A particle is moving around in a circle and its position is given in polar coordinates as x = Rcosθ, and y = Rsinθ, where R is the radius of the circle, and θ is in radians. From these equations derive the equation for centripetal acceleration.
Answers
, We only need to look at the equation for the x-position, since we know that centripetal acceleration points towards the center of the circle. Thus, when θ = 0, the second derivative of x with respect to time must be the centripetal acceleration.
The first derivative of x with respect to time t is:
dx/dt = -Rsinθ(dθ/dt)
The second derivative of x with respect to time t is:
d^2x/dt^2 = -Rcosθ(dθ/dt)^2−Rsinθ(d^2θ/dt^2)
In both of the above equations the chain rule of Calculus is used and by assumption θ is a function of time. Therefore, θ can be differentiated with respect to time.
Now, evaluate the second derivative at θ = 0.
We have,
d^2x/dt^2 = -R(dθ/dt)^2
dθ/dt is usually called the angular velocity, which is the rate of change of the angle θ. we can set w ≡ dθ/dt.
Therefore,
d^2x/dt2 = -Rw^2
This is the well-known form for the centripetal acceleration equation.
hope it helps...
Answer:
Consider the equation for the x-position, since we know that centripetal acceleration points towards the center of the circle. Thus, when θ = 0, the second derivative of x with respect to time must be the centripetal acceleration.
The first derivative of x with respect to time t is:
dx/dt = -Rsinθ(dθ/dt)
The second derivative of x with respect to time t is:
d^2x/dt^2 = -Rcosθ(dθ/dt)^2−Rsinθ(d^2θ/dt^2)
In both of the above equations the chain rule of Calculus is used and by assumption θ is a function of time. Therefore, θ can be differentiated with respect to time.
Now, evaluate the second derivative at θ = 0.
We have,
d^2x/dt^2 = -R(dθ/dt)^2
dθ/dt is usually called the angular velocity, which is the rate of change of the angle θ.
we can set w ≡ dθ/dt.
Therefore, d^2x/dt2 = -Rw^2