Chapter 6.2, Problem 23SSC

To determine

Centripetal acceleration and velocity of the rider.

Answer to Problem 23SSC

Centripetal acceleration of rider is $54.46{\text{ m/s}}^2$ and velocity is $14.7\text{6 m/s}$ .

Explanation of Solution

Given:

Radius of horizontal circle $r=4\text{ m}$

Time taken to complete one revolution $T=1.7\text{ s}$

Formula used:

  $a_c={\omega }^{2}r$

Here, $a_c$ centripetal acceleration, $\omega$ is angular velocity and r is radius.

  $v=\omega r$

Here, $\omega$ is angular velocity and r is radius.

Calculation:

Angular velocity can be obtained by:

  $\begin{array}{c}\omega =\frac{2\pi }{T}\\ =\frac{2\times 3.14}{1.7}\\ =3.69\text{ rad/s}\end{array}$

Now, centripetal acceleration can be obtained by:

  $\begin{array}{c}{a}_c={\omega }^{2}r\\ ={3.69}^2\times 4\\ =54.46{\text{ m/s}}^2\end{array}$

Now, velocity of rider can be obtained by:

  $\begin{array}{l}v=\omega r\\ =3.69\times 4\\ =14.76\text{ m/s}\end{array}$

Conclusion:

Centripetal acceleration of rider is $54.46{\text{ m/s}}^2$ and velocity is $14.7\text{6 m/s}$ .