A stone tied to the end of a string 80cm long is whirled in a horizontal circle with a constant speed. If the same stone makes 14 evolutions in 25 sec, what is the magnitude of the acceleration of the stone?

Explanation

The following is the solution:The formula for centripetal acceleration can be written as:a = v2/ r = rω2where v is the velocity of the object, r is the radius of the circular path, and ω is the angular velocity of the object.We know that the object is making 14 revolutions in 25 seconds, so the time for one revolution is:t = 25 s / 14 = 1.79 s/revThe angular velocity of the object can be calculated using the formula:ω = 2π / twhere t is the time for one revolution.Substituting the given values, we get:ω = 2π / 1.79 s/rev = 3.51 rad/sNow we can calculate the centripetal acceleration using the formula:a = rω2Substituting the given values, we get:a = (0.8 m)(3.51 rad/s)2 = 9.91 m/s2Therefore, the magnitude of the acceleration is 9.91 m/s2, directed along the radius towards the centre of the circular path.