Chapter 8, Problem 53P

(a)

To determine

The angular speed of discuss at a moment just before the release.

Answer to Problem 53P

The angular speed is $13\text{rad}/\text{s}$.

Explanation of Solution

Number of revolutions are $1.5$ , time taken is $1.4\text{s}$, radius of circular path is $0.90\text{m}$, and the mass of discuss is $2.0\text{kg}$.

Write the equation for average angular velocity and the angular displacement.

${\omega }_{av}=\frac{\Delta \theta }{\Delta t}$

Here, ${\omega }_{av}$ is the average angular velocity, $\Delta \theta$ is the displacement, and $\Delta t$ is the time.

Write the equation to find ${\omega }_{av}$.

$\alpha$

Here,${\omega }_i$ is the initial angular speed and ${\omega }_f$ is the final angular speed.

Equate the right hand sides of above two equations.

$\frac{\Delta \theta }{\Delta t}=\frac{{\omega }_i+{\omega }_f}{2}$

Conclusion:

Substitute $1.5\text{rev}$ for $\Delta \theta$, $0\text{rad}/\text{s}$ for ${\omega }_i$, and $1.4\text{s}$ for $\Delta t$ in the above equation to find ${\omega }_f$.

$\begin{array}{c}\frac{\left(1.5\text{rev}\left(\frac{2\pi \text{rad}/\text{rev}}{1\text{rev}}\right)\right)}{1.4\text{s}}=\frac{0\text{rad}/\text{s}+{\omega }_f}{2}\\ {\omega }_f=13\text{rad}/\text{s}\end{array}$

Therefore, the angular speed is $13\text{rad}/\text{s}$.

(b)

To determine

The torque that must be applied on discuss.

Answer to Problem 53P

The torque is $16\text{N}/\text{m}$.

Explanation of Solution

Number of revolutions are $1.5$ , time taken is $1.4\text{s}$, radius of circular path is $0.90\text{m}$, and the mass of discuss is $2.0\text{kg}$.

Consider the discus as a point object.

Write the equation for torque.

$\tau =I\alpha$ (I)

Here, $\tau$ is the torque, $I$ is the moment of inertia, and $\alpha$ is the angular velocity.

Write the equation for $I$.

$I=M{R}^2$ (II)

Here, $M$ is the mass of discus and $R$ is the

Write the equation for $\alpha$.

$\alpha =\frac{{\omega }_f-{\omega }_i}{\Delta t}$ (III)

Rewrite equation (I) by substituting equations (II) and (III).

$\tau =\left(M{R}^2\right)\left(\frac{{\omega }_f-{\omega }_i}{\Delta t}\right)$

Write the equation for ${\omega }_f-{\omega }_i$.

${\omega }_f-{\omega }_i=\frac{2\Delta \theta }{\Delta t}$

Rewrite the previous equation for $\tau$ by substituting the above equation for ${\omega }_f-{\omega }_i$.

$\begin{array}{c}\tau =\left(M{R}^2\right)\left(\frac{2\Delta \theta /\Delta t}{\Delta t}\right)\\ =\frac{2M{R}^2\Delta \theta }{{\left(\Delta t\right)}^2}\end{array}$

Conclusion:

Substitute $2.0\text{kg}$ for $M$, $0.90\text{m}$ for $R$, $1.5\text{rev}$ for $\Delta \theta$, and $1.4\text{s}$ for $\Delta t$ in the above equation to find ${\omega }_f$.

$\begin{array}{c}\tau =\frac{2\left(2.0\text{kg}\right){\left(0.90\text{m}\right)}^2\left(1.5\text{rev}\left(\frac{2\pi \text{rad}/\text{rev}}{1\text{rev}}\right)\right)}{{\left(1.4\text{s}\right)}^2}\\ =\frac{31.36}{1.96}\text{N}\cdot \text{m}\\ =16\text{N}\cdot \text{m}\end{array}$

Therefore, the torque is $16\text{N}/\text{m}$.

(c)

To determine

The distance from athlete to the landing point of discus.

Answer to Problem 53P

The distance is $15\text{m}$.

Explanation of Solution

Number of revolutions is $1.5$ , time taken is $1.4\text{s}$, radius of circular path is $0.90\text{m}$, the mass of discuss is $2.0\text{kg}$, and the angle of projection is $45°$ with respect to the horizontal.

The discus will execute a parabolic path. Write the equation to find the initial velocity of discus.

$v_i=R\left(\frac{2\Delta \theta }{\Delta t}\right)$

Here, $v_i$ is the initial velocity of discus.

Write the equation to find the distance to landing place from the athlete.

$\Delta x=\frac{2v_i^2\sin \theta \cos \theta }{g}$

Here, $\Delta x$ is the distance to landing place from the athlete, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity.

Conclusion:

Substitute $0.90\text{m}$ for $R$, $1.5\text{rev}$ for $\Delta \theta$, and $1.4\text{s}$ for $\Delta t$ in the above equation to find $v_i$.

$\begin{array}{c}{v}_i=\left(0.90\text{m}\right)\left(\frac{2\left(1.5\text{rev}\left(\frac{2\pi \text{rad}/\text{rev}}{1\text{rev}}\right)\right)}{1.4\text{s}}\right)\\ =\left(0.90\text{m}\right)\left(13.33{\text{s}}^{-1}\right)\\ =12\text{m}/\text{s}\end{array}$

Substitute $12\text{m}/\text{s}$ for $v_i$, $45°$ for $\theta$, and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above equation to find $\Delta x$.

$\begin{array}{c}\Delta x=\frac{2{\left(12\text{m}/\text{s}\right)}^2\left(\sin 45°\right)\left(\cos 45°\right)}{9.8\text{m}/{\text{s}}^{\text{2}}}\\ =\frac{143.96{\text{m}}^2/{\text{s}}^2}{9.8\text{m}/{\text{s}}^{\text{2}}}\\ =15\text{m}\end{array}$

Therefore, the distance is $15\text{m}$.