From §4.4 we know that an object in uniform circular motion has centripetal acceleration a_c = v^2/r toward the center. The key question now: which force provides that acceleration?
Newton's second law for circular motion
\[ \vec F_\text{net} = m\vec a_c \Rightarrow F_\text{net,radial} = \frac{mv^2}{r} \]
Important: "centripetal force" is not a new kind of force — it's a role played by an existing force (friction, rope tension, gravity, normal, electromagnetism).
Example 1 — car in a curve
Car at speed v on a horizontal curve of radius r. Friction between tires and road provides the centripetal force:
\[ f_s = \frac{mv^2}{r} \]
Maximum speed without slipping: when static friction hits its ceiling (\mu_s N = \mu_s mg):
\[ \mu_s mg = \frac{mv_\text{max}^2}{r} \Rightarrow v_\text{max} = \sqrt{\mu_s g r} \]
Numerical example: \mu_s = 0.7, r = 50 m:
\[ v_\text{max} \approx 18.5\ m/s \approx 66.6\ km/h \]
Faster than that, friction can't provide the centripetal acceleration — car slides.
Example 2 — banked turn
Road banked at angle \theta. Without friction (design speed), only the normal force provides the centripetal acceleration.
Decompose normal force:
- Vertical:
N\cos\theta = mg - Horizontal (radial):
N\sin\theta = mv^2/r
Divide:
\[ \tan\theta = \frac{v^2}{gr} \]
\[ v_\text{design} = \sqrt{gr\tan\theta} \]
Result: at a banked turn, without friction you can navigate at v_\text{design}. Faster or slower requires friction.
Example 3 — conical pendulum
A ball on a string of length L swings in a horizontal circle. String makes angle \theta with vertical. Circle radius: r = L\sin\theta.
Decompose tension:
- Vertical:
T\cos\theta = mg - Horizontal:
T\sin\theta = mv^2/r
\[ \tan\theta = \frac{v^2}{gr} = \frac{v^2}{g L\sin\theta} \]
\[ v^2 = gL\sin\theta\tan\theta \]
Example 4 — vertical loop
Roller coaster on a vertical loop. At the top of the loop, gravity AND normal both point down (toward center):
\[ N + mg = \frac{mv^2}{r} \]
Minimum speed at top for the coaster not to lose contact (N = 0):
\[ v_\text{min} = \sqrt{gr} \]
At the bottom: normal points up (toward center), gravity down:
\[ N - mg = \frac{mv^2}{r} \Rightarrow N = m\left(g + \frac{v^2}{r}\right) \]
Rider feels "heavier" — several times body weight.
Example 5 — satellite in orbit
Gravity plays the centripetal role:
\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \Rightarrow v = \sqrt{\frac{GM}{r}} \]
Moon: r = 3.84 \times 10^8 m, GM = 4 \times 10^{14}:
\[ v \approx 1\ km/s \]
Key points
1. "Centripetal force" is not a distinct kind. Always sourced from an ordinary force. Friction, rope, gravity, ...
2. Centripetal always points to the center. Otherwise it's not circular motion.
3. Constant speed doesn't mean zero force. In uniform circular motion, force is always present.
4. Only the radial component matters for circular motion. Tangential components appear in §6.3 — when speed varies.
What you should be able to do
- Second law in circular motion —
F_\text{net,radial} = mv^2/r - Identify the force playing the centripetal role in various scenarios
- Compute maximum speed for a horizontal turn
- Derive the design angle for a banked turn
- Vertical loop dynamics — normal force at top and bottom
Preview of §6.3
What if the speed also changes? A tangential acceleration adds. §6.3.
📚 See also: Halliday Vol 1, Ch 6, §6.5.
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