## Related questions with answers

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

Verifieda) 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):

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

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