Chapter 10, Problem 74P

(a)

To determine

To Calculate:

The horizental component of angular momentum and the vertical componeent of angular momentum.

Answer to Problem 74P

  $\begin{array}{l}{L}_x=\left[\left(5.36\right)\left(\cos \omega t\widehat{i}+\sin \omega t\widehat{j}\right)\right]J\cdot s\\ L_y=\left(3.09\widehat{k}\right)J\cdot s\end{array}$

Explanation of Solution

Given data:

Mass of the ball is $m$

Formula Used:

Angular momentum: $\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$

Where, $\overrightarrow{p}$ is the linear momentum and $\overrightarrow{r}$ is the distance to rotational axis.

Newton’s second law of motion:

  $\sum F=ma$

Where, m is the mass and a is the acceleration.

  $$

Calculation:

  

By the expression of the angular momentum of the ball about the point of support as

  $\begin{array}{c}\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}\\ =\overrightarrow{r}\times m\overrightarrow{v}\\ =m\left(\overrightarrow{r}\times \overrightarrow{v}\right)\mathrm{.........}\left(1\right)\end{array}$

Then, by applying Newton’s second law to the ball to get

  $F\left(x\right)=T\sin \theta =m\frac{v^2}{r\sin \theta }\mathrm{.........}\left(2\right)$

And $F\left(z\right)=T\cos \theta -mg=0\mathrm{.........}\left(3\right)$

Now, eliminate $T$ fromequation (3):

  $T=\frac{mg}{\cos \theta }$

By substituting the $T$ value in equation (2).

  $\left(\frac{mg}{\cos \theta }\right)\sin \theta =m\frac{v^2}{r\sin \theta }$

Then solve for the velocity of the ball

  $v=\sqrt{rg\sin \theta \tan \theta }$

Substitute numerical values to get

  $\begin{array}{c}v=\sqrt{rg\sin \theta \tan \theta }\\ =\sqrt{\left(\text{1}\text{.5}\right)\left(9.8\right)\sin {30}^{\text{o}}\tan {30}^{\text{o}}}\\ =2.06\text{m/s}\end{array}$

Express the position vector of the ball as

  $\overrightarrow{r}=\left(1.5\text{m}\right)\sin {30}^{\text{o}}\left(\cos \omega t\widehat{i}+\sin \omega t\widehat{j}\right)-\left(1.5\text{m}\right)\cos {30}^{\text{o}}\widehat{k}$

Find the velocity of the ball:

  $\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=\left(0.75\omega \right)\left(-\sin \omega t\widehat{i}+\cos \omega t\widehat{j}\right)$

Evaluate for $\omega$ :

  $\begin{array}{c}\omega =\frac{2.06\text{m/s}}{\left(\text{1}\text{.5}\text{m}\right)\sin {30}^{\text{o}}}\\ =2.75\text{rad/s}\end{array}$

Substitute for $\omega$ to obtain

  $\overrightarrow{v}=\left(2.06\text{m/s}\right)\left(-\sin \omega t\widehat{i}+\cos \omega t\widehat{j}\right)$

Substitute into equation $\left(1\right)$ and evaluate $L$ to get

  $\begin{array}{c}L=\left(2\text{kg}\right)\left[\left(\text{1}\text{.5}\text{m}\right){\sin }^{\text{o}}\left(\cos \omega t\widehat{i}+\sin \omega t\widehat{j}\right)-\left(1.5\text{m}\right)\cos {30}^{\text{o}}\widehat{k}\right]\times \left(2.06\text{m/s}\right)\left(-\sin \omega \widehat{t}+\cos \omega t\widehat{j}\right)\\ =\left[\left(5.36\right)\left(\cos \omega t\widehat{i}+\sin \omega t\widehat{j}\right)+3.09\widehat{k}\right]\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}\end{array}$

Therefore, the horizontal component of angular momentum $\overrightarrow{L}$ is

  $L_x=\left[\left(5.36\right)\left(\cos \omega t\widehat{i}+\sin \omega t\widehat{j}\right)\right]\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}$

And the vertical component of angular momentum $\overrightarrow{L}$ is

  $L_y=\left(3.09\widehat{k}\right)\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}$

Conclusion:

  $\begin{array}{l}{L}_x=\left[\left(5.36\right)\left(\cos \omega t\widehat{i}+\sin \omega t\widehat{j}\right)\right]\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}\\ L_y=\left(3.09\widehat{k}\right)\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}\end{array}$

(b)

To determine

To Calculate:The magnitude of the torque exerted by gravity about the point of support.

Answer to Problem 74P

  $14.7\text{ N}\cdot \text{m}$

Explanation of Solution

Given data:

From the previous part:

  $\begin{array}{l}{L}_x=\left[\left(5.36\right)\left(\cos \omega t\widehat{i}+\sin \omega t\widehat{j}\right)\right]\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}\\ L_y=\left(3.09\widehat{k}\right)\text{kg}\cdot {\text{m}}^{\text{2}}\text{/s}\end{array}$

Formula Used:

The magnitude of the torque exerted by gravity about the point of support as

  $\tau =mgr\sin \theta$

Calculation:

By differentiating the angular momentum:

  $\begin{array}{c}\frac{d\overrightarrow{L}}{dt}=\frac{d}{dt}\left[\left(5.36\right)\left(\cos \omega t\widehat{i}+\sin \omega t\widehat{j}\right)+3.09\widehat{k}\right]\\ =\left(5.36\right)\left(-\sin \omega t\widehat{i}+\cos \omega t\widehat{j}\right)\text{N}\cdot \text{m}\end{array}$

Therefore, the magnitude of above derivative

  $\begin{array}{c}\left|\frac{d\overrightarrow{L}}{dt}\right|=\left(5.36\right)\left(2.75\right)\\ =14.7\text{N}\cdot \text{m}\end{array}$

Substitute the values:

  $\begin{array}{c}\tau =mgr\sin \theta \\ =\left(\text{2}\text{kg}\right)\left(9.81{\text{m/s}}^{\text{2}}\right)\left(\text{1}\text{.5}\text{m}\right)\sin {30}^{\text{o}}\\ =14.7\text{N}\cdot \text{m}\end{array}$

Conclusion:

Torque exerted by gravity about the point of support is $14.7\text{ N}\cdot \text{m}$ .