SATHEE: Kinematics Question 655

Kinematics Question 655

Question: A stone tied to the end of a string of 1 m long it’s whirled in a horizontal circle with a constant speed. If the stone makes 22 revolution in 44 seconds, What is the magnitude and direction of acceleration of the stone?

Options:

A) $π^{2} m s^{- 2}$ and direction along the radius towards the center.

B) $π^{2} m s^{- 2}$ and direction along the radius away from the center.

C) $π^{2} m s^{- 2}$ and direction along the tangent to the circle.

D) $π^{2} /4m s^{- 2}$ and direction along the radius towards the center.

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Answer:

Correct Answer: A

Solution:

[a] $a_{r} - ω^{2} R$ $a_{r} = {(2 π 2)}^{2} R = 4 π^{2} 2^{2} R$

$= 4 π^{2} {(\frac{22}{44})}^{2} (1) [∵ v = \frac{22}{44}]$ .

$a_{t} = \frac{dv}{dt} = 0$

$a_{n e t} = a_{t} = π^{2} m s^{- 2}$ and direction along the radius towards the center.

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Kinematics Question 655

Question: A stone tied to the end of a string of 1 m long it’s whirled in a horizontal circle with a constant speed. If the stone makes 22 revolution in 44 seconds, What is the magnitude and direction of acceleration of the stone?

Options:

Answer:

Solution:

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