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The motor of an engine is rotating about its axis with angular velocity of 120 r.p.m. it comes to rest in 10s, after being switched off. Assuming constant decleration, calculate the number of revolutions made by it before coming to rest. |
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Answer» Solution :Here, `omega_(0)=120 ` r.p.m. = 2 r.p.s. `= 4pi rad s^(-1)` `omega = 0`and `t=10 s "" omega = omega_(0)+at` `0=4pi+axx10` or `a=-0.4 PI rad s^(-2)` Also, the angle covered by the MOTOR, `THETA = omega_(0)t+(1)/(2)at^(2)` `because theta = 4pi xx10+(1)/(2)xx(-0.4 pi)xx10^(2)=40 pi-20pi` `= 20 pi` rad Hence, the number of revolutions completed, `n =(theta)/(2PI)=(20pi)/(2pi)=10` |
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