A stone is tied to the end of a string 1 m long is whirled in a horizontal circle with a constant speed....
Answers
Complete Question: A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 seconds, what is the magnitude and direction of acceleration of the stone?
Answer:
Direction = π² m/s²
Magnitude = Along the radius towards the center.
Explanation:
[This is a case of Uniform Circular Motion]
Given;-
Radius, R = 1 m
Frequency, f= 22/44
f= 1/2 Hz
Now;-
We know that, by formula;-
Ac = ω² R ______(1)
where, Ac = Centripetal Acceleration
And, ω = 2π/t [By formula]
ω = 2π f ___(2) [where, f= frequency] & [Since f= 1/t]
On putting (2) in (1), we get;-
Ac = (2π f)² × R
Ac = 4π² × 1/4 × 1 [since, R= 1m]
Ac = π² m/s²
Now; Note that in a uniform circular motion, the centripetal acceleration always acts in the direction along the radius and towards the centre.
Hence,
Direction = π² m/s²
Magnitude = Along the radius towards the center.
Hope it helps! ;-))