Question

If a cycle wheel of radius $0.5m$ completes one revolution in $1s$, then the acceleration of a point on the circumference of the wheel will be

• A. $19.7m{s}^{-2}$
• B. $10.7m{s}^{-2}$
• C. $9.8m{s}^{-2}$
• D. $14.2m{s}^{-2}$

Answer 1

The correct option is A

$19.7m{s}^{-2}$

Step 1: Given data

The radius of the cycle wheel, $r=0.5m$

Frequency of revolution, $f=1rad/s$

Step 2: Formula used

Acceleration, $a=r{\omega }^2$

where $r$ is radius and $\omega$ is angular frequency.

Angular frequency, $\omega =2\mathrm{\pi f}$

where $f$ is the frequency

Step 3: Calculation

Angular frequency,

$$\begin{gathered} \omega =2\mathrm{\pi f} \\ \Rightarrow \omega =2\times 3.14\times 1 \\ \Rightarrow \omega =6.28rad/s \end{gathered}$$

Acceleration,

$$\begin{gathered} a=r{\omega }^2 \\ \Rightarrow a=0.5\times 6.{28}^2 \\ \Rightarrow a=19.7m{s}^{-2} \end{gathered}$$

Therefore, acceleration of the point on the circumference of the wheel will be $19.7m{s}^{-2}$