A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If th

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 s, what is the magnitude and direction of acceleration of the stone?

Accepted Answer

1.  **Identify Given Values:** The problem input contains repeated digits due to formatting errors. Based on the options and standard physics problems (PMT 2005), the values are interpreted as: Radius ($r$) = 1 m, Number of revolutions ($n$) = 22, Time ($t$) = 44 s.
2.  **Calculate Angular Velocity ($\omega$):** The angular velocity is given by $\omega = \frac{2\pi n}{t}$.
    Substituting values: $\omega = \frac{2\pi (22)}{44} = \frac{44\pi}{44} = \pi 	ext{ rad/s}$ .
3.  **Calculate Acceleration Magnitude:** The acceleration in uniform circular motion is the centripetal acceleration, given by $a_c = \omega^2 r$.
    $a_c = (\pi)^2 (1) = \pi^2 	ext{ m/s}^2$.
4.  **Determine Direction:** By definition, centripetal acceleration is always directed along the radius towards the center of the circle .

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