Centripetal Force

**saxyliz** · Dec 4th 2008, 02:55 PM

A body of mass 9 kg moves in a (counterclockwise) circular path of radius 5 meters, making one revolution every 9 seconds. You may assume the circle is in the xy-plane, and so you may ignore the third component.

Compute the centripetal force acting on the body. (As a vector of 2 components.)

I know the first part of the two components, but I need help solving for the last parts.

$<[(-9(10\pi/9)^2) / 5][cos((10\pi/9)t) / ... ], [(-9(10\pi/9)^2) / 5][sin((10\pi/9)t) / ... ]>$

Please help!

**skeeter** · Dec 4th 2008, 03:25 PM

uniform circular motion with angular speed $\omega = \frac{2\pi}{9} \, rad/sec$

position ...

$x = 5\cos(\omega t)$

$y = 5\sin(\omega t)$

velocity ...

$\frac{dx}{dt} = -5\omega\sin(\omega t)$

$\frac{dy}{dt} = 5\omega\cos(\omega t)$

acceleration ...

$\frac{d^2x}{dt^2} = -5\omega^2\cos(\omega t)$

$\frac{d^2y}{dt^2} = -5\omega^2\sin(\omega t)$

**saxyliz** · Dec 4th 2008, 08:19 PM

Originally Posted by skeeter

uniform circular motion with angular speed $\omega = \frac{2\pi}{9} \, rad/sec$

position ...

$x = 5\cos(\omega t)$

$y = 5\sin(\omega t)$

velocity ...

$\frac{dx}{dt} = -5\omega\sin(\omega t)$

$\frac{dy}{dt} = 5\omega\cos(\omega t)$

acceleration ...

$\frac{d^2x}{dt^2} = -5\omega^2\cos(\omega t)$

$\frac{d^2y}{dt^2} = -5\omega^2\sin(\omega t)$

Thank you so much!! I'm so rusty with physics, I didn't know how to use the information given.

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