Question

1) If a cycle wheel of radius 0.4 m completes one revolution in 2 seconds, the acceleration of the cycle is:
a) 0.4 pi m/s2
by 0.4 pi^2 m/s2
c) pi^2/0.4 m/s2
d) 0.4/pi^2 m/s2​

Answer 1

Answer:

we have T=2sec ,r=0.4m

we know ,

a=rw^2= r(2πn)^2= 0.4 (2π ×1/2)^2=0.4 π^2m/s^2

Answer 2

Answer:

The acceleration of the cycle is $0.4\pi ^{2}m/s^{2}$.

Explanation:

Given a cycle wheel of radius,  $r=0.4 m$

         time for one revolution, $t= 2 s$

Cycle completes one revolution in 2s,

         therefore frequency, $f=\frac{1}{T}$

                                                $=\frac{1}{2}Hz$

and hence angular frequency,

                                            $\omega=2\pi f$

                                               $=2\pi \times \frac{1}{2} =\pi rad/s$

Then, the acceleration of the cycle,

                                            $a=r\omega^{2}$

                                               $=0.4\pi ^{2}m/s^{2}$

Therefore, the acceleration of the cycle is $0.4\pi ^{2}m/s^{2}$.