(a) Angular acceleration
The initial angular speed of the wheel is [tex] \omega_i=0 [/tex], the final angular speed is [tex] \omega_f=25 rad/s [/tex], and the time taken is t=10 s. Therefore, the angular acceleration of the wheel is given by:
[tex] \alpha=\frac{\omega_f-\omega_i}{t}=\frac{25 rad/s-0}{10 s}=2.5 rad/s^2 Â [/tex]
(b1) Tangential acceleration
The tangential acceleration is given by the product between the angular acceleration [tex] \alpha [/tex] and the distance from the wheel's center r. In this case, [tex] \alpha=2.5 rad/s^2 Â [/tex] Â and [tex] r=10 cm=0.1 m [/tex], therefore the tangential acceleration is
[tex] a_t=\alpha r=(2.5 rad/s^2)(0.1 m)=0.25 m/s^2 [/tex]
b2) Radial acceleration
The radial acceleration (also called centripetal acceleration) is given by:
[tex] a_c =\omega^2 r [/tex]
where [tex] \omega=25 rad/s [/tex] is the final angular speed while r=0.1 m is the distance from the center of the wheel. Substituting numbers, we get
[tex] a_c=\omega^2 r=(25 rad/s)^2(0.1 m)=62.5 m/s^2 [/tex]
c) Number of revolutions
First of all, we need to find the angle covered during this time interval, which is given by:
[tex] \theta=\omega_i t+ \frac{1}{2}\alpha t^2=\frac{1}{2} \alpha t^2=\frac{1}{2}(2.5 rad/s^2)(10 s)^2=125 rad [/tex]
And keeping in mind that [tex] 1 rev=2 \pi rad [/tex], the number of revolutions made is:
[tex] \theta = \frac{125 rad}{2 \pi rad/rev}=19.9 rev [/tex]
d) Deceleration
In this last part of the problem, we are told that the wheel comes to a stop after [tex] \theta=5 rev=31.4 rad [/tex]. We also know the initial angular speed, [tex] \omega_i =25 rad/s [/tex], and the final angular speed, [tex] \omega_f=0 Â [/tex], so we can find the new angular (de)celeration by using the equation:
[tex] 2\alpha \theta=\omega_f^2-\omega_i^2 [/tex]
Substituting numbers, we get
[tex] \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}=\frac{0-(25 rad/s)^2}{2(31.4 rad)}=-9.95 rad/s^2 [/tex]
