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lakkis [162]
3 years ago
5

Someone help with this also

Physics
1 answer:
Mumz [18]3 years ago
3 0
I’m pretty sure it’s average speed= total distance and total time which is A.
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Two children are riding on the edge of a merry-go-round that has a mass of 100.kg and radius of 1.60m and is rotating at 20.0rpm
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Here since both children and merry go round is our system and there is no torque acting on this system

So we will use angular momentum conservation in this

I_1\omega_1 = I_2\omega_2

now here we have

I_1 = \frac{MR^2}{2} + m_1R^2 + m_2R^2

I_1 = \frac{100(1.60)^2}{2} + (22 + 28)(1.60)^2

I_1 = 256

Now when children come to the position of half radius

then we will have

I_2 = \frac{MR^2}{2} + m_1(\frac{R}{2})^2 + m_2(\frac{R}{2})^2

I_2 = \frac{100(1.6)^2}{2} + (28 + 22)(0.8)^2

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now from above equation we have

256 (20.0 rpm) = 160(\omega_2)

\omega_2 = 32 rpm

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