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An object moves at a constant speed in a circular path of radius $r$ at a rate of 1 revolution per second. What is its acceleration?(A) $\mathrm{o}$(B) 2$\pi^{2} r$(C) 2$\pi^{2} r^{2}$(D) 4$\pi^{2} r$

Physics 101 Mechanics

Chapter 7

Uniform Circular Motion, Newton’s Law of Gravitation, and Rotational Motion

Gravitation

Dynamics of Rotational Motion

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for this problem On the topic off rotational motion, we're told that an object is moving at a constant speed in a circular path off radius R. And we know that it covers one revolution every second we want to find its acceleration. So we're given the angular speed and the endless BB noise. Omega, we're told Omega is equal to one revolution a second now. One revolution per second is 360 degrees per second, which is to buy radiance. So remember, we made the conversion that one revolution is to I radiance. So if we have the angular speed in the proper units, which is civilians per second, we can now calculate the linear speed. V and V is equal to omega tens I So we is only good times are where r is the radius off the motion. And that's two pi times are and that's units off meters per second. Now that we have the velocity, we could find the acceleration off these interpret emotion A and we know for these interpret emotion is integral acceleration. A is equal to V squared over our and the squared from above. So we have V we squared instantly for pi squared r squared. So a is for pi squared r squared over art, which gives us the acceleration to be for Paice. Crib times are, which is solution deep.

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The number 2 is also the smallest & first prime number (since every other even number is divisible by two).

If you write pi (to the first two decimal places of 3.14) backwards, in big, block letters it actually reads "PIE".

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