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

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

ωf=ωi+αt(1)\omega_{f} = \omega_{i} + \alpha t \qquad(1)

From givens\textbf{givens} 3 rev/s=3×60=180 rev/min3\ \mathrm{rev/s} = 3 \times 60 = 180\ \mathrm{rev/min}

[180   revmin2π radrev1 min60 s]=18.8495 rad/s\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:

  • ωf\omega_{f} is the final angular velocity of the body.
  • ωi\omega_{i} is the initial angular velocity of the body.
  • α\alpha is the angular acceleration of the body.
  • ata_{t} is the tangential acceleration .
  • aca_{c} is the centripetal acceleration .
  • tt is the time .

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

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

ωf=ωi+αtαt=ωfωiα=ωfωit=18.849505=3.769\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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