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
Messages
70
Reaction score
0
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.
 
Physics news on Phys.org
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:
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Let there be a person in a not yet optimally designed sled at h meters in height. Let this sled free fall but user can steer by tilting their body weight in the sled or by optimal sled shape design point it in some horizontal direction where it is wanted to go - in any horizontal direction but once picked fixed. How to calculate horizontal distance d achievable as function of height h. Thus what is f(h) = d. Put another way, imagine a helicopter rises to a height h, but then shuts off all...
Back
Top