Question

any one taking university physics

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.

Answer 1

does physical science count or no?

Answer 2

not really

Answer 3

Then im not sure i can help, but ill try. What do u need help on?

Answer 4

It depends on what you need help on

Answer 5

Do you have any school-related questions??

Answer 6

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.

Answer 7

Ok, i think i can help.

Answer 8

kk

Answer 9

wait is this the right answer d2x/dt2 = -Rw2

Answer 10

We only need to look at the equation for the x-position, as we know that centripetal acceleration points towards the center of the circle.

Answer 11

do you have a ss of this all being seperated apart

Answer 12

So, when θ = 0, the second derivative of x with respect to time has to be the centripetal acceleration.

Answer 13

So, what do you think the first derivative of x is?

Answer 14

dx/dt = -Rsinθ(dθ/dt)

Answer 15

Correct, and what you think the second is?

Answer 16

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

Answer 17

That is correct, because we usethe chain rule of Calculus and by assumption θ is a function of time.

Answer 18

So, θ can be differentiated with respect to time.

Now we have to evaluate the second derivative at θ = 0.

Answer 19

Do you know wat comes next?

Answer 20

is it this am not sure d2x/dt2 = -R(dθ/dt)2

Answer 21

You're close.

Answer 22

dθ/dt is called the angular velocity, which is the rate of change of the angle θ. So, we'll let
w ≡ dθ/dt.

Answer 23

So, your final answer is d^2x/dt2 = -Rw^2

Answer 24

kk ty

Answer 25

yw anyday

Answer 26

gj