Question

A bicycle wheel has a radius r = 0.26 m and rotates at a constant frequency of f = 51 rev/min. calculate the period of the wheel T in seconds?

Answer 1

Givens.

• Radius = 0.26 m.

,

• Frequency = 51 rev/min.

The period is the inverse of the frequency

`\begin{gathered} T=(1)/(f)=\frac{1}{51\cdot\frac{\text{rev}}{\min }} \\ T=0.02\cdot\frac{\min }{\text{rev}} \end{gathered}`

Then, we transform it into seconds.

`\begin{gathered} T=0.02\cdot\frac{\min}{\text{rev}}\cdot(60\sec )/(1\min ) \\ T=1.2\cdot(\sec )/(rev) \end{gathered}`

The period in seconds is 1.2.

Then, the tangential speed formula is

`v=\omega\cdot r`

Where omega represents the angular speed, which in this case can be found using the frequency.

`\begin{gathered} v=2\pi f\cdot r \\ v=2\pi\cdot51\cdot\frac{\text{rev}}{\min}\cdot(1\min)/(60\sec)\cdot0.26m \\ v=1.39\cdot(m)/(s) \end{gathered}`

Therefore, the tangential speed is 1.39 meters per second.