OpenStudy (anonymous):

a stone tied to the end of a string is swung with a constant speed around a horizontal circle with a radius of 1.5m. It makes two complete revolutions each second. what is its acceleration?

5 years ago
OpenStudy (anonymous):

in circular motion acceleration is given by $a= v ^{2}/r$ where v is the velocity and r is the radius and the velocity can be calculated from $v=x/t=2\pi r/ t$ here x is the circumference t is the time of one revolution that's one way to solve this problem

5 years ago
OpenStudy (anonymous):

let me solve this for u $a=2(2\times3.14\times1.5)^{2}/1.5$ a=118.31m/s^2 if i ddnt do anything wrong :P

5 years ago
OpenStudy (anonymous):

I got the answer as 236.6304 m/s^2. Here's how: we know that it completes 2 revolutions in one sec. So, its time period T = 1/2 sec we also know the equation, T = 2 pi / w ..............w is the angular velocity from this eq., we get w = 2pi/T = 4pi now, another formula for centripetal acceleration is a = r . w^2 = 1.5[(4) (3.14)]^2 = 236.6304 m/s^2

5 years ago
OpenStudy (anonymous):

@theyatin u missed out squaring the t in velocity equation. then the lone 2 in the numerator will become 4.

5 years ago
OpenStudy (anonymous):

well i think i forgot multiplying by 2 as a= 2 (...)/r sorry for the mistake

5 years ago