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pav-90 [236]
4 years ago
8

A playground merry-go-round of radius R = 1.20 m has a moment of inertia I = 240 kg · m2 and is rotating at 9.0 rev/min about a

frictionless vertical axle. Facing the axle, a 26.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round?
Physics
1 answer:
Vesnalui [34]4 years ago
3 0

Answer:

ω' = 0.815 rad/s

Explanation:

Given,

R = 1.20 m

Inertia of merry-go- round= 240 kg.m²

Rotating speed  = 9 rpm = 9\times \dfrac{2\pi}{60}

                           =0.9424 rad/s

mass of the child, m = 26 kg

angular speed of the merry-go-round=?

we know

Angular momentum, L = I ω

Moment of inertia of the child

I' = m  r² = 26 x 1.2² = 37.44 kgm²

Conservation of angular momentum

initial angular momentum = Final angular momentum

I ω = (I+I')ω'

240 x 0.9424 = (240+37.44) ω'

226.176= 277.44 ω'

ω' = 0.815 rad/s

new angular speed of the merry-go- round is equal to 0.815 rad/s

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Answer:

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If for a given pair of media CR
nika2105 [10]

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<h3>Which factors affect the critical angle for a given pair of media?</h3>

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(i) Effect of colour of light: The critical angle for a pair of media is less for the violet light and more for the red light. Thus the critical angle increases with the increase in wavelength of light.

(ii) Effect of temperature: The critical angle increases with increase in temperature because on increasing temperature of medium, its refractive index decreases.

According to the question,

μ 1​ sinCR​ =1

μ 2​ sinCY =1

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