Force on an object in circular motion

In summary, a hammer thrower whirls a 18.2 kg iron ball on the end of a 1.2 m string at 1.0 revolutions per second in a horizontal circle. Using the formula F=m(v^2/r), the force in the string can be calculated by first converting the angular velocity to meters per second and then plugging in the values. The correct formula to use is v=wr, where w is the angular velocity in radians per second and r is the radius of the circle. After plugging in the values, the force is calculated to be 7.54 m/s, which is the correct result.
  • #1
BuBbLeS01
602
0

Homework Statement


A hammer thrower whirls a 18.2 kg iron ball on the end of a 1.2 m string at 1.0 revolutions per second in a horizontal circle. Calculate the force in the string.


Homework Equations


F=m(v^2/r)


The Attempt at a Solution


I am not sure how to do this...
I set up a table of forces in the x and y direction...
x = T & centripetal force
y = W
I am not even sure I am starting off right.
 
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  • #2
What type of force is acting through the string? (i.e. What type of motion is it causing?)
 
  • #3
Since the circle is horizontal, you know that F=m(v^2/r) will be equal to the tension in the string.

Can you obtain that value from the known information?
 
  • #4
No I need to get the revolutions/sec in radians/sec
 
  • #5
So I multiplied 1 rev/sec by 2pi to get the velocity...
F= 18.2 kg(2pi^2/1.2) = 598.76 N
That seems high?
 
  • #6
Ok, you have the velocity in radians per second, so you have the angular velocity. The formula you are using assumes that the velocity is in meters per second. So, you'll either have to convert or use the version of the formula where the angular velocity is involved. Do you know this form of the centripetal force formula?
 
Last edited:
  • #7
Hmm, still missing something.

The velocity in your formula needs units of m/s, not radian/s. You're almost there, can you change it to that?
 
  • #8
I think the formula is v= wr to get it into m/s so that would be 598.76rad/s * 1.2 m = 718.51 m/s
 
  • #9
BuBbLeS01 said:
I think the formula is v= wr to get it into m/s so that would be 598.76rad/s * 1.2 m = 718.51 m/s

This is not correct.

Check you math when computing the angular frequency.

Remember:

[tex]\omega = 2 \pi f[/tex]
 
Last edited:
  • #10
ok if I am doing this right I am getting a huge number of 7818780.67 N
 
  • #11
I think I did something wrong maybe its supposed to be...
w=2pi * f = 6.28 rad/s
v=wr
6.28 rad/s * 1.2 m = 7.54 m/s
 
  • #12
BuBbLeS01 said:
I think I did something wrong maybe its supposed to be...
w=2pi * f = 6.28 rad/s
v=wr
6.28 rad/s * 1.2 m = 7.54 m/s

Yes this is correct. Sorry, I apparently didn't notice the error either.
 
  • #13
thanks!
 
  • #14
BuBbLeS01 said:
thanks!

No problem!
 

What is the force on an object in circular motion?

The force on an object in circular motion is called centripetal force. It is the force that keeps an object moving in a circular path and is always directed towards the center of the circle.

How is the magnitude of centripetal force determined?

The magnitude of centripetal force can be determined by the mass of the object, the speed of the object, and the radius of the circular path. The formula for centripetal force is Fc = mv²/r, where m is the mass, v is the speed, and r is the radius.

What happens if the centripetal force is not enough?

If the centripetal force is not enough, the object will not be able to maintain its circular motion and will either move in a larger circle or move in a straight line.

Is centripetal force a real force?

Yes, centripetal force is a real force. It is often referred to as a "fictitious" or "pseudo" force because it only exists in a rotating frame of reference, but it is still a real force that acts on the object in circular motion.

Can centripetal force change the speed of an object in circular motion?

No, centripetal force does not change the speed of an object in circular motion. It only changes the direction of the object's velocity, keeping it moving in a circular path.

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