Question

A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of-5.00 rad/s2·orng a 3.40-s time interval, the wheel rotates through 80.4 rad. What is the angular speed of the wheel at the end of the 3.40-s interval? rad/s Need Help? Read it Next

Answer 1

Given is :-

Angular acceleration  $\alpha = -5 rad/s^2$

time taken  $t = 3.40s$

Angular displacement = 80.4 radians

Now,

by using the first equation of motion the angular displacement is given by

$\theta = w_o t + \frac{1}{2}\alpha t^2$

by plugging all the values we get

$80.4= w_o (3.40) - \frac{1}{2}(5) (3.40)^2$

which gives us initial angular velocity which is

$w_o = 32.15 rad/s$

Now by using third equation of motion we get

$w^2 = w_o^2 + 2 \alpha \theta$

by plugging all the values we get

$w^2 = (32.15)^2 - 2 (5) (80.4)$

which gives us the final angular velocity after the breaking of 3.40 seconds

$w = 15.15 rad/s$

Hence the angular speed of the wheel at the end of 3.40 seconds is 15.15 rad/s