Dynamics and uniform circular motion

I am working on the unit review questions in my textbook in order to prepare for a major assessment this week. Can someone please help me with question 1 - 3? And for questions 4 to 6, my answers are significantly different from those provided by the textbook. I am not sure it is my mistake or theirs. Can anyone check on that if you have time? If my answers are wrong, I would re-attempt on them. Thank you very much!

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1) Obtain the value of g at the surface of Earth using the motion of the Moon Assume that the Moon's period around Earth is 27 d 8 h and that the radius of its orbit is 60.1 times the radius (6.38 x 10^6 m) of Earth.

2) In a disaster film, an elevator full of people falls freely when the cable snaps. The film depicts the people pressed upward against the ceiling. Is this good physics? Why or why not?

3) A 1.12 m string pendulum has a bob of mass 0.200kg. What is the magnitude of tension at the bottom of the swing if the pendulum is moving at 1.20m/s.
[Currently, I have only learnt uniform circular motion where there is constant speed and constant radius throughout. But for this pendulum question, the bob is definitely not in uniform circular motion because the speed is changing at every point. Can I still use the formula Fc = mv^2 / r ? Why or why not?]

 

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4) The orbit of Venus is approximately circular. The masses of the Sun and Venus are 1.99 x 10^30 kg and 4.83 x 10^24 kg, respectively. The Sun-Venus distance is 1.08 x 10^8 km. Determine the centripetal acceleration of Venus.
[I got 1.14 x 10^(-2) m/s^2, but the answer in my textbook is 1.3 x 10^(-2) m/s^2]

5)

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[My answer is 4.83 x 10^26 N [65.7 degrees from the line to the Earth], the answer in my textbook is 4.82 x 10^20 N at an angle of 24.4 degrees from the line to the Sun.]

6)

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Determine the frequency of the bob.
[My answer is 0.493 Hz, the answer in my textbook is 0.567 Hz]

 

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I will answer #2 since I'm too lazy to figure out the others:

People pressed against the ceiling of the elevator is bad physics because all bodies fall, i.e. the people and elevator, with the same acceleration.

 

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2) So in that case the normal force on the people would be zero because they are falling in 9.8m/s^2[down], right? As a result, they will just stand (or float) on the floor of the elevator, I think...

In the International Space Station, it is free falling towards (and orbiting around) the Earth, the astronauts also experiences an apparent weight, or normal force of zero. Why? The acceleration due to gravity up there is definitely not 9.8m/s^2 [down] because it is much further from the center of the Earth.

 

At what rate are the astronauts (and the ISS) accelerating?

 

I get 2.pi.√( g/L.cosθ ) = 0.49 Hz for g=9.8m/s²

Looks like another bad mark for the textbook

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This could be a good test of your physics intuition.
What is your gut feeling for this one?

Things to consider:
- What is the bob's angular acceleration at the bottom of the swing?
- Does it matter what the angular acceleration is anyway? ie Does the string "care" if the bob's angular velocity is changing, or only what its angular velocity is right now?

 

What does the Moon's motion tell you about Earth's gravity?
What is the relationship between acceleration due to gravity and distance from Earth?

 

g = G.M/r² = 1.14e-2 m/s²
Text wrong again, unless I'm making the same mistake as you

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I agree with the text (near enough) for this one. You might have forgotten to convert kilometres to metres?

 

Definitely far less than 9.8m/s^2[down] because it is far from the earth's surface. Then why would the normal force on the astronauts be 0 and they apparent to be floating?

 

What is angluar acceleration? I've neither learnt that in my physics course.
The only thing I have learnt is centripetal acceleration and uniform circular motion, so I would approach this for all the questions I encounter. However, in this case where the speed of the object in circular motion is not constant, I wonder if the formula Fc=mv^2/r is still valid......

 

Keep thinking.

Go back to the elevator for a minute.
Why, exactly, is the normal force on the elevator occupants zero?
What two quantities are the same?

 

Fc=Fg
4(pi^2)mr / T^2 = GmM / r^2
4(pi^2) / T^2 = GM / r^3
I can find the mass of the earth this way. But that doesn't help...

The moon is far from the earth's surface, how can I obtain the value of g=9.8m/s^2 at the "surface" of Earth using the motion of the Moon, I really don't get this...

 

Is your gut feeling that it is probably valid, or probably not valid?

Angular acceleration is the rate of change in angular velocity. It's proportional to the rate of change in the magnitude of v, so...

1 - Is the magnitude of v changing at the bottom of the swing? Is it increasing, decreasing, or level?

2 - Do you think that the string "cares" how the magnitude of v is changing anwyay, or do you think it's only the value of v right now that matters?

 

Thanks Pete, you're so nice.

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My textbook is driving my crazy...way too many mistakes that I am starting to lose confidence on it...

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OK, I will look over my work now.

 

What is the value of r at the Earth's surface?

What is the gravitational force between the Earth and an object of mass 1kg at the surface of the Earth?

 

Keep at it, kingwinner! You're doing fine.

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If you are standing on a flat ground, Fn=Fg
If you're falling at less than 9.8m/s^2[down], Fn < Fg

If you're falling at 9.8m/s^2[down], Fn=0. Only Fg acts on you so your acceleration is 9.8m/s^2[down]