Question:
A clock has a continuously moving second's hand of 0.1 m length. The average acceleration of the tip of the hand (in units of \(\mathrm{ms^{-2}}\)) is of the order of :
A clock has a continuously moving second's hand of 0.1 m length. The average acceleration of the tip of the hand (in units of \(\mathrm{ms^{-2}}\)) is of the order of :
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The second hand turns through \(2\pi\) in 60 s, so \(\omega = 2\pi/60\). Compute the tip speed and acceleration scale \(\omega^2 r\) (or \(\Delta v/\Delta t\) over a fraction of a turn).
Updated On: Nov 5, 2026
- \(10^{-3}\)
- \(10^{-1}\)
- \(10^{-2}\)
- \(10^{-4}\)
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The Correct Option is A
Solution and Explanation
Period \(T = 60\) s, so \(\omega = \frac{2\pi}{60} = 0.105\) rad/s.
Speed of the tip: \(v = \omega r = 0.105\times 0.1 = 1.05\times 10^{-2}\) m/s.
Average acceleration over a quarter turn (15 s): \(|\Delta \vec v| = \sqrt2\,v\), so \(a_{avg} = \frac{\sqrt2\times 1.05\times 10^{-2}}{15} \approx 9.9\times 10^{-4}\approx 10^{-3}\) m/s\(^2\).
This is the same order as the instantaneous magnitude \(\omega^2 r = (0.105)^2\times 0.1 = 1.1\times 10^{-3}\) m/s\(^2\).
Order of magnitude: \(10^{-3}\) m/s\(^2\).
Answer: (1) \(10^{-3}\)
Speed of the tip: \(v = \omega r = 0.105\times 0.1 = 1.05\times 10^{-2}\) m/s.
Average acceleration over a quarter turn (15 s): \(|\Delta \vec v| = \sqrt2\,v\), so \(a_{avg} = \frac{\sqrt2\times 1.05\times 10^{-2}}{15} \approx 9.9\times 10^{-4}\approx 10^{-3}\) m/s\(^2\).
This is the same order as the instantaneous magnitude \(\omega^2 r = (0.105)^2\times 0.1 = 1.1\times 10^{-3}\) m/s\(^2\).
Order of magnitude: \(10^{-3}\) m/s\(^2\).
Answer: (1) \(10^{-3}\)
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