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

500kg?m2 is freely rotating with an angular velocity of 2.0rad/s . Ryan, whose mass is 70kg , 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?

Respuesta :

Answer:

[tex] 7.1[/tex] ms⁻¹

Explanation:

[tex]d[/tex] = diameter of merry-go-round = 4 m

[tex]r[/tex] = radius of merry-go-round = [tex]\frac{d}{2}[/tex] =  [tex]\frac{4}{2}[/tex] = 2 m

[tex]I[/tex] = moment of inertia = 500 kgm²

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

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

[tex]v[/tex] = velocity of ryan before jumping onto the merry-go-round

[tex]m[/tex] = mass of ryan = 70 kg

Using conservation of angular momentum

[tex]Iw_{i} - m v r = (I + mr^{2})w_{f}[/tex]

[tex](500)(2.0) - (70) v (2) = (I + mr^{2})(0)[/tex]

[tex]1000 = 140 v[/tex]

[tex]v = 7.1[/tex] ms⁻¹

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