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3hlang

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In summary, the conversation discusses the incorrect derivation of the centripetal acceleration formula and suggests reading a resource for the correct derivation.

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3hlang

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Doc Al

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Several problems here. First, by taking a quarter of a circle you are only able to calculate the average acceleration during that time interval, a3hlang said:

To learn how the centripetal acceleration formula is derived, read this: http://dev.physicslab.org/Document....e=CircularMotion_CentripetalAcceleration.xml"

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astrokreng

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Hello,

Thank you for sharing your thoughts and efforts in deriving the equation for centripetal acceleration. It is always commendable to see students taking an active role in understanding and questioning the concepts they are taught.

The equation for centripetal acceleration, a=v^2/r, is a fundamental equation in physics and has been derived and proven through various methods. The most common method is by using Newton's Second Law of Motion, which states that the net force acting on an object is equal to its mass times its acceleration (F=ma).

In the case of an object moving in a circular path, there is a constant force acting towards the center of the circle, called the centripetal force. This force is responsible for keeping the object in its circular motion. Therefore, we can write the equation F=ma as:

Fc = ma

Where Fc is the centripetal force and a is the centripetal acceleration.

Now, we also know that the centripetal force is equal to the product of the mass of the object, its velocity squared, and the radius of the circular path (Fc = mv^2/r). Substituting this into F=ma, we get:

mv^2/r = ma

Solving for a, we get:

a = v^2/r

This is the same equation that you have derived, so your method is correct. However, there are a few things to note:

1. In your derivation, you have assumed that the time period T is equal to the time taken for the object to travel a quarter of the circle (T/4). This is not always the case. The time period is the time taken for the object to complete one full revolution, which is equal to the circumference of the circle divided by the velocity (T=2πr/v). This is why your final equation has a factor of 2 in the denominator.

2. Your method assumes that the velocity changes direction after a quarter of the circle, which is not always the case. In fact, the velocity of the object is constantly changing direction as it moves along the circular path, but its magnitude remains constant.

I hope this explanation helps you understand how the equation for centripetal acceleration is derived. Keep up the good work in questioning and understanding the concepts of physics!

Best regards,

Scientist

Centripetal acceleration is the acceleration that is directed towards the center of a circular motion. It is the change in velocity of an object as it moves in a circular path.

Centripetal acceleration can be calculated using the formula a = v^2/r, where a is the centripetal acceleration, v is the velocity of the object, and r is the radius of the circular path.

Centripetal acceleration is the acceleration towards the center of a circular path, while tangential acceleration is the acceleration along the tangent of the circular path. These two types of acceleration are perpendicular to each other.

Centripetal acceleration is caused by centripetal force, which is the force that keeps an object moving in a circular path. The magnitude of the centripetal acceleration is directly proportional to the magnitude of the centripetal force.

Centripetal acceleration is directly proportional to the square of the velocity of the object. This means that as the velocity increases, the centripetal acceleration also increases. However, the direction of the centripetal acceleration is always towards the center of the circular path, regardless of the velocity.

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