Worked example 8.1: Balancing tires

Question: A tire placed on a balancing machine in a service station starts from rest and turns through $5.3$ revolutions in $2.3 {\rm s}$ before reaching its final angular speed. What is the angular acceleration of the tire (assuming that this quantity remains constant)? What is the final angular speed of the tire?

Answer: The tire turns through $\phi=5.3\times 2 \pi = 33.30 {\rm rad.}$ in $t=2.3 {\rm s}$. The relationship between $\phi$ and $t$ for the case of rotational motion, starting from rest, with uniform angular acceleration $\alpha$ is

$$\phi = \frac{1}{2} \alpha t^2.$$

Hence,

$$\alpha = \frac{2 \phi}{t^2} = \frac{2\times 33.30}{2.3^2} = 12.59 {\rm rad./s^2}.$$

Given that the tire starts from rest, its angular velocity after $t$ seconds takes the form

$$\omega = \alpha t = 12.59\times 2.3 =28.96 {\rm rad./s}.$$