Question

A bicycle wheel is rotating at 49 rpm when the cyclist begins to pedal harder, giving the wheel a constant angular acceleration of 0.45 rad/s2 . what is the wheel's angular velocity, in rpm, 9.0 s later?

Answer 1

In 9 sec, it increases the angular velocity by (0.45 x 9) rad/s which will give us 4.05 rad/s 
Now get the angular velocity and divide it 2pi = 4.05 by 2(pi) to give how many revolutions 4.05 rad is equivalent to = 0.6446 rps 
Them, multiply this by 60 to get it from rps to rpm increase (0.6446 x 60)=38.676 rpm 
Add this and the starting revolution frequency of 49 rpm to give: (49 rpm + 38.676 rpm) = 87.6760 rpm

Answer 2

Answer:

Angular velocity, $\omega_f=9.180\ rad/s$

Explanation:

It is given that

Initial angular velocity of the wheel, $\omega_o=49\ rpm$

Since, 1 revolutions per minute =  0.10471 radians per second

$\omega_o=49\ rpm=5.1307\ rad/s$

Angular acceleration of the wheel, $\alpha =0.45\ rad/s^2$

Time taken, t = 9 s

Let $\omega_f$ is the final velocity of the wheel 9 seconds later. Using the equation of kinematics to find it :

$\omega_f=\omega_o+\alpha t$

$\omega_f=5.1307+0.45\times 9$

$\omega_f=9.180\ rad/s$

So, the final angular velocity of the wheel is 9.180 rad/s. Hence, this is the required solution.