Make sure that you have all of the uniform circle motion equations handy as you complete the following problems that involve centripetal force.
Uniform Circular Motion
\(\mathsf{ T = \frac{1}{f} }\)
\(\mathsf{ f = \frac{1}{T} }\)
\(\mathsf{ v_c = \frac{C}{T} = \frac{\pi \times 2r}{T} }\)
\(\mathsf{ a_c = \frac{v^2}{r} }\)
\(\mathsf{ F_c = ma_c = \frac{mv^2}{r} }\)
| Problem | Picture | Given/Find | Equation | Solution |
|---|---|---|---|---|
| A 63 kg softball player is running at 5.5 m/s as she rounds second base on curved path with a radius of 15 m. What is the centripetal force? |
|
\(\small\mathsf{ v_c = 5.5 \text{ m/s} }\) \(\small\mathsf{ m = 63 \text{ kg} }\) \(\small\mathsf{ r = 15 \text{ m}}\) \(\mathsf{ F_c = ? \text{ N} }\) |
\(\small\mathsf{ F_c = ma_c = \frac{mv^2}{r} }\) | \(\mathsf{ F_c = \frac{(63 \text{ kg})(5.5 \text{ m/s})^2}{15 \text{ m}} }\) \(\mathsf{ F_c = 130 \text{ N} }\) |
| A 0.015 kg ball is shot from the plunger of a pinball machine. The ball follows a circular arc with a radius of 0.25 m. If the centripetal force is 0.028 N, what is the speed of the ball? |
|
\(\small\mathsf{ v_c = ? \text{ m/s} }\) \(\small\mathsf{ m = 0.015 \text{ kg} }\) \(\small\mathsf{ r = 0.25 \text{ m}}\) \(\mathsf{ F_c = 0.028 \text{ N} }\) |
\(\small\mathsf{ F_c = ma_c = \frac{mv^2}{r} }\) | \(\small\mathsf{ 0.028 \text{ N} = \frac{(0.015 \text{ kg})(v)^2}{0.25 \text{ m}} }\) \(\small\mathsf{ v_c = \sqrt{\frac{0.028 \text{ N} \times 0.25 \text{ m}}{0.015 \text{ kg}}} }\) \(\small\mathsf{ v_c = 0.68 \text{ m/s} }\) |
| A 72.0 kg stunt pilot takes his plane into a steep dive. The plane follows a curved path with a radius of 360. meters while traveling at 120. m/s. What centripetal force does the pilot feel? |
|
\(\small\mathsf{ v_c = 120 \text{ m/s} }\) \(\small\mathsf{ m = 72.0 \text{ kg} }\) \(\small\mathsf{ r = 360 \text{ m}}\) \(\mathsf{ F_c = ? \text{ N} }\) |
\(\small\mathsf{ F_c = ma_c = \frac{mv^2}{r} }\) | \(\mathsf{ F_c = \frac{(72.0 \text{ kg})(120. \text{ m/s})^2}{360. \text{ m}} }\) \(\mathsf{ F_c = 2880 \text{ N} }\) |
| A person is flying a 0.0822 kg model airplane in a horizontal circular path on the end of a string 10.5 m long. The string is horizontal and exerts a force of 3.22 N on the hand of the person holding it. What is the speed (in m/s) of the plane? |
|
\(\small\mathsf{ v_c = ? \text{ m/s} }\) \(\small\mathsf{ m = 0.0822 \text{ kg} }\) \(\small\mathsf{ r = 10.5 \text{ m}}\) \(\mathsf{ F_c = 3.22 \text{ N} }\) Note: the force the plane exerts on the hand is equal and opposite to the force the hand exerts on the plane. |
\(\small\mathsf{ F_c = ma_c = \frac{mv^2}{r} }\) | \(\small\mathsf{ 3.22 \text{ N} = \frac{(0.0822 \text{ kg})(v)^2}{10.5 \text{ m}} }\) \(\small\mathsf{ v_c = \sqrt{\frac{3.22 \text{ N} \times 10.5 \text{ m}}{0.0822 \text{ kg}}} }\) \(\small\mathsf{ v_c = 20.3 \text{ m/s} }\) |
