Question

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?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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