Vertical Circular Motion

Vertical circular motion happens when an object travels around a circular path that stands upright, such as a ball on a string swinging in a loop, a bucket of water spun overhead, or a cart going through a roller coaster loop. This is different from horizontal circular motion, where the circle lies flat and gravity acts perpendicular to the plane of motion. In a vertical circle, gravity acts along the same plane as the motion, so it speeds the object up on the way down and slows it down on the way up. That means the speed is almost never constant, unlike in simple horizontal circular motion.

At every point on a vertical circle, the net force pointing toward the center must equal the required centripetal force, \(F_c = \frac{mv^2}{r}\). Two forces usually combine to create this net inward force: gravity, \(mg\), which always points straight down, and a contact force such as tension \(T\) or a normal force \(N\), which points toward the center of the circle (for a string) or away from it (for a track from the inside). Because the direction of gravity relative to the center changes as the object moves around the loop, the size of \(T\) or \(N\) also changes from point to point.

Since gravity has a component along the direction of motion everywhere except at the very top and bottom of the circle, it constantly does work on the object, speeding it up as it falls and slowing it down as it rises. The cleanest way to track this is with energy conservation instead of trying to track a changing tangential acceleration. If \(v_0\) is the speed at the bottom of the loop and \(h(\theta)\) is the height above the bottom at angle \(\theta\) measured from the bottom, then:

\(v^2 = v_0^2 - 2gh(\theta)\), where \(h(\theta) = r(1 - \cos\theta)\)

The graph below shows how \(v^2\) drops as the object rises from the bottom \((\theta = 0)\) to the top \((\theta = \pi)\) of a loop, using \(r = 5\) m\(\), \(g = 9.8\) m/s\(^2\), and a bottom speed of \(16\) m/s\(\).

The top and bottom are the two positions where gravity acts entirely along the radial direction, so they give the simplest equations.

At the top, both gravity and the contact force point toward the center: \(N + mg = \frac{mv_{top}^2}{r}\), so \(N = \frac{mv_{top}^2}{r} - mg\).

At the bottom, gravity points away from the center while the contact force points toward it: \(T - mg = \frac{mv_{bottom}^2}{r}\), so \(T = \frac{mv_{bottom}^2}{r} + mg\).

Notice that the contact force is always larger at the bottom than at the top for the same speed, which is why a swing or a roller coaster restraint feels tightest at the lowest point of the ride.

For an object held by a string or riding on the inside of a track, the contact force can only push or pull in one direction and cannot go negative. At the top of the loop, if the speed is too slow, \(N\) would need to be negative to keep the object on the circular path, which is impossible for a string (it would go slack) or a track (the object would fall away from it). The smallest possible speed at the top occurs when \(N = 0\), so gravity alone supplies the centripetal force:

\(mg = \frac{mv_{min}^2}{r}\)

\(v_{min} = \sqrt{gr}\)

Any speed at the top less than this means the object cannot maintain the circular path and will fall inward before completing the loop.

A roller coaster loop has a radius of \(r = 8\) m\(\). What is the minimum speed a cart must have at the top so that it does not lose contact with the track? Using \(g = 9.8\) m/s\(^2\):

\(v_{min} = \sqrt{gr} = \sqrt{9.8 \times 8} = \sqrt{78.4} \approx 8.85\) m/s\(\)

Any speed at the top of \(8.85\) m/s\(\) or higher keeps the cart pressed against the track.

A ball of mass \(0.5\) kg\(\) is attached to a \(1.2\) m\(\) string and swung in a vertical circle. At the lowest point, its speed is \(6\) m/s\(\). Find the tension in the string at that instant.

\(T = \frac{mv^2}{r} + mg = \frac{0.5 \times 6^2}{1.2} + (0.5 \times 9.8) = \frac{18}{1.2} + 4.9 = 15 + 4.9 = 19.9\) N\(\)

Vertical circular motion shows up any time an object loops, swings, or spins in an upright circle: roller coaster loops, a bucket of water swung overhead without spilling, a ball on a string, a pilot pulling out of a dive, or a car cresting and dipping on a hilly, curved road. In every case, the same two ideas apply: use energy conservation to find the speed at the point you care about, then apply the centripetal force equation at that point to find the tension, normal force, or minimum speed required.

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