Centripetal Force and proportionality

AI Thread Summary
Centripetal force is directly proportional to mass and radius, and to the square of frequency, as described by the equation Fc = (mv^2)/r. This relationship indicates that a heavier object requires more force to maintain circular motion due to greater inertia. Additionally, as frequency increases, the object's velocity rises, necessitating a larger centripetal force to sustain its path. The force is inversely proportional to the radius; a larger radius results in a lower required force since the object moves at a slower velocity. Understanding these relationships is crucial for explaining the dynamics of circular motion.
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Hi,

can someone please point me in a direction. I need to explain WHY "The centripetal force is proportional to the mass and to the radius and proportional to the square of the frequency."

We did the spinning lab, and now I need to do a discussion.

Thanks.
 
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MathHelp said:
Hi,

can someone please point me in a direction. I need to explain WHY "The centripetal force is proportional to the mass and to the radius and proportional to the square of the frequency."
You have to work out the change in velocity as a function of its tangential speed, v or \omega r.

Draw a diagram of the velocity vector of the mass at time 0. Then draw its velocity vector after a time dt. The mass turns through an angle d\theta = ds/r = \frac{vdt}{r} in that time.

Also remember that v = 2\pi r/T = \omega r and d\theta = \omega dt where \omega = 2\pi f is the angular frequency in radians/sec.

Now, the new velocity vector at t=dt is the same length as at t=0 but pointed d\theta to the original. The difference is the change in velocity or dv and is directed toward the centre of the circle along the radius. You can see from the diagram that:

dv = vsin(d\theta) which approaches the limit of dv = vd\theta as d\theta \rightarrow 0.

This means: dv = vd\theta = \omega r d\theta = \omega^2r dt so

dv/dt = a_{centripetal} = \omega^2r so:

F_{centripetal} = m\omega^2r

AM
 


Sure! The reason why the centripetal force is proportional to the mass and the radius, and also to the square of the frequency, is because of the mathematical relationship between these variables and the centripetal force equation.

The centripetal force, Fc, is given by the equation Fc = (mv^2)/r, where m is the mass of the object, v is its velocity, and r is the radius of its circular motion. This equation shows that the force required to keep an object moving in a circular path is directly proportional to its mass and the square of its velocity, and inversely proportional to the radius of its motion.

When considering the relationship between the centripetal force and the mass, we can see that the greater the mass of the object, the greater the force needed to keep it moving in a circular path. This is because a heavier object has more inertia, or resistance to change in motion, and therefore requires a greater force to overcome this inertia and keep it moving in a circular path.

Similarly, the centripetal force is also proportional to the square of the frequency, which is related to the velocity of the object. As the frequency increases, the velocity of the object also increases, which means the object is moving faster in a circular path. This increase in velocity requires a greater force to maintain the circular motion, as seen in the centripetal force equation.

Finally, we can see that the centripetal force is also inversely proportional to the radius of the circular path. This means that as the radius increases, the force required to keep the object in circular motion decreases. This is because a larger radius means the object has to cover a greater distance in its circular path, which requires a lower velocity and therefore a lower force according to the centripetal force equation.

In conclusion, the proportionality between the centripetal force, mass, radius, and frequency can be explained by the mathematical relationship between these variables and the centripetal force equation. I hope this helps with your discussion.
 
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