Physics, asked by PhysicsHelper, 1 year ago

A wheel rotating at a speed of 600 rpm (revolutions per minute) about its axis is brought to rest by applying a constant torque for 10 seconds. Find the angular deceleration and the angular velocity 5 seconds after the application of the torque. ?

Answers

Answered by tiwaavi
56

Given in the question :-


Rotation of the wheel(Initial) = 600 rpm or  \omega  = 10 revolution / second

Final velocity  \omega' = 0

Time = 10 second

now we have to find  \alpha  = ?


We know the formula ,


 \omega' = \omega - \alpha t

put all the values in the given formula

0 = 10 - α × 10

α = 1 revolution /second ²


Now  \omega'' the angular velocity at time change t = 5 sec.


 \omega'' = \omega - \alpha t

ω'' = 10 - 1 × 5

ω'' = 5 revolution/second.



Hope it Helps :-)

Answered by Deepsbhargav
17
» Initial angular velocity of the wheel :-

 = > v _{0} = 600 \: \: rpm \\ \\ = > v _{0} = \frac{600}{60} = 10 \: \: \frac{rev}{sec}
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» After 10 sec final angular velocity of the wheel :-

=> v = 0

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So

 = > v _{0} = - at \\ \\ = > a = - \frac{v _{0} }{t} = - \frac{10}{10} \\ \\ = > a = - 1 \: \: rev. {sec}^{ - 2}
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Now,

=> t = 5 sec

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» We know that :-

 = > v' = v _{0} + at \\ \\ = > v' = 10 - 1 \times 5 \\ \\ = > v' = 5 \: \: rev. {sec}^{ - 1}
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Deepsbhargav: theku Dimpy
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