Centripetal force on table.

[cruisx]

Homework Statement

hi guys i need help solvign the following problem.

A mass 1kg on a frictionless table is attached to a hanging mass 4kg by a cord through a hole in the table. Find the frequency with which 1KG mass must move for 4kg to stay at rest, if the radius of the circle is 75m.



http://img32.imageshack.us/img32/9601/phyp.jpg

Homework Equations

m(v²/R

v = 2piR/T

T = 1/f

ac = 4pi * R/ T²

The Attempt at a Solution

Well i don't really know where to start as there is no angle given so i found the circumference first. which is 4.71 or 471 cm.

I think that is useless because i don't know what to do with it. I know i need to find ac first before i can find the frequency right?

So can someone give me a hit on the first step or where to start?

thanks.

 

[ideasrule]

This is more of a force problem than a harmonic motion problem. Try writing out Newton's second law for both the 1 kg mass and the 4 kg mass. Remember that for an object moving in a circle at speed v, acceleration=v^2/r. Also note that the tension in any massless rope is constant throughout the rope.

 

[tomcue]

To solve this problem, you will need to use the concept of centripetal force. This is the force that keeps an object moving in a circular path. In this case, the 1kg mass is moving in a circular path due to the tension of the cord attached to the 4kg mass. The 4kg mass is also experiencing an equal and opposite force, which keeps it at rest. In order for the 4kg mass to stay at rest, the frequency at which the 1kg mass moves must be equal to the natural frequency of the system.

To find the frequency, you can use the equation T = 1/f, where T is the period (time for one complete revolution) and f is the frequency. The period can be found using the equation v = 2piR/T, where v is the velocity of the 1kg mass and R is the radius of the circle (given to be 75m).

To find the velocity, you can use the equation m(v^2/R) = ac, where m is the mass of the 1kg mass, v is the velocity, R is the radius, and ac is the centripetal acceleration. Rearranging this equation, we get v = sqrt(acR/m).

Now, we can substitute this value of v into the equation v = 2piR/T to get T = 2piR/sqrt(acR/m). Substituting this value of T into the equation T = 1/f, we get f = 1/(2piR/sqrt(acR/m)).

Finally, we can substitute the given values of mass, radius, and acceleration to get the frequency. I will leave it to you to plug in the numbers and solve for f. Remember to use the correct units (meters for length, seconds for time, and meters per second squared for acceleration).

I hope this helps! Remember to always start with the relevant equations and substitute the given values before attempting to solve the problem. Good luck!