Difficulty: Intermediate | Prerequisites: 1D and 2D kinematics, free-body diagrams, basic understanding of forces (gravity, normal force).
Newton's second law (ΣF = ma) is the bridge between kinematics (describing motion) and dynamics (explaining why things move the way they do). In PHYS 114, circular motion is one of the first non-trivial applications: an object moving in a circle is constantly accelerating toward the centre, so there must be a net force pointing inward. The roller coaster loop problem is the classic test of whether you can draw a free-body diagram, identify which direction the net force must point, and solve for the normal force. If you are comfortable with free-body diagrams and Newton's second law in a straight line, the new piece here is recognising that circular motion demands a centripetal acceleration, and therefore a net centripetal force.
Net force equals mass times acceleration, and acceleration is always in the direction of the net force (not necessarily the direction of motion). For circular motion, the acceleration points toward the centre of the circle, so the net force must also point inward. At the bottom of a loop the normal force must exceed the weight; at the top, gravity alone can provide the centripetal force, so no additional "holding" force is needed.
Net force (ΣF)
The vector sum of all forces acting on an object. This is what determines the acceleration via Newton's second law. Think of it as the total push or pull after all individual forces are combined.
Newton's second law
ΣF = ma. The net force on an object equals its mass times its acceleration. The acceleration is always in the same direction as the net force, regardless of which way the object is moving.
Centripetal acceleration (a_c)
The acceleration directed toward the centre of the circular path, with magnitude v²/r. Any object moving in a circle must have this inward acceleration. In simple terms, this is the acceleration that keeps bending the object's path into a circle rather than letting it fly off in a straight line.
Normal force (N)
The contact force exerted by a surface perpendicular to the surface. On a roller coaster track, this is the push the track exerts on the car. Its magnitude adjusts depending on the situation; it is not always equal to the weight.
Weight (W = mg)
The gravitational force on an object, always directed straight downward. It does not depend on motion.
Centripetal force
Not a new type of force. It is the name given to whatever net force provides the centripetal acceleration. It could be gravity, normal force, tension, friction, or a combination. Think of "centripetal force" as a role, not a separate force to add to your free-body diagram.
Newton's second law tells you that acceleration is parallel to the net force: ΣF = ma, so a = ΣF / m.
The direction of the net force determines the direction of the acceleration, not the direction of the velocity.
An object can be moving in one direction while accelerating in a completely different direction. This is the case in circular motion, where velocity is tangent to the circle but acceleration points inward.
Example from the exam: if the net force on a box is in the positive x-direction, then its acceleration must be in the positive x-direction. Its velocity can point anywhere.
At the bottom of a vertical loop, the centre of the circle is above the object.
The centripetal acceleration points upward (toward the centre).
Two forces act on the coaster: weight (mg, downward) and the normal force from the track (N, upward).
Newton's second law in the radial direction: N - mg = mv²/r.
Solving for N: N = m(g + v²/r).
The normal force is always greater than the weight at the bottom of a loop.
Example: m = 270 kg, v = 26 m/s, r = 15 m. N = 270 × (9.81 + 26²/15) = 270 × (9.81 + 45.07) = 270 × 54.88 ≈ 1.5 × 10⁴ N.
At the top of the loop, the centre of the circle is below the object.
Both the weight (mg) and the normal force (if any) point downward, toward the centre.
Newton's second law: mg + N = mv²/r.
There is no force holding the coaster up at the top. Both gravity and the normal force push the coaster inward (downward), and together they provide the centripetal force.
If the coaster moves fast enough, N can be large. If it moves at just the right minimum speed, N = 0 and gravity alone provides the centripetal force.
There is no separate force "holding the coaster up" at the top of the loop. The answer to the exam question is simply: no.
Newton's second law:
\sum \vec{F} = m\vec{a}Centripetal acceleration:
a_c = \frac{v^2}{r}Normal force at the bottom of a vertical loop:
N = m\left(g + \frac{v^2}{r}\right)Normal force at the top of a vertical loop:
N = m\left(\frac{v^2}{r} - g\right)Minimum speed at the top of a loop (N = 0):
v_{\min} = \sqrt{g\,r}Roller coaster designers use exactly these equations to ensure riders feel exciting but safe g-forces. At the bottom of a loop, riders feel heavier because the normal force exceeds their weight. Pilots experience the same physics in banked turns and vertical manoeuvres, where the "apparent weight" they feel depends on the centripetal acceleration.
Students often think the net force must be in the same direction as the velocity. It does not. In circular motion, the net force is perpendicular to the velocity (pointing inward), which changes the direction of motion without changing the speed.
Students often add "centripetal force" as a separate force on a free-body diagram. It is not a separate force. It is the net result of real forces (gravity, normal force, tension, friction) that happen to point toward the centre.
Students often think the normal force always equals mg. It does not. At the bottom of a loop, N > mg. At the top, N < mg (and can be zero).
Students often think something must push the coaster upward at the top of the loop to keep it from falling. Nothing does. At the top, both gravity and the normal force point downward (toward the centre). The coaster stays on the track because it is moving fast enough that its inertia carries it through the curve.
⚠️ The exam will test whether you know that acceleration is parallel to the net force, not to the velocity. Be ready for a question where the object moves in one direction but accelerates in another.
⚠️ Loop-the-loop problems at the bottom and the top are high-frequency exam questions. Know the free-body diagram for each position and the sign convention (which direction is "toward the centre").
⚠️ At the bottom: N - mg = mv²/r. At the top: mg + N = mv²/r (if the track pushes inward) or mg - N = mv²/r (if the track pushes outward, which happens at low speed on the outside of a curve). Know which applies.
⚠️ The "is there a force holding the coaster up at the top" question tests conceptual understanding. The answer is no: both forces point downward at the top.
True or false: if the net force on an object points east, the object must be moving east.
False. The acceleration points east, but the velocity can point in any direction.
Fill in the blank: centripetal acceleration has magnitude ______ and points toward ______.
v²/r; the centre of the circular path.
True or false: at the bottom of a vertical loop, the normal force on the coaster equals its weight.
False. N = m(g + v²/r), which is greater than mg.
True or false: "centripetal force" is a new type of force, separate from gravity and normal force.
False. It is the label for whatever net force provides the inward acceleration.
Fill in the blank: the minimum speed at the top of a loop of radius r is v = ______.
√(gr).
Q: The net force on a box is in the positive x-direction. Which statement is true?
A: Its acceleration must be in the positive x-direction. Its velocity can point in any direction. (Newton's second law: ΣF = ma, so a is parallel to ΣF.)
Q: A 270 kg roller coaster goes around a vertical loop of radius 15 m at 26 m/s. What is the normal force at the bottom of the loop?
A: N = m(g + v²/r) = 270 × (9.81 + 26²/15) = 270 × (9.81 + 45.07) = 270 × 54.88 ≈ 1.5 × 10⁴ N.
Q: Is there a force holding the coaster up at the top of the loop so it does not fall?
A: No. At the top, both gravity and the normal force (if any) point downward toward the centre. There is no upward force. The coaster maintains its circular path because its inertia (from its speed) keeps it moving through the curve, while gravity and the track both push inward.
Q: What is the minimum speed the coaster must have at the top of a 15 m radius loop to maintain contact with the track?
A: At minimum speed, N = 0, so mg = mv²/r, giving v = √(gr) = √(9.81 × 15) = √147 ≈ 12.1 m/s.
Newton's second law is the backbone of the entire mechanics portion of PHYS 114. Circular motion connects to gravitational orbits (planets follow circular or elliptical paths with gravity as the centripetal force) and to rotational dynamics later in the course. The idea that acceleration need not be in the direction of velocity also appears in projectile motion, where gravity accelerates the projectile downward while it moves diagonally.
Newton's second law, F = ma, net force, centripetal acceleration, centripetal force, normal force, weight, circular motion, vertical loop, roller coaster physics, loop-the-loop, minimum speed at top, free-body diagram, radial direction, PHYS 114, general physics, University of Washington