Rusty view
Well-Known Member
What's it like to be able to walk on water?
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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.
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
Is the answer.........2?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.
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
Google search, CTRL - V, CTRL - C.Considering your 'disability' that's remarkable!
What does he anticipate will be the biggest challenge in stepping up to the Championship?I've managed to arrange an interview with Chris tomorrow morning for my BFBS radio show. If any of you have any questions you'd like me to put to him please share.
I'll let you know when you can hear the interview.
Thanks.
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