Question

The angular acceleration of a fly wheel is given by α = 12 – t . If the angular velocity of fly wheel is 60 rad / sec at the end of 4 sec.
Determine the angular velocity at the end of 6 sec. How many revolutions takes place in these 6 seconds?​

Answer 1

Answer:

74 rad/s,

Explanation:

74 rad/s

is alpha=12-t

Answer 2

Answer:

In 6 seconds the angular velocity will become 74 rad/s.

Explanation:

Angular acceleration is given as,  α = 12 – t

If ω = angular velocity, then $\frac{d\omega}{dt}$ = 12 - t

                                              dω = (12 - t) dt

Now integrating both sides, ∫dω =  ∫(12 - t) dt

                                                ω = 12t - $\frac{t^2}{2}$

Suppose at end of 6 second the angular velocity is ω'. In left side putting the upper limit  ω' and lower limit 60 rad/s, in right hand side putting upper limit 6 second and lower limit 4 second we get,

                                           ω' - 60 = 12(6 - 4) - $\frac{1}{2}$(6²- 4²)

                                            ω' - 60 = 24 - 10 = 14

                                            ω' = 14 + 60 = 74

In 6 seconds the angular velocity will become 74 rad/s.

In 6th second 74 rads revolution will take place.

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