Question

The angular velocity of a particle moving in a circle of radius 50 cm is increased in 5 min from 100 revolutions per minute to 400 revolutions per minute. Find tangential acceleration of the particle.

Answer 1

r= 0.5m = 1÷2

T =5min =300 s

V1= 2 pi ×400÷60

v2 = 2 pi ×100÷60

Angular velocity = tv

V 1 = 2 pi ×100 ÷120 = 15 pi ÷3

V2= 2 pi ×400 ÷120 = 20 pi ÷ 3

A = v2 _ v1 ÷ t

=( 20 pi ÷3 - 15 pi ÷ 3 ) ÷ 300

= pi ÷ 60 m /s^2

Answer 2

Answer:

Tangential acceleration of the particle corresponds to π/60ms⁻².

Explanation:

At time $t=0$, the angular velocity of particle ω₁ $=\frac{200\pi }{60 }rads^{-1}$

After time $t=5min$, angular velocity becomes ω₂  $=\frac{800\pi }{60 }rads^{-1}$

We know  angular acceleration $\alpha = \frac{(\omega _{2} -\omega _{1})}{T}$

where T = total time and $T=5*60=300sec$

Here $\alpha =\frac{1}{300sec} (\frac{800\pi }{60 }rads^{-1}-\frac{200\pi }{60 }rads^{-1})=\frac{\pi }{30} rads^{-2}$

For a particle  the tangential acceleration  $\alpha=\alpha r$

where $r = 0.5m=$ radius of the circle

  ∴             $a=(\frac{\pi }{30}) (0.5)ms^{-2}\\ \\ a=\frac{\pi }{60} ms^{-2}$