Question

A 5.1-m-diameter merry-go-round is initially turning with a 4.1 s period. It slows down and stops in 25 s . Part A Before slowing, what is the speed of a child on the rim? Express your answer to two significant figures and include the appropriate units. vv = nothing nothing Request Answer Part B How many revolutions does the merry-go-round make as it stops? Express your answer using two significant figures. ΔθΔ θ = nothing rev

Answer 1

Answer:

a)

$1.53$ rad s⁻¹

b)

$3$ rev

Explanation:

$d$ = diameter of the merry-go-round = 5.1 m

$r$ = radius of the merry-go-round = $\frac{d}{2}$ =  $\frac{5.1}{2}$ = 2.55 m

$T$ = time period of the merry-go-round = 4.1 s

$w_{i}$ = initial angular velocity of merry-go-round

Initial angular velocity is given as

$w_{i}$ = $\frac{2\pi }{T}$

$w_{i}$ =  $\frac{2\pi }{4.1}$

$w_{i}$ = 1.53 rad s⁻¹

b)

$w_{f}$ = final angular velocity of merry-go-round = 0 rad s⁻¹

$t$ = time taken to stop = 25 s

$\theta$ = angular displacement = ?

$n$ = Number of revolutions made before stopping

Angular displacement is given as

$\theta =\frac{(w_{f}+ w_{i})t}{2}$

$\theta =\frac{(0 + 1.53)(25)}{2}$

$\theta =19.125$ rad

Number of revolutions are given as

$n = \frac{\theta}{2\pi }$

$n = \frac{19.125}{2\pi }$

$n = 3$ rev