Question

A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of

−4.60 rad/s².

During a 2.20-s time interval, the wheel rotates through 52.9 rad. What is the angular speed of the wheel at the end of the 2.20-s interval?
rad/s

Answer 1

We can use the equation of motion:

$\theta=\theta_{0}+w_{0}\cdot t+a\cdot t^{2}$

when   

$\theta=\theta_{0}+w_{0}\cdot (0[s])+(-4.60[\frac{rad}{s^{2}}])\cdot (0[s])^{2}=\theta_{0}$

During the 2.20 seg time interval the wheel rotates 52.9 rad, we substitute the values on the equation:

$52.9[rad]+\theta_{0}=\theta_{0}+w_{0}\cdot (2.20[s])+(-4.60[\frac{rad}{s^{2}}])\cdot (2.20[s])^{2}$

$52.9[rad]=w_{0}\cdot (2.20[s])+(-4.60[\frac{rad}{s^{2}}])\cdot (2.20[s])^{2}$

Solving for $w_{0}$

$w_{0}=\frac{52.9[rad]+(4.60[\frac{rad}{s^{2}}])\cdot (2.20[s])^{2}}{2.20[s]}=34.16 [\frac{rad}{s}]$

Now we use the equation of angular speed:

$w=w_{0}+a\cdot t$

$w=34.16[\frac{rad}{s}]-4.60[\frac{rad}{s^{2}}]\cdot (2.2 [s])=24.04[\frac{rad}{s}]$

$w=24.04[\frac{rad}{s}]$