2 years ago

I need the answer please

Physics

1 answer:

Tamiku [17]2 years ago

3 0

Answer:

,1) point B

2) point A

I hope it's helps you

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A playground merry-go-round of radius R = 1.40 m has a moment of inertia I = 265 kg · m2 and is rotating at 11.0 rev/min about a

Tom [10]

Answer:

The value of new value of angular speed of merry go round. $\omega_{2}$ = 0.96 $\frac{rad}{sec}$

Explanation:

Given data

r = 1.4 m

Moment of inertia $I_{1}$ = 265 kg - $m^{2}$

$N_{1} =$ 11 RPM

$\omega_{1} = \frac{2 \pi N}{60}$

$\omega_{1} = \frac{2 \pi (11)}{60}$

$\omega_{1}$ = 1.15 $\frac{rad}{sec}$

From conservation of momentum principal

$I_{1} \omega_{1} = I_{2} \omega_{2}$ ------- (1)

$I_{2} = m r^{2} + 265$

$I_{2} = 27 (1.4)^{2} + 265$

$I_{2} = 317.92 \ kg m^{2}$

Put all the values in equation (1)

265 × 1.15 = 317.92 × $\omega_{2}$

$\omega_{2}$ = 0.96 $\frac{rad}{sec}$

This is the value of new value of angular speed of merry go round.

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