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Old Jan 18th 2015, 10:10 PM   #1
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finding the angle of velocity

This is not homework. This problem has been made by me. One day ago, I asked help about how to determine the direction of velocity if force act with velocity at right angle. After asking help, some comments helped me to give some better idea. Then I made this problem with some values to realize that fact easily. But I found some problems for which I want to post this mathematical problem at this site. If it is necessary, please edit this post as necessary.

Here we will try to solve two mathematical problems related to centripetal force. This two problems are interrelated to each other. Suppose, a particle is travelling through a circular path with uniform value of velocity. Assume that, value of velocity, v= 4 m/s, mass of particle, m= 2 kg, radius of circle, r= 2 m. Hence by calculating we get the value of centripetal force, F= 16 N, acceleration, a= 8 m/sē, angular frequency, w= 2 rad/second.

Let, at t=0, particle is at point A of circle. and, at t=2, particle is at point B of circle. The angle produced by arc AB at the centre of circle is= w*t =4 rad= 229.2 degree.

Now we will determine the angle between the angle of direction of initial velocity of particle (when t=0) and the direction of final velocity (at t=2). This angle equals to the angle produced by arc AB at the centre of circle. Geometrically it can be proved easily that these two angles are equal. Let, at t=0, the direction of velocity of particle is parallel to the direction of X axis. As a result, along X axis, the value of initial velocity is 4 m/s, and along Y axis, it is 0. Since the centripetal force acts at right angle with the direction of initial velocity, along X axis the value of velocity will not change. It will change along X axis only. After 2 seconds, the value of velocity along Y axis will be 16 m/s. And at this moment the direction of resultant velocity with X axis can be determined by the following way. tan∅= (16/4)= 4, ∅= 75.96 degree. This angle is not equal to the angle produced by arc AB at the centre of circle. Please help me to find out the error. If it is possible to you, please show the correct way to solve it.
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Old Jan 19th 2015, 05:47 AM   #2
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Originally Posted by Nousher View Post
Suppose, a particle is travelling through a circular path with uniform value of velocity. Assume that, value of velocity, v= 4 m/s, mass of particle, m= 2 kg, radius of circle, r= 2 m. Hence by calculating we get the value of centripetal force, F= 16 N, acceleration, a= 8 m/sē, angular frequency, w= 2 rad/second.
The angular velocity is actually v/(2 pi r) = 1/pi rad/second, not 2 rad/second.

Originally Posted by Nousher View Post
As a result, along X axis, the value of initial velocity is 4 m/s, and along Y axis, it is 0.
Yes, this is the instantaneous velocity at t=0.

Originally Posted by Nousher View Post
Since the centripetal force acts at right angle with the direction of initial velocity, along X axis the value of velocity will not change.
This is the source your error. For a particle traveling in a circle at constant speed the centripetal acceleration is always in the direction towards the center of rotation. As the particle moves around the circle the direction of acceleration continually changes - its x-axis components does not remain constant. For example for a satellite in a circular orbit acceleration is always towards the center of the planet it is revolving about. At one particular moment its acceleration is in the +/- x-direction only (no y-direction component), and 1/4 of an orbit later its acceleration is purely n the +/- y-direction with no x-direction component.

Originally Posted by Nousher View Post
After 2 seconds, the value of velocity along Y axis will be 16 m/s. And at this moment the direction of resultant velocity with X axis can be determined by the following way. tan∅= (16/4)= 4, ∅= 75.96 degree. .
You have just calculated the velocity of a particle subjected to constant acceleration in the y-direction. The resulting path of motion is a parabola - not a circle - similar to a projectile that's given an initial velocity in the x-direction and then falls under gravity.

Last edited by ChipB; Jan 19th 2015 at 09:04 AM.
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