A pulley of radius 2m is rotated about its axis by a force F = (20t – 5t2)
newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2, the number of rotations made by the pulley before its direction of motion if reversed, is :
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To reverse the direction òtdq = 0 (work done is zero)
t = (20 t – 5t2) 2 = 40t – 10t2
\(\alpha = \frac{\tau }{I} = \frac{{40t - 10{t^2}}}{{10}} = 4t - {t^2}\)
\(\omega = \int_0^t {\alpha dt = 2{t^2} - \frac{{{t^3}}}{3}} \)
w is zero at
\(2{t^2} - \frac{{{t^3}}}{3} = 0\)
t3 = 6t2
t = 6 sec.
q = ò wdt
\( = \int_0^6 {\left( {2{t^2} - \frac{{{t^3}}}{3}} \right)dt} \)
\(\left[ {\frac{{2{t^3}}}{3} - \frac{{{t^4}}}{{12}}} \right]_0^6 = 216\left[ {\frac{2}{3} - \frac{1}{3}} \right] = 36\,rad.\)
No or revolution \(\frac{{36}}{{2\pi }}\) Less than 6
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