R7 · inward force
A 0.50 kg puck moves at 4.0 m/s in a circle of radius 1.2 m on a frictionless horizontal table. Find the string tension.
Circular motion · AP/F=ma · 6 min
Constant speed does not mean zero acceleration
Acceleration measures velocity change, and velocity includes direction.
01 The trap
Where the wrong model begins.
Wrong path
Students see constant speed and conclude the net force must be zero.
Why it feels right
In straight-line motion, constant speed usually means no acceleration. Circular motion breaks that shortcut because direction changes continuously.
02 Interactive clinic
Predict the arrows. Run the model. Prove the transfer.
Follow the Q04 → NL-05 → R7 repair path from direction choice to calculation. The final check moves the puck to a new position so you must rebuild the vectors instead of recalling a diagram.
Circular motion clinic / Q04 · NL-05 · R7 PREDICT
01 · Predict before motion
The puck is at the rightmost point, moving counterclockwise.
Choose each arrow direction before the trace reveals the model.
02 · Live vector model
Velocity stays tangent. Net force stays inward.
- aᶜ
- —
- Fᵣ
- —
- direction
- —
Changing mass leaves acceleration unchanged. Changing speed has a squared effect.
03 · Worked example
Build the 6.7 N result one decision at a time.
Reveal only the next move you need.
- Start by naming the system and the real forces on it.
04 · Checked practice
Three checks. Numbers and units both matter.
0 / 3 cleared
Speed-squared transfer
Mass and radius stay fixed while speed doubles. By what factor does the required inward force change?
Vertical-circle transfer
At the top of a vertical circle, a 0.20 kg ball moves at 5.0 m/s on a 0.80 m string. Find the string tension.
05 · Transfer check
Move to the leftmost point. Does the model survive?
The puck still moves counterclockwise with the original values. Enter the net-force magnitude, then choose velocity and force directions.
Clear all three practice checks to unlock transfer.
03 Correct model
The first-principles repair.
model repair
For curved motion, draw the acceleration toward the center of curvature first. Then ask which real force or force component points inward.
- 01Separate speed from velocity.
- 02For any curved path, mark inward acceleration.
- 03Do not add an outward centripetal force.
- 04Write sum F_radial = mv²/r toward the center.
04 Mini-example
Same problem, cleaner model.
worked trap check
Prompt
A car rounds a flat curve at constant speed. What direction is the net force?
Common wrong answer
Zero, because speed is constant.
Correct reasoning
Toward the center. Static friction supplies the centripetal force.
Diagnostic cue
If the path curves, acceleration can be nonzero even when the speed number stays fixed.
05 Guided practice
Try it before the solution.
Warm-up isolates the principle. Bridge changes the context. Contest-style requires a complete setup on less familiar geometry.
warm-upAP
Flat curve force
A car moves around a flat circular turn at constant speed. What direction is its acceleration?
Hint
Acceleration follows change in velocity, not change in speed only.
Solution
The acceleration points toward the center of the circle. The velocity's direction changes continuously even though its magnitude stays fixed.
Trap: Saying acceleration is zero because speed is constant.
bridgeF=ma
Minimum friction on a curve
A 900 kg car rounds a flat curve of radius 40 m at 12 m/s. What minimum friction force is needed?
Hint
Static friction supplies the inward net force.
Solution
F_f = mv²/r = 900(12²)/40 = 3240 N inward. It is static friction if the tires do not slide.
Trap: Adding an outward force or setting net force to zero.
contest-styleF=ma
Top of a vertical circle
A 0.20 kg ball on a string moves through the top of a vertical circle of radius 0.80 m at 5.0 m/s. Find the string tension at the top.
Hint
At the top, inward is downward. Both gravity and tension may point inward.
Solution
T + mg = mv²/r, so T = m(v²/r - g) = 0.20(25/0.80 - 9.8) = 4.29 N.
Trap: Putting gravity on the wrong side because it points down.
contest-styleF=ma
Banked curve design speed
A frictionless road is banked at angle θ for a circular turn of radius r. Derive the speed v that needs no friction.
Hint
Use N cos(θ) = mg and N sin(θ) = mv²/r.
Solution
Divide the inward equation by the vertical equation: tan(θ) = v²/(rg), so v = √(rg tan(θ)).
Trap: Inventing a separate centripetal force instead of using the horizontal component of the normal force.
contest-styleF=ma
Tire-grip threshold
A car rounds a level curve of radius 35 m at 14 m/s without slipping. What minimum coefficient of static friction is required?
Hint
Set μsmg = mv²/r.
Solution
μs = v²/(rg) = 14²/[35(9.8)] = 0.571, so the minimum coefficient is about 0.57.
Trap: Including the car's mass in the final answer even though it cancels.
06 Mock checks
When the drills feel clean.
AP → F=ma readiness25 min
Mechanics foundation check
A compact check for the force-direction, normal-force, friction, and circular-motion habits that decide early contest mechanics.
6 questions · 15 pts
F=ma bridge35 min
Forces and constraints mock
A bridge set for pulleys, system boundaries, friction direction, and accelerating frames.
6 questions · 19 pts