Question

Without loss of generality, 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. 


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Answer 1

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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: 

d2x/dt2 = -Rcosθ(dθ/dt)2−Rsinθ(d2θ/dt2) 

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

d2x/dt2 = -R(dθ/dt)2 

The term dθ/dt is usually called the angular velocity, which is the rate of change of the angle θ. It has units of radians/second. 

For convenience we can set w ≡ dθ/dt. 

Therefore, 

d2x/dt2 = -Rw2 

This is the well-known form for the centripetal acceleration equation. 


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Answer 2

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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: 

d2x/dt2 = -Rcosθ(dθ/dt)2−Rsinθ(d2θ/dt2) 

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

d2x/dt2 = -R(dθ/dt)2 

The term dθ/dt is usually called the angular velocity, which is the rate of change of the angle θ. It has units of radians/second. 

For convenience we can set w ≡ dθ/dt. 

Therefore, 

d2x/dt2 = -Rw2 

This is the well-known form for the centripetal acceleration equation.