Q61P - University Physics with Modern Physics

Question

You must design a device for shooting a small marble vertically upward. The marble is in a small cup that is attached to the rim of a wheel of radius $\mathbf{0}\mathbf{.}\mathbf{260}\mathbf{\text{ m}}$; the cup is covered by a lid. The wheel starts from rest and rotates about a horizontal axis that is perpendicular to the wheel at its center. After the wheel has turned through $\mathbf{20}\mathbf{\text{ rev}}$, the cup is the same height as the center of the wheel. At this point in the motion, the lid opens and the marble travels vertically upward to a maximum height   $\mathit{h}$ above the center of the wheel. If the wheel rotates with a constant angular acceleration  , what value of $\mathit{a}$ is required for the marble to reach a height of $\mathit{h}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{\text{ m}}$?

Step-by-Step Solution

Answer

The angular acceleration is$13.8 \raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.$.

Step 1: Definition:

Angular velocity is measured in angles per unit time or in radians per second. The rate of change of angular velocity is angular acceleration.

Step 2: Given data:

Wheel of radius, $r=0.260\text{ m}$

A maximum height, $h=12.0\text{ m}$

Angular displacement, $\theta =20.0\text{ rev}$

Step 3: Apply the law of conservation:

Apply the law of conservation of energy to define the angular acceleration to reach the given maximum height.

$$\begin{gathered} K_{\text{i}}+U_{\text{i}}=K_{\text{f}}+U_{\text{f}} \\ \frac{1}{2}m{v}^2=mgh \\ v=\sqrt{2gh} \end{gathered}$$

Here, $K_{\text{i}}$ and $U_{\text{i}}$ are the initial kinetic and potential energy respectively. $K_{\text{f}}$ and $U_{\text{f}}$ are the final kinetic and potential energy respectively. The velocity is $v$, the height is $h$, and $g$ is the acceleration due to gravity has a value $9.8 \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.$.

Substitute known values in the above equation.

$\begin{array}{rcl}v& =& \sqrt{2\times 9.8 \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.\times 2.0\text{ m}}\\ & =& \sqrt{235.2 \raisebox{1ex}{${\text{m}}^2$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.}\\ & =& 15.336 \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\end{array}$

Step 4: Define angular acceleration:

Now, calculate the angular acceleration by using the following equation.

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

Substitute $\left(\frac{v}{r}\right)$ for  $\omega$ and $0$ for ${\omega }_0$ in the above equation.

${\left(\frac{v}{r}\right)}^2=0+2\alpha \theta$

$\alpha =\frac{v^2}{2\theta r^2}$                                                                                                            ..... (1)

Convert the angular displacement into radian,

$\begin{array}{rcl}\theta & =& \left(20.0\text{ rev}\right)\left(\frac{2\pi \text{ rad}}{1\text{ rev}}\right)\\ & =& 20\times 2\pi \text{ rad}\\ & =& 40\pi \text{ rad}\end{array}$

Substitute $40\pi \text{ rad}$ for $\theta$, $15.336 \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$ for $v$, $0.260\text{ m}$ and   for $r$ into equation (1).

$\begin{array}{rcl}\alpha & =& \frac{{\left(15.336 {\displaystyle \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}\right)}^2}{2\times \left(40\pi \text{ rad}\right)\text{×}\left(\text{0.260 m}\right)}\\ & =& \frac{235.193 {\left({\displaystyle \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}\right)}^2}{\left(5.408\pi \text{ rad}·{\text{m}}^2\right)}\\ & =& 13.8 \raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^2$}\right.\end{array}$

Hence, the required angular acceleration is $\begin{array}{rcl}& & 13.8 \raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^2$}\right.\end{array}$.