Question

A 4.0-m-diameter playground merry-go-round, with a moment of inertia of 350 kg⋅m2 is freely rotating with an angular velocity of 1.5 rad/s . Ryan, whose mass is 60 kg , runs on the ground around the outer edge of the merry-go-round in the opposite direction to its rotation. Still moving, he jumps directly onto the rim of the merry-go-round, bringing it (and himself) to a halt.How fast was Ryan running when he jumped on?

Answer 1

Answer:

$v = 4.375\,\frac{m}{s}$

Explanation:

The situation of the system Ryan - merry-go-round is modelled after the Principle of the Angular Momentum Conservation:

$(350\,kg\cdot m^{2})\cdot (1.5\,\frac{rad}{s} ) - (2\,m)\cdot (60\,kg)\cdot v = 0\,kg\cdot \frac{m^{2}}{s}$

The initial speed of Ryan is:

$v = 4.375\,\frac{m}{s}$

Answer 2

Answer:

4.375 m/s

Explanation:

Given that:

diameter of the playground merry-go-round = 4.0 m

then the radius will be = d/2 = 4/2 = 2.0 m

moment of Inertia (I) = 350 kg.m ²

angular velocity (ω) = 1.5 rad/s

mass (m) = 60 kg

Using conservation of angular momentum;

I ω = m V r

350 × 1.5 = 60 × V × 2

525 = 120 V

V = $\frac{525}{120}$

V = 4.375 m/s

Thus, Ryan was running 4.375 m/s when he jumped on