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 ms–2) is of the order of :
Solution
R = 0.1 m
<br><br>$\omega$ = ${{2\pi } \over T}$ = ${{2\pi } \over {60}}$ = 0.105 rad/sec
<br><br>a = ${\omega ^2}R$
<br><br>= (0.105)<sup>2</sup>(0.1)
<br><br>= 0.0011
<br><br>= 1.1 $\times$ 10<sup>-3</sup>
<br><br>Average acceleration is of the order of 10<sup>–3</sup>.
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Torque and Angular Momentum
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