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A stone tied at the end of string 80cm long is whriled in a horizontal circle with a constant speed. If the stone makes 25 revolutions in 14s, what is the magnitude of acceleration of the stone?
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Calculating the magnitude of acceleration of the stone


  1. Identify the given values


    • Length of string (l) = 80cm = 0.8m

    • Number of revolutions made by stone (n) = 25

    • Time taken for n revolutions (t) = 14s



  2. Calculate the frequency (f) of the stone

  3. Frequency (f) = Number of revolutions/time taken = n/t = 25/14 Hz


  4. Calculate the angular velocity (ω) of the stone

  5. Angular velocity (ω) = 2πf = 2π(25/14) rad/s = 11.28 rad/s


  6. Calculate the tangential velocity (v) of the stone

  7. Tangential velocity (v) = ωl = 11.28 x 0.8 m/s = 9.024 m/s


  8. Calculate the centripetal acceleration (a)

  9. Centripetal acceleration (a) = v^2/l = (9.024)^2/0.8 m/s^2 = 101.8 m/s^2


  10. Final answer

  11. Therefore, the magnitude of acceleration of the stone is 101.8 m/s^2.



Explanation

This problem can be solved using the formula for centripetal acceleration, which is a = v^2/l, where v is the tangential velocity and l is the length of the string. To apply this formula, we first need to calculate the tangential velocity of the stone, which can be obtained using the formula ωl, where ω is the angular velocity of the stone. The angular velocity can be calculated using the frequency of the stone, which is the number of revolutions made by the stone per unit time. Once we have the tangential velocity, we can use it to calculate the centripetal acceleration using the formula a = v^2/l.
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A stone tied at the end of string 80cm long is whriled in a horizontal circle with a constant speed. If the stone makes 25 revolutions in 14s, what is the magnitude of acceleration of the stone?
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