Chapter 8, Problem 73QAP

To determine

  $(a)$

Average angular acceleration of the merry-go-round.

Answer to Problem 73QAP

  $\alpha =0.042rad{s}^{-2}$

Explanation of Solution

Given:

Initial angular velocity of roulette wheel $={\omega }_{0z}=0rpm$

Final angular velocity of roulette wheel $={\omega }_z=18rpm$

Time $=43s$

Formula used:

Equation for constant angular acceleration,

  ${\omega }_z={\omega }_{0z}+\alpha t$$\mathrm{...}(1)$

( ${\omega }_z=$ Angular velocity at time $=t$, ${\omega }_{0z}=$ Angular velocity at time=0, $\alpha =$ constant angular acceleration, $t=$ time)

Calculation:

Consider the rotational motion of roulette wheel,

Applying the $(1)$ equation,

  ${\omega }_z={\omega }_{0z}+\alpha t$

Let's substitute the values,

  $(18rpm)=\left(0rpm\right)+(\alpha )43s$

  $\alpha =\frac{18rpm}{43s}$

  $\alpha =0.042rad{s}^{-2}$

Conclusion:

Average angular acceleration of the merry-go-round. $=$$0.042rad{s}^{-2}$

To determine

  $(b)$

Angular displacement of the merry-go-round during this time interval.

Answer to Problem 73QAP

Angular displacement $=27rad$

Explanation of Solution

Given:

Initial angular velocity of roulette wheel $={\omega }_{0z}=0rpm$

Final angular velocity of roulette wheel $={\omega }_z=18rpm$

Time $=43s$

Angular acceleration of roulette wheel $=\alpha$ = $0.42rad{s}^{-2}$

Formula used:

Equation for constant angular acceleration,

  $\theta ={\theta }_o+{\omega }_{0z}t+\frac{1}{2}\alpha t^2$$\mathrm{...}(2)$

( $\theta =$ Angular position at time= t, ${\theta }_o=$ Angular position at time=0, ${\omega }_{0z}=$ Angular velocity at time=0, $\alpha =$ constant angular acceleration, $t=$ time)

Calculation:

Consider the rotational motion of roulette wheel,

Applying the $(2)$ equation,

  $\theta ={\theta }_o+{\omega }_{0z}t+\frac{1}{2}\alpha t^2$

Let's substitute the values,

  $\theta ={\theta }_o+{\omega }_{0z}t+\frac{1}{2}\alpha t^2$

  $\theta =(0)+(0rpm)(43s)+\frac{1}{2}(\frac{18}{43}rad{s}^{-2})(43s)(43s)$

  $\theta =\frac{1}{2}*18(43)rad$

  $\theta =387rad$

It means merry-go-round goes 1 round and 27 radians in distance.

Accordingly, angular displacement is $27rad$.

Conclusion:

Angular displacement of the merry-go-round during this time interval. $=27rad$

To determine

  $(c)$

Maximum tangential speed of the child if she rides on the edge of platform

Answer to Problem 73QAP

Maximum tangential speed = $36m{s}^{-1}$

Explanation of Solution

Given:

Initial angular velocity of roulette wheel $={\omega }_{0z}=0rpm$

Final angular velocity of roulette wheel $={\omega }_z=18rpm$

Time $=43s$

Angular acceleration of roulette wheel $=\alpha$ = $0.42rad{s}^{-2}$

Radius of the merry-go-round = $=2.00m$

Formula used:

Equation for maximum tangential speed of an object if it on the edge of disk platform,

  $v_{\max imum}=r{\omega }_{\max imum}$$\mathrm{...}(1)$

( $v_{\max imum}=$ Maximum linear velocity, $r=$ radius of circle, ${\omega }_{\max imum}=$ maximum angular velocity which taken in that given time)

Calculation:

Consider the rotational motion of child,

Applying the $(1)$ equation,

  $v_{\max imum}=r{\omega }_{\max imum}$

Let's substitute the values,

  $v_{\max imum}=r{\omega }_{\max imum}$

  $v_{\max imum}=(2.00m)(18rad{s}^{-1})$

  $v_{\max imum}=36m{s}^{-1}$

Conclusion:

Maximum tangential speed of the child if she rides on the edge of platform $=36m{s}^{-1}$