Vertical Circles & Universal Gravitation
- Apply ΣF = mv²/r at the top and bottom of a vertical circle, where gravity changes its role
- Find the minimum speed for an object to maintain contact at the top of a loop
- Use Newton's law of universal gravitation to derive orbital speed and period
Centripetal force is a role, not a new force
There is no such thing as "the centripetal force" to add to a free-body diagram. Centripetal describes a direction — toward the center. Whatever real forces happen to point that way (tension, gravity, normal force, friction) are what sum to mv²/r. Adding a separate centripetal force to a diagram double-counts, and it is the most common way to get a vertical-circle problem wrong.
The same loop, two very different equations
For a ball on a string swung in a vertical circle, gravity always points down but "toward the center" changes. At the top, down is toward the center, so both tension and gravity help: T + mg = mv²/r. At the bottom, toward the center is up, so gravity now opposes: T − mg = mv²/r. The tension is therefore largest at the bottom and smallest at the top — which is where a string breaks if it is going to, and why a bucket of water swung overhead does not spill.
The minimum speed at the top
At the top, tension can go to zero but no lower — a string can pull, never push. Setting T = 0 gives mg = mv²/r, so v_min = √(gr). Below that speed the object leaves the circular path. The mass cancels, so a heavy bucket and a light one need the same minimum speed. The same result holds for a car cresting a hill or a roller-coaster loop, with the normal force in place of tension.
A 0.50 kg ball on a 1.2 m string is swung in a vertical circle at a constant speed of 6.0 m/s. Find the tension at the top and at the bottom. (g = 10 m/s²)
- 1.The centripetal requirement is the same everywhere: mv²/r = (0.50)(6.0)² ÷ 1.2 = 18 ÷ 1.2 = 15 N.
- 2.Weight is mg = (0.50)(10) = 5.0 N.
- 3.Top — gravity points toward the center, so T + mg = 15 N → T = 15 − 5.0 = 10 N.
- 4.Bottom — gravity points away from the center, so T − mg = 15 N → T = 15 + 5.0 = 20 N.
Gravitation: the force that supplies the orbit
Every pair of masses attracts along the line joining them with F = Gm₁m₂/r². Because the force falls off as 1/r², doubling the separation quarters the force. For a satellite in circular orbit that gravitational attraction is the only force acting, so it is the entire centripetal sum — and setting the two expressions equal gives the orbital speed with no extra assumptions.
Why astronauts float
Not because gravity is absent. At the height of the space station, g is still about 89% of its surface value. Astronauts float because they are in free fall — the station and everyone in it accelerate toward Earth together, so there is no normal force between them and the floor. Apparent weightlessness is the absence of a contact force, not the absence of gravity.
Speed is constant in uniform circular motion but velocity is not — the direction changes continuously, so the object is accelerating the entire time. "Constant speed, therefore no acceleration" is wrong here and is the reason circular motion belongs to dynamics rather than kinematics.
What is the minimum speed a roller-coaster car needs at the top of a vertical loop of radius 8.0 m to stay on the track? (g = 10 m/s²)
A ball is swung in a vertical circle at constant speed. Where is the string tension greatest?
A satellite orbits at radius r. To what radius must it move for its orbital speed to be halved?
On any circular-motion free-response, write "toward the center is positive" and then sum only the real forces. If your equation contains a term called F_c alongside tension and weight, you have counted the same force twice.
Answer the 3 checkpoints as you read.
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