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A stone tied to the end of string 80cm long is whirled in a horizontal circle with a constant speed . If the stone makes 14 revolutions in 25 seconds what is the magnitude of acceleration of the stone?
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A stone tied to the end of string 80cm long is whirled in a horizontal...
Given data:
- Length of string (r) = 80 cm = 0.8 m
- Number of revolutions (n) = 14
- Time taken (t) = 25 s

Calculating the radius:
- The length of the string is equal to the circumference of the circle.
- So, 2πr = L
- 2πr = 0.8 m
- r = 0.8 / (2π) m
- r ≈ 0.127 m

Calculating the time period:
- The time period (T) is the time taken for one complete revolution.
- T = t / n
- T = 25 s / 14
- T ≈ 1.786 s

Calculating the angular velocity:
- The angular velocity (ω) is the rate of change of angle with respect to time.
- ω = 2π / T
- ω = 2π / 1.786 s
- ω ≈ 3.527 rad/s

Calculating the linear velocity:
- The linear velocity (v) is the rate of change of displacement with respect to time.
- v = ωr
- v = 3.527 rad/s * 0.127 m
- v ≈ 0.448 m/s

Calculating the acceleration:
- The acceleration (a) is the rate of change of velocity with respect to time.
- a = v^2 / r
- a = (0.448 m/s)^2 / 0.127 m
- a ≈ 1.581 m/s^2

Result:
The magnitude of acceleration of the stone is approximately 1.581 m/s^2.
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