Question #123984

Awheel rotating about a fixed axis at 24 r.p.m. is uniformly accelerated for 70 seconds,

during which time it makes 50 revolution. Find (i) angular velocity at the end of this

interval, and (ii) time required for the speed to reach 150 r.p.m.

(Ans. - 61.4 r.p.m. ;3 min 55.6 s)

Expert's answer

Explanations & Calculations


  • Equivalent forms of the equations used in linear motion under uniform acceleration are used in rotational motion.

1).

  • Apply θ=(ωfinal+ωinitial2)t\small \theta = \Big(\frac{\omega_{final}+\omega_{initial}}{2} \Big)\small t to calculate the angular velocity.

50rev.=(ωfinal+24 r.p.m2)×(7060min)ωfinal=61.71 rpm\qquad\qquad \begin{aligned} \small 50 rev. &= \small \Big(\frac{\omega_{final}+24 \,r.p.m}{2}\Big)\times \Big(\frac{70}{60} min\Big)\\ \small \omega_{final}&= \small \bold{61.71 \,rpm } \end{aligned}


  • Before stepping forward, angular acceleration (α\alpha )should be calculated. Apply ωfinal=ωinitial+αt\small \omega_{final} = \omega_{initial} +\alpha t to calculate it.

α=ωf−ωIt=61.71−24(7060)⋯⋯(1)\qquad\qquad \begin{aligned} \small \alpha &= \small \frac{\omega_f -\omega_I}{t}\\ \small &= \small \frac{61.71 -24}{(\frac{70}{60})}\cdots\cdots(1)\\ \end{aligned}

  • And there is no need to calculate it as a final figure since it's not asked to calculate.

2)

  • Apply again the above relationship from start to when it makes 150 rpm s.

α=150rpm−24rpmt⋯⋯(2)\qquad\qquad \begin{aligned} \small \alpha &= \small \frac{150 rpm -24rpm}{t} \cdots\cdots(2) \end{aligned}

  • By (1) = (2), time take to reach 150 rpm s,

61.71−24(7060)=150rpm−24rpmtt=3.898min⋯(3min+0.898min)=3min+(0.898×60s)=3min 53.9s\qquad\qquad \begin{aligned} \small \frac{61.71 -24}{(\frac{70}{60})} &= \small \frac{150 rpm -24rpm}{t} \\ \small t &= \small 3.898min\cdots(3min+0.898min)\\ &= \small 3min +(0.898\times60s)\\ &= \small \bold{3min\,53.9s} \end{aligned}

\qquad\qquad \begin{aligned} \end{aligned}


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