Kinematics of uniform circular motion — practice questions

1. Convert period to seconds: $T = 90 \text{ min} \times 60 \text{ s/min} = 5400 \text{ s}$.
2. Calculate angular velocity: $\omega = \frac{2\pi}{T} = \frac{2\pi}{5400} \approx 0.00116 \text{ rad s}^{-1}$.
3. Calculate linear speed: $v = \omega r = (0.00116)(7.0 \times 10^6) \approx 8120 \text{ m s}^{-1}$.
4. Calculate centripetal acceleration: $a_c = \frac{v^2}{r} = \frac{(8120)^2}{7.0 \times 10^6} \approx 9.43 \text{ m s}^{-2}$. Alternatively, $a_c = \omega^2 r = (0.00116)^2 (7.0 \times 10^6) \approx 9.41 \text{ m s}^{-2}$ (slight difference due to rounding).

First, list the known quantities: Mass, $m = 1200$ kg Linear speed, $v = 20$ m s⁻¹ Radius, $r = 50$ m

(a) Calculate angular velocity ($\omega$): We use the relationship $v = \omega r$. Rearranging for $\omega$: $\omega = \frac{v}{r}$ $\omega = \frac{20 \text{ m s}^{-1}}{50 \text{ m}} = 0.40 \text{ rad s}^{-1}$

(b) Calculate centripetal acceleration ($a_c$): We can use the formula $a_c = \frac{v^2}{r}$. $a_c = \frac{(20 \text{ m s}^{-1})^2}{50 \text{ m}} = \frac{400}{50} = 8.0 \text{ m s}^{-2}$ Alternatively, using $\omega$ from part (a): $a_c = \omega^2 r = (0.40 \text{ rad s}^{-1})^2 \times 50 \text{ m} = 0.16 \times 50 = 8.0 \text{ m s}^{-2}$.

(c) Calculate the minimum frictional force ($F_f$): The centripetal force required to keep the car on the circular path is provided by the static friction between the tyres and the road. Therefore, the frictional force must be equal to the centripetal force. $F_f = F_c = ma_c$ $F_f = (1200 \text{ kg}) \times (8.0 \text{ m s}^{-2}) = 9600 \text{ N}$ The minimum frictional force required is 9600 N.