A particle is moving around in a circle and its position is given in polar coordinates as x = Rcos?, and y = Rsin?, where R is the radius of the circle, and ? is in radians. From these equations derive the equation for centripetal acceleration.

Question

Sher Mohammad · Answer

we only need to look at the equation for thex-position, since we know that centripetal acceleration points towards the center of the circle. Thus, when?= 0, the second derivative ofxwith respect to time must be the centripetal acceleration.
The first derivative ofxwith respect to timetis:
dx/dt= —Rsin?(d?/dt)
The second derivative ofxwith respect to timetis:
d²x/dt²= —Rcos?(d?/dt)²—Rsin?(d²?/dt²)
In both of the above equations the chain rule of Calculus is used and by assumption?is a function of time. Therefore,?can be differentiated with respect to time.
Now, evaluate the second derivative at?= 0.
We have,
d²x/dt²= —R(d?/dt)²
The term d?/dtis usually called the angular velocity, which is the rate of change of the angle?. It has units of radians/second.
For convenience we can setw= d?/dt.
Therefore,
d²x/dt²= —Rw²

this is centripital acceleration

sher mohammad

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A particle is moving around in a circle and its position is given in polar coordinates as x = Rcos?, and y = Rsin?, where R is the radius of the circle, and ? is in radians. From these equations derive the equation for centripetal acceleration.

A particle is moving around in a circle and its position is given in polar coordinates as x = Rcos?, and y = Rsin?, where R is the radius of the circle, and ? is in radians. From these equations derive the equation for centripetal acceleration.

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