Space station rotation problem

[Purduenub03]

Quick physics help please

First off, I'm new to the board and want to say hi to everyone! I hope i can give and take from this site, i should be helpful in the chem forums, but physics is not my best. I have 2 quick questions if you could help me that'd be great.

A space station is shaped like a ring and rotates to simulate gravity. If the radius of the space station is 120 m, at what frequency must it rotate so that it simulates Earth's gravity? [Hint: The apparent weight of the astronauts must be the same as their weight on Earth.] - in rev/s

I did the following and got it wrong
9.8 = V^2/120, found v = 34.29m/s
34.29 / 2*pi*120 = .045 rev / s

Where have i gone wrong?/

A coin is placed on a record that is rotating at 33.3 rpm. If the coefficient of static friction between the coin and the record is 0.3, how far from the center of the record can the coin be placed without having it slip off?

I'm plain lost on this one.

 

[Gokul43201]

First off, I'm new to the board and want to say hi to everyone! I hope i can give and take from this site, i should be helpful in the chem forums, but physics is not my best. I have 2 quick questions if you could help me that'd be great.

A space station is shaped like a ring and rotates to simulate gravity. If the radius of the space station is 120 m, at what frequency must it rotate so that it simulates Earth's gravity? [Hint: The apparent weight of the astronauts must be the same as their weight on Earth.] - in rev/s

I did the following and got it wrong
9.8 = V^2/120, found v = 34.29m/s
34.29 / 2*pi*120 = .045 rev / s

Where have i gone wrong?/

Looks okay to me. How do you know you're wrong ? Do you know what the answer should be ?

A coin is placed on a record that is rotating at 33.3 rpm. If the coefficient of static friction between the coin and the record is 0.3, how far from the center of the record can the coin be placed without having it slip off?

I'm plain lost on this one.

How much force would you need to exert on a body to overcome friction, and cause it to slide? In the rest frame of the coin, this force would be provided by the centrifugal force. Remember, the linear velocity depends on the distance of the coin from the record center.

PS : Welcome to PF. It's a great place to be !

 

[wais]

Hi there! Welcome to the board. In regards to your first question about the space station rotation problem, you have the right idea but there are a few things you need to consider. First, the equation you used (9.8 = V^2/120) is the equation for centripetal acceleration, not frequency. To find the frequency, you need to use the equation w = v/r, where w is angular velocity, v is linear velocity, and r is the radius. So in this case, w = 34.29/120 = 0.285 rev/s. This is the angular velocity, not the frequency. To find the frequency, you need to divide the angular velocity by 2*pi, so the frequency would be 0.045 rev/s, just like you calculated. However, you need to convert this to Hz (Hertz), which is the unit for frequency. To do this, you need to multiply by 60, since there are 60 seconds in a minute. So the final answer would be 0.045*60 = 2.7 Hz.

For your second question about the coin on the record, you need to use the equation for centripetal force, which is Fc = mv^2/r. In this case, the force is equal to the coefficient of static friction (0.3) times the normal force (which is the weight of the coin, mg). So the equation would be 0.3mg = mv^2/r. You can rearrange this to solve for r, which would be r = mv^2/0.3mg. Since you are given the angular velocity (33.3 rpm), you can use the equation v = wr, where w is the angular velocity in radians per second. So w = 33.3*2*pi/60 = 3.49 rad/s. Plug this into the equation and you should be able to solve for r.

I hope this helps! Don't hesitate to ask for clarification if needed. Good luck!