A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ?
Given: The length of string is 80 m and a stone tied to it makes 14 revolutions in 25 s .
Frequency of revolutions is given as,
f = n t
Where, the number of revolution made is n and time taken to complete the revolutions is t .
By substituting the given values in the above expression, we get
f = 14 25 Hz
Angular frequency of revolutions is given as,
ω = 2 π f
By substituting the given values in the above expression, we get
ω = 2 π 14 25 = 88 25 rad s − 1
Centripetal acceleration is given as,
a e = ω 2 l
Where, l is the length of string.
By substituting the given values in the above expression, we get
a e = ( 88 25 ) 2 × 0.8 = 9.91 ms − 2
Thus, the magnitude of centripetal acceleration is 9.91 ms − 2 and its direction is always directed along the string, toward the centre, at all points.
Given: The length of string is
Frequency of revolutions is given as,
Where, the number of revolution made is
By substituting the given values in the above expression, we get
Angular frequency of revolutions is given as,
By substituting the given values in the above expression, we get
Centripetal acceleration is given as,
Where,
By substituting the given values in the above expression, we get
Thus, the magnitude of centripetal acceleration is
