Question

A 4.0-m-diameter playground merry-go-round, with a moment of inertia of

Answer 1

Answer:

$7.1$ ms⁻¹

Explanation:

$d$ = diameter of merry-go-round = 4 m

$r$ = radius of merry-go-round = $\frac{d}{2}$ =  $\frac{4}{2}$ = 2 m

$I$ = moment of inertia = 500 kgm²

$w_{i}$ = angular velocity of merry-go-round before ryan jumps = 2.0 rad/s

$w_{f}$ = angular velocity of merry-go-round after ryan jumps = 0 rad/s

$v$ = velocity of ryan before jumping onto the merry-go-round

$m$ = mass of ryan = 70 kg

Using conservation of angular momentum

$Iw_{i} - m v r = (I + mr^{2})w_{f}$

$(500)(2.0) - (70) v (2) = (I + mr^{2})(0)$

$1000 = 140 v$

$v = 7.1$ ms⁻¹