Chapter 8, Problem 8P

(a)

To determine

The linear speed of the child seated on wheel.

Answer to Problem 8P

  $5.3\times {10}^{-4}\text{m/s}$

Explanation of Solution

Given:

rpm of the wheel, $\text{rpm}=\frac{1}{4\times 60}=4.1\times {10}^{-3}$

Distance of child from centre, $r=1.2\text{m}$

Formula used:

  $\text{rad/s = rpm}\times \frac{\pi }{30}$

  Linear speed, $v=r\times \omega$

  Where, $\omega$ is angular velocity.

Calculation:

Substituting the given values

Velocity in rad/s = $4.1\times {10}^{-3}\times \frac{\pi }{30}=0.44\times {10}^{-3}\text{rad/s}$ .

Therefore,

  $\begin{array}{l}v=r\times \omega \\ v=1.2\text{m}\times 0.44\times {10}^{-3}\text{rad/s}\\ v=5.3\times {10}^{-4}\text{m/s}\end{array}$ .

Conclusion:

Thus, the linear speed of the child on the wheel is $5.3\times {10}^{-4}\text{m/s}$ .

(b)

To determine

The acceleration of the child on the wheel.

Answer to Problem 8P

  $0.21\times {10}^{-6}{\text{m/s}}^{\text{2}}$

Explanation of Solution

Given:

Rpm of the wheel $=4.1\times {10}^{-3}$

Distance of the child from the centre, $r=1.2\text{m}$

Formula used:

  $\text{rad/s}=rpm\times \frac{\pi }{30}$

  $a=\frac{v^2}{r}$

  Linear speed , $v=r\times \omega$

  where, $\omega$ is angular velocity and a is the acceleration.

Calculation:

Substituting the given values

  $\begin{array}{c}v=r\times \omega \\ =1.2\times 0.44\times {10}^{-3}\\ =5.3\times {10}^{-3}\text{m/s}\end{array}$

  $\alpha =\frac{v^2}{r}$

  $\begin{array}{l}a=\frac{{\left(5.3\times {10}^{-4}\text{m/s}\right)}^2}{1.2}\\ a=2.34\times 0^{-7}{\text{m/s}}^{\text{2}}\end{array}$

  $$

Conclusion:

Thus, the acceleration of the child on the wheel is $2.34\times 0^{-7}{\text{m/s}}^{\text{2}}$ .