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Alexxx [7]
3 years ago
13

A playground merry-go-round has a mass of 50 kg and a diameter of 4.0 m. There are 4 children who want to ride on it. They have

masses of 15 kg, 18 kg, 22 kg, and 25 kg. They start out on the edge of the merrygo-round but later move toward the center until they are halfway between the edge and the center.
A. Sketch the arrangement of the children on the merry-go-round when they are at the edge and halfway to the middle.
B. What is the total moment of inertia for both arrangements?
C. The children are standing at the edge when their parents get the ride turning at 0.21 radians/sec. What is their tangential velocity and the period of rotation?
D. While the ride rotates the children move to their positions halfway to the center. What is the angular velocity when they get to their new spots? Explain.

Physics
1 answer:
mixer [17]3 years ago
5 0

Answer:

B) I1 = 1680 kg.m^2          I2 = 1120 kg.m^2

C) V = 0.84m/s      T = 29.92s

D) ω2 = 0.315 rad/s

Explanation:

The moment of inertia when they are standing on the edge:

I1 = 1/2*M*R^2 + (m1+m2+m3+m4)*R^2   where M is the mass of the merry-go-round.

I1 = 1680 kg.m^2

The moment of inertia when they are standing half way to the center:

I2 = 1/2*M*R^2 + (m1+m2+m3+m4)*(R/2)^2

I2 = 1120 kg.m^2

The tangencial velocity is given by:

V = ω1*R = 0.84m/s

Period of rotation:

T = 2π / ω1 = 29.92s

Assuming that there is no friction and their parents are not pushing anymore, we can use conservation of the angular momentum to calculate the new angular velocity:

I1*ω1 = I2*ω2    Solving for ω2:

ω2 = I1*ω1 / I2 = 0.315 rad/s

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If the gymnast mass were doubled, her height (h) from the top of the board would be as follows,

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Explanation:

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3 years ago
What is the frequency of a clock waveform whose period is 750 microseconds?
Allushta [10]
Use this formula to find your answer...

Determine the frequency of a clock waveform whose period is 2us or (micro) and 0.75ms

frequency (f)=1/( Time period).

Frequency of 2 us clock =1/2*10^-6 =10^6/2 =500000Hz =500 kHz.

Frequency of 0..75ms clock =1/0.75*10^-3 =10^3/0.75 =1333.33Hz =1.33kHz.

6 0
3 years ago
An observer sitting at a bus stop sees the bus drive by at 30 m/s to the east. He also sees a
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Answer:

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Explanation:

The speed of a bus is 30 m/s due East wrt the passenger

He also sees a  passenger on the bus walking to the back at 2 m/s.

We need to find the passenger's velocity relative  to the bus. As the observer sees that the bus and the passenger are moving in opposite direction. Let v is the relative velocity. So,

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Hence, the passenger's velocity relative  to the bus is 32 m/s.

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1. A large turbine has an initial angular momentum of 6700 kgm^2/s. A storm is rolling in and the wind picks up. 8 seconds later
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Answer:

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Explanation:

Torque is the rate of change of angular momentum.

Hence, we have

\tau = \dfrac{\Delta L}{t}

Δ<em>L</em> is the change in angular momentum.

Using values in the question,

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