Circular motion and centripetal force

Example 1: a car on a curve

Problem A 1200 kg car goes around a flat curve of radius 50.0 m at 15.0 m/s. Find its centripetal acceleration and the friction force needed.

1. Acceleration toward the center.
   $$a_c = \frac{v^2}{r} = \frac{(15.0\ \text{m/s})^2}{50.0\ \text{m}} = 4.50\ \text{m/s}^2$$
2. Friction is the only inward force, so it must equal $ma_c$.
   $$F_c = (1200\ \text{kg})(4.50\ \text{m/s}^2) = 5400\ \text{N}$$

Answer 4.50 m/s² toward the center, needing $5.40 \times 10^3$ N of friction

Example 2: a ball on a string

Problem A 0.20 kg ball is whirled in a horizontal circle of radius 0.80 m, making one turn every 0.50 s. Ignoring gravity's small effect on the string, what is the tension?

1. Speed from the period.
   $$v = \frac{2\pi r}{T} = \frac{2\pi(0.80\ \text{m})}{0.50\ \text{s}} = 10.05\ \text{m/s}$$
2. Tension provides the centripetal force.
   $$T_{\text{string}} = m\frac{v^2}{r} = (0.20\ \text{kg})\frac{(10.05\ \text{m/s})^2}{0.80\ \text{m}} = 25.3\ \text{N}$$

Answer 25 N

Example 3: top speed on a flat curve

Problem Tires and road have $\mu_s = 0.60$. What is the fastest a car can take a flat curve of radius 40. m without skidding?

1. At the limit, maximum static friction supplies the centripetal force. On flat ground $F_N = mg$.
   $$\mu_s mg = m\frac{v^2}{r}$$
2. Mass cancels. Solve for $v$.
   $$v = \sqrt{\mu_s g r} = \sqrt{(0.60)(9.8\ \text{m/s}^2)(40.\ \text{m})} = 15.3\ \text{m/s}$$

Answer 15 m/s (about 34 mph)

Example 4: the top of a loop

Problem A roller coaster goes through a vertical loop of radius 10.0 m. What is the slowest it can go at the top and still stay on the track?

1. At the top, weight and the track's normal force both point down, toward the center. The slowest speed is when the normal force just reaches zero, so weight alone provides the centripetal force.
   $$mg = m\frac{v^2}{r}$$
2. Solve.
   $$v = \sqrt{gr} = \sqrt{(9.8\ \text{m/s}^2)(10.0\ \text{m})} = 9.90\ \text{m/s}$$

Answer 9.90 m/s