Question

The angular velocity of a 755-g wheel 15.0 cm in diameter is given by the equation .

w(t)= (2.00 rad/s^2)t + (1.00rad/s^4)t^3

(a) Through how many radians does the wheel turn during the first 2.00 s of its motion?

(b) What is the angular acceleration (in rad/s2) of the wheel at the end of the first 2.00 s of its motion?

Answer 1

a)

$$\theta=\displaystyle\intop_{0}^2\omega(t)dt,$$ $$\theta=\displaystyle\intop_{0}^2(2t+t^3)dt=(2\ s)^2\cdot \dfrac{rad}{s^2}+\dfrac{(2\ s)^4}{4}\cdot \dfrac{rad}{s^4}=8\ rad.$$

b) The angular acceleration is the derivative of angular velocity with respect to time:

$$\alpha=\dfrac{d\omega(t)}{dt},$$ $$\alpha=\dfrac{d}{dt}(2t+t^3)=2+3t^2.$$

Finally, substituting $t=2$ into the equation for $\alpha$ we get:

$$\alpha=2\ \dfrac{rad}{s^2}+3\ \dfrac{rad}{s^4}\cdot(2\ s)^2=14\ \dfrac{rad}{s^2}.$$