A 5.0-m-diameter merry-go-round is initially turning with a $4.0\text{s}$period. It slows down and stops in $20\text{s}$.

a. Before slowing, what is the speed of a child on the rim?

b. How many revolutions does the merry-go-round make as it stops?

The diameter of the merry-go-round is$5.0\text{m}$ , the merry-go-round is initially turning with a period of $4.0\text{s}$, and it requires $20\text{s}$to slows down and stop.

The speed of the child on the rim before slowing down is,

$v_1=\left(\frac{d}{2}\right){\omega }_1\dots \dots \left(\text{I}\right)$

Here, $v_1$is the speed of the child on the rim before slowing down, $d$is the diameter of the rim, and ${\omega }_1$is the angular velocity of the rim before slowing down.

The angular velocity of the child before slowing down is,

${\omega }_1=\frac{2\pi }{T_1}$$T_1$is the initial turning period of the rim before it slows down.

Substitute for $T$in the above equation to find ${\omega }_1$.

${\omega }_1=\frac{2\pi }{4.0\text{s}}=1.57\text{rad}/\text{s}$Thus, the angular velocity of the rim before it slows down is $1.57\text{rad}/\text{s}$.

Substitute $1.57\text{rad}/\text{s}$for ${\omega }_1$and $5.0\text{m}$for$d$ in the equation (I) to find $v_1$.

$v_1=\left(\frac{5.0\text{m}}{2}\right)(1.57\text{rad}/\text{s})=3.925\text{m}/\text{s}$Conclusion:

Therefore, the speed of the child on the rim before the slowing down is$3.925\text{m}/\text{s}$ .