Question

# A bicycle wheel with radius 0.3m rotates from rest to 3 rev/s in 5 s. What is the magnitude and direction of the total acceleration vector at the edge of the wheel at 1.0 s?

Solution

Verified
Step 1
1 of 3

a) From $\textbf{the kinematics of the rotational motion }$ we know that :

$\omega_{f} = \omega_{i} + \alpha t \qquad(1)$

From $\textbf{givens}$ $3\ \mathrm{rev/s} = 3 \times 60 = 180\ \mathrm{rev/min}$

$\left[\dfrac{180\ \; \mathrm{rev}}{\mathrm{min}} \cdot \dfrac{2 \pi \ \mathrm{rad}}{\mathrm{rev}} \cdot \dfrac{1 \ \mathrm{min}}{60\ \mathrm{s}}\right] =18.8495 \ \mathrm{rad/s}$

where:

• $\omega_{f}$ is the final angular velocity of the body.
• $\omega_{i}$ is the initial angular velocity of the body.
• $\alpha$ is the angular acceleration of the body.
• $a_{t}$ is the tangential acceleration .
• $a_{c}$ is the centripetal acceleration .
• $t$ is the time .

Form $\textbf{givens}$ we know that : $\omega_{i} = 0$ rad/s , $\omega_{f}=18.8495\ \mathrm{ rad/s}$ , $r = 0.3\ \mathrm{m}$ and $t = 5$ s.

$\textbf{ Rearrange and plugging}$ known information into Eq. (1):

\begin{align*} \omega_{f}& = \omega_{i} + \alpha t \\\\ \alpha t &=\omega_{f} - \omega_{i}\\\\ \alpha&= \dfrac{\omega_{f} - \omega_{i}}{t}\\\\ &= \dfrac{18.8495 - 0}{5}\\\\ &= 3.769 \end{align*}

## Recommended textbook solutions #### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th EditionISBN: 9780133942651 (5 more)Randall D. Knight
3,508 solutions #### Mathematical Methods in the Physical Sciences

3rd EditionISBN: 9780471198260Mary L. Boas
3,355 solutions #### Fundamentals of Physics

10th EditionISBN: 9781118230718 (4 more)David Halliday, Jearl Walker, Robert Resnick
8,950 solutions #### University Physics, Volume 1

1st EditionISBN: 9781938168277Jeff Sanny, Samuel J Ling, William Moebbs
1,471 solutions