Slightly confused regarding centripetal force, radius and frequency

In summary, the conversation discusses the relationship between frequency of rotation and the changes in radius and centripetal force. The formula for centripetal force is mentioned, along with the rearranged formula for radius. It is stated that as frequency increases, the radius decreases and this affects the acceleration and centripetal force. The conversation also mentions the method of achieving circular motion and clarifies that it involves an object spinning around an axis.
  • #1
chris_0101
65
0

Homework Statement


Hello everybody,

I have a question that is tied into my lab report regarding centripetal force. The question asks if the frequency of rotation were to increase how would the radius r and the centripetal force change?


Homework Equations



2[itex]\pi[/itex]f = v/r

rearrange for r:

r = v/(2[itex]\pi[/itex]f)

a = r(ω^2)

F_c = m*a

The Attempt at a Solution



By observing the formula rearranged for r, I believe that the radius is inversely proportional to frequency - so in this case as frequency increases, the radius will decreases.
Is this statement correct?

Since acceleration is dependent on radius and if the radius decreases so does the acceleration due to the direct proportionality between acceleration and radius. This ultimately means that the centripetal force is decreased as well when the frequency is increased
Is this statement also correct?

Your comments are greatly appreciated.

Thanks
 
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  • #2
chris_0101 said:

Homework Statement


Hello everybody,

I have a question that is tied into my lab report regarding centripetal force. The question asks if the frequency of rotation were to increase how would the radius r and the centripetal force change?


Homework Equations



2[itex]\pi[/itex]f = v/r

rearrange for r:

r = v/(2[itex]\pi[/itex]f)

a = r(ω^2)

F_c = m*a

The Attempt at a Solution



By observing the formula rearranged for r, I believe that the radius is inversely proportional to frequency - so in this case as frequency increases, the radius will decreases.
Is this statement correct?

Since acceleration is dependent on radius and if the radius decreases so does the acceleration due to the direct proportionality between acceleration and radius. This ultimately means that the centripetal force is decreased as well when the frequency is increased
Is this statement also correct?

Your comments are greatly appreciated.

Thanks

How is this circular motion being achieved? - what is going in a circle.

The question can be easily interpreted if it is a satellite around the Earth.

Is this the standard prac where you rotate a rubber stopper on a fishing line with a bunch of washers hanging from the fishing line?
 
  • #3
I should have been more specific. Circular motion is achieved by an object spinning around an axis.
 
  • #4
chris_0101 said:
I should have been more specific. Circular motion is achieved by an object spinning around an axis.

how is it attached to the axis? What is the object? What is it made of?
 
  • #5




Hello,

Your understanding of the relationship between frequency, radius, and centripetal force is correct. As frequency increases, the radius decreases due to the inverse relationship between the two. This in turn leads to a decrease in centripetal force, as it is directly proportional to the acceleration, which is dependent on the radius. Your reasoning and use of the equations are also correct. Keep up the good work! If you have any further questions, don't hesitate to ask.
 

Related to Slightly confused regarding centripetal force, radius and frequency

1. What is centripetal force?

Centripetal force is the force that acts towards the center of a circular motion, keeping an object moving in a circular path. It is a necessary force for any object to maintain circular motion, and its magnitude is directly proportional to the mass and speed of the object, as well as the radius of the circular path.

2. How does the radius affect centripetal force?

The radius of a circular path has a direct relationship with the centripetal force. As the radius increases, the centripetal force decreases, and vice versa. This means that a larger circular path requires less centripetal force to keep an object in motion, while a smaller circular path requires a greater centripetal force.

3. What is the relationship between centripetal force and frequency?

The frequency of a circular motion is the number of revolutions an object makes in a given time period. The relationship between centripetal force and frequency is that they are inversely proportional. This means that as the frequency increases, the centripetal force decreases, and vice versa. This is because a higher frequency implies a smaller radius, and therefore, a greater centripetal force is needed to keep the object in motion.

4. How does centripetal force affect the speed of an object?

The centripetal force affects the speed of an object by keeping it in circular motion. Without this force, the object would move in a straight line at a constant speed. However, with the presence of centripetal force, the object is constantly changing direction, and therefore, its speed remains constant. The magnitude of the centripetal force also affects the speed, with a greater force resulting in a higher speed.

5. Can centripetal force be calculated?

Yes, centripetal force can be calculated using the equation Fc = mv^2/r, where Fc is the centripetal force, m is the mass of the object, v is the speed, and r is the radius of the circular path. This equation shows that the centripetal force is directly proportional to the mass and speed of the object, and inversely proportional to the radius of the circular path.

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