and the radius is ${\theta }^1={1.136}^0$ .

Explanation of Solution

Given information:

A hammer thrower accelerates the hammer (mass $m=7.30\text{kg}$ ) from rest within four full turns (revolutions) and releases it at a speed of $28.0\text{m/s}$ . Assuming a uniform rate of increase in angular velocity and a horizontal circular path of radius $1.20\text{m}$ .

Formula used:

  $\theta ={\tan }^{-1}\left(\frac{a_T}{a_C}\right)$

Here,

  $a_C$ is the centripetal acceleration,

  $a_T$ is the tangential acceleration,

  $\theta$ is the angle between force and radius.

Calculation:

Angle that the force makes with the radius $\theta ={\tan }^{-1}\left(\frac{a_{\Upsilon }}{a_C}\right)$

$\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{12.96}{653.3}\right)\\ \theta =1.136°\end{array}$

Conclusion:

The angle between the force and the radius is ${\theta }^1={1.136}^0$ .