The angular speed of truck wheel is increased from 900 rpm to 2460 rpm in 26 seconds. The number of revolutions by the truck engine during this time is _____________. (Assuming the acceleration to be uniform).
Answer (integer)
728
Solution
${\omega _f} = 2460 \times {{2\pi } \over {60}}$<br><br>$= 82\pi$<br><br>${\omega _i} = {{900 \times 2\pi } \over {60}} = 30\pi$<br><br>$\alpha = {{{\omega _f} - {\omega _i}} \over t}$<br><br>$= {{82\pi - 30\pi } \over {26}}$<br><br>= 2 $\pi$ rad/sec<sup>2</sup><br><br>$\theta = {{\omega _f^2 - \omega _i^2} \over {2\alpha }}$<br><br>$= {{(82\pi + 30\pi )(82\pi - 30\pi )} \over {2 \times 2\pi }}$<br><br>$= {{(112 \times 52){\pi ^2}} \over {4\pi }}$<br><br>No. of revolution $= {{(112 \times 13)\pi } \over {2\pi }}$<br><br>= 728
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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