why centripetal acceleration a=v²/r ?
I know roughly the meaning of that, but what does, say, a centripetal acceleration of 1m/s² means?
- gintableLv 71 decade agoFavorite Answer
but what does, say, a centripetal acceleration of 1m/s² means?
It means that the net force per unit mass must act on the object of magnitude 1 Newton/kilogram toward the center of the circular motion, if it is to travel in a circle at constant speed.
Ok, if you aren't ready to think about forces, it means that the velocity vector is changing at the rate given by the acceleration vector.
The acceleration vector points inward, and constantly drags the velocity vector to follow it, but never changes its speed (only if there is tangential acceleration does speed change).
Where does the formula come from? Let's derive it.
Consider a body on an x-y coordinate system with speed v traveling in the +y direction and it is at a location of x=r, y=0. It is circling around the origin as you would likely guess.
At this location its velocity vector is V = <0, V>
And its displacement vector is D = <r, 0>
Now, as a function of time, it is traveling in a circle.
Its displacement is after traveling an angle of theta:
D = <r*cos(theta), r*sin(theta)>
Recall how nicely the radian is defined, 1 radius wrapped around the circle?
d = theta*r
where d is arc-length traveled.
D = <r*cos(d/r), r*sin(d/r)>
Take the derivative to find the velocity vector:
V = dD/dt = <-v*sin(v*t/r), v*cos(v*t/r)>
Take derivative again to find acceleration vector:
A = dV/dt = <-v*v/r*cos(v*t/r), -v*v/r*sin(v*t/r)>
a = |A|
a = sqrt((-v*v/r*cos(v*t/r))^2 + (-v*v/r*sin(v*t/r))^2)
a = sqrt((v*v/r)^2*(cos(v*t/r))^2 + (sin(v*t/r))^2))
a = (v^2/r)*sqrt((cos(v*t/r))^2 + (sin(v*t/r))^2))
a = (v^2/r)*sqrt(1)
a = v^2/r
- Randy PLv 71 decade ago
Take the velocity vector at a point in time t = 0. Take the velocity vector at a slightly later time, say t = 0.1 sec. This vector will be in a different direction as circular motion involves constantly deflecting the direction of the velocity vector.
The vector difference between these two velocity vectors, measured over a time interval of 0.1 seconds, would be 0.1 m/sec. That is, the change of velocity per unit time is (0.1 m/sec)/(0.1 sec) = 1 m/sec^2.
- Anonymous5 years ago
Let w = angular speed, rad/sec Then, at some time t = T, w^2*r = r*(dw/dt), r = radius of bit So, w^2 = 2dw/dt Let a = angular accel. = constant, rad/s^2 w = aT Then, (aT)^2 = 2*(dw/dt) = 2a a = 2/T^2 Let total angle traveled = theta = (a*T^2)/2 Then theta = (a*T^2)/2 = (T^2/2)*(2/T^2) = 1 radian Answer