Circular Motion: Perpendicular Force & Velocity Change Explained Quickly

AI Thread Summary
A force acting on a moving object in a direction other than its motion alters both the magnitude and direction of its velocity. In uniform circular motion, the applied force is always perpendicular to the velocity, resulting in a change in direction without affecting the speed. This is because only forces with a parallel component to the velocity can change its magnitude. The energy transferred by a perpendicular force does not contribute to a change in speed, as shown through mathematical derivations. Thus, a force that remains perpendicular to motion maintains constant velocity magnitude while altering direction.
tasnim rahman
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A force on a moving object, in any direction other than direction of motion causes an overall change in velocity(both in magnitude and direction). Then in circular motion why does a perpendicular force applied change only direction and not magnitude. Is this because the force produces 0 velocity change towards the center at any instant, but overall circular velocity change? Someone please explain quickly.
 
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Only a force with a component parallel to an object's velocity can cause a change in the magnitude of the velocity. In uniform circular motion, the force is always perpendicular to the velocity, so only the direction changes.
 
A force that is always perpendicular to the direction of motion does not change the magnitude of the velocity.

One way of seeing it is considering the energy a force insert to the system (or the energy per unit time):
P=\vec{f}*\vec{}v = 0

Another way is simply taking the derivative of the magnitude of the velocity (assume 2-D case):
d(v^2)\dt= d(v_x)^2\dt + d(v_y)^2\dt = 2(a_x*v_x + a_y*v_y) = 2\vec{a}*\vec{v}= 2\m(\vec{f}*\vec{v}) = 0
 
ibc said:
A force that is always perpendicular to the direction of motion does not change the magnitude of the velocity.

One way of seeing it is considering the energy a force insert to the system (or the energy per unit time):
P=\vec{f}*\vec{}v = 0

Another way is simply taking the derivative of the magnitude of the velocity (assume 2-D case):
d(v^2)\dt= d(v_x)^2\dt + d(v_y)^2\dt = 2(a_x*v_x + a_y*v_y) = 2\vec{a}*\vec{v}= 2\m(\vec{f}*\vec{v}) = 0

I believe that should be perfectly clear to everyone. :eek:
 
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
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