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Honors Physics Unit 6 Review: Circular and Rotational Motion

Unit 6 takes the motion ideas you already know and applies them to circles and spinning objects. After it, you should be able to describe rotation with angles and angular velocity, find the inward force that keeps something moving in a circle, and use torque to predict how an object's spin changes.

Start with the overview for the big picture, then work through each topic's notes and answer the checkpoint question before you read its answer.

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What is Honors Physics unit 6?

Wheels and fans spin around an axis, while cars on curves and satellites in circular orbits follow curved paths. This unit gives you the tools to describe that motion and explain what causes it. Most of the math is the kinematics and Newton's laws you already know, rewritten with angular variables.

Honors Physics Unit 6 covers circular and rotational motion. You measure rotation in radians, connect angles to arc length with s=rθs = r\theta, and connect angular velocity to linear speed with v=rωv = r\omega. You analyze uniform circular motion using ac=v2/ra_c = v^2/r and a net inward force of mv2/rmv^2/r. The unit finishes with rotational kinematics, torque (τ=rFsin⁡θ\tau = rF\sin\theta), moment of inertia, and the rotational version of Newton's second law, ∑τ=Iα\sum\tau = I\alpha. Order and depth vary by school, and some courses move torque to a later unit.

Every linear quantity has a rotational partner.

Displacement becomes angle θ\theta, velocity becomes angular velocity ω\omega, and acceleration becomes angular acceleration α\alpha. The radius links each pair, as in s=rθs = r\theta and v=rωv = r\omega. The constant-acceleration equations carry over with the symbols swapped.

Moving in a circle at constant speed still means accelerating.

Velocity includes direction, and direction changes constantly on a circle. That change is centripetal acceleration, ac=v2/ra_c = v^2/r, pointing toward the center. Newton's second law then requires a net force pointing toward the center, supplied by real forces like tension, friction, or gravity.

Torque, not force alone, decides how a spin changes.

The same push can open a door easily or barely move it, depending on where and at what angle you push. Torque, τ=rFsin⁡θ\tau = rF\sin\theta, captures that. How much a given torque speeds up a spin depends on the moment of inertia, through ∑τ=Iα\sum\tau = I\alpha.

Rotation follows the same rules as straight-line motion, with angles in place of distances.

Once you see that θ\theta, ω\omega, and α\alpha behave like xx, vv, and aa, the unit becomes a translation exercise. Torque acts like force and moment of inertia acts like mass, so ∑τ=Iα\sum\tau = I\alpha mirrors ∑F=ma\sum F = ma. Circular motion also follows Newton's second law. Its acceleration has a component toward the center and can also have a tangential component when speed changes.

Honors Physics unit 6 topics

6.2

Uniform Circular Motion

In uniform circular motion the speed stays constant but the direction keeps changing, so the velocity changes and the object accelerates.

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Unit 6 review notes

6.1

Angle of Rotation and Angular Velocity

The angle of rotation θ\theta tells you how far something has turned, the way displacement tells you how far it has moved. Physics uses radians because they make the link to distance clean. One full turn is 2π2\pi rad, or 360°, so 1 rad is about 57.3°.

A point at radius rr that turns through an angle of magnitude ∣Δθ∣|\Delta\theta| in one direction travels an arc length s=r∣Δθ∣s = r|\Delta\theta|, with the angle in radians. Average angular velocity is ωavg=Δθ/Δt\omega_{\text{avg}} = \Delta\theta/\Delta t, measured in rad/s. At any instant, the point's linear speed is v=r∣ω∣v = r|\omega|, where ω\omega is its instantaneous angular velocity.

Every point on a rigid spinning object has the same ω\omega. Points farther from the axis still move faster, because they cover a bigger circle in the same time. For constant angular speed, the period TT is the time for one turn and the frequency is f=1/Tf = 1/T. The angular speed is then ∣ω∣=2π/T=2πf|\omega| = 2\pi/T = 2\pi f.

Worked example. A ceiling fan spins at 120 revolutions per minute, and its blade tips are 0.60 m from the axis. Convert to rad/s first: 120 rev/min×2π rad/rev÷60 s/min=4π≈12.6120 \text{ rev/min} \times 2\pi \text{ rad/rev} \div 60 \text{ s/min} = 4\pi \approx 12.6 rad/s. The tip speed is v=rω=(0.60)(12.6)≈7.5v = r\omega = (0.60)(12.6) \approx 7.5 m/s. The period is T=2π/ω=0.50T = 2\pi/\omega = 0.50 s, which matches 120 turns in 60 seconds.

  • Radian: The angle that makes an arc length equal to the radius, with 2π2\pi rad in one full turn.
  • Angular velocity ($\omega$): The rate at which an object's angle changes, measured in rad/s.
  • Arc length ($s$): The distance a point travels along a circle, found with s=rθs = r\theta when θ\theta is in radians.
  • Period and frequency: The period TT is the time for one revolution and the frequency f=1/Tf = 1/T is the number of revolutions per second.

Check yourself. Two kids ride a merry-go-round, one 1 m from the center and one 2 m from the center. Who has the larger angular velocity, and who has the larger speed? (Answer: They have the same angular velocity because the whole platform turns together, but the kid at 2 m has twice the linear speed since v=rωv = r\omega.)

6.2

Uniform Circular Motion

In uniform circular motion the speed stays constant but the direction keeps changing, so the velocity changes and the object accelerates. The velocity always points along the tangent to the circle. The acceleration points toward the center and is called centripetal acceleration, with magnitude ac=v2/r=rω2a_c = v^2/r = r\omega^2.

By Newton's second law, an inward acceleration needs an inward net force of size mv2/rmv^2/r. That net inward force is what people call the centripetal force. It is not a new kind of force. Tension, friction, gravity, or the normal force supplies it, alone or in combination. If the net force becomes zero, the object continues in a straight line along the tangent.

Because aca_c depends on v2v^2, doubling your speed on the same curve takes four times the inward force.

Worked example. A 1000 kg car rounds a flat curve of radius 50 m at 15 m/s. The required inward force is F=mv2/r=(1000)(15)2/50=4500F = mv^2/r = (1000)(15)^2/50 = 4500 N, and static friction from the road supplies it. Friction can provide at most μsmg\mu_s mg, so setting μsmg=mv2/r\mu_s mg = mv^2/r gives a minimum coefficient of μs=v2/(rg)=225/(50×9.8)≈0.46\mu_s = v^2/(rg) = 225/(50 \times 9.8) \approx 0.46. Mass cancels, so a heavier car needs the same μs\mu_s to make this turn.

  • Uniform circular motion: Motion in a circle at constant speed, where only the direction of velocity changes.
  • Centripetal acceleration: The center-pointing component of acceleration in circular motion, with magnitude v2/rv^2/r or rω2r\omega^2.
  • Centripetal force: The net force toward the center, mv2/rmv^2/r, supplied by real forces such as tension, friction, or gravity.

Check yourself. A 0.50 kg puck on a 0.80 m string moves in a horizontal circle on a frictionless table at 4.0 m/s. What is the tension in the string? (Answer: Tension is the only inward force, so T=mv2/r=(0.50)(4.0)2/0.80=10T = mv^2/r = (0.50)(4.0)^2/0.80 = 10 N.)

SituationReal force pointing toward the center
Puck on a string on a frictionless tableTension in the string
Car turning on a flat roadStatic friction from the road on the tires
Satellite in circular orbitEarth's gravitational pull
Car on a banked curve with no frictionHorizontal component of the normal force
6.3

Rotational Motion

Rotational kinematics works like linear kinematics with the symbols swapped. For constant angular acceleration, use ω=ω0+αt\omega = \omega_0 + \alpha t, Δθ=ω0t+12αt2\Delta\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2, and ω2=ω02+2αΔθ\omega^2 = \omega_0^2 + 2\alpha\Delta\theta. A point on a wheel that is speeding up has tangential acceleration at=rαa_t = r\alpha, which changes its speed, plus centripetal acceleration rω2r\omega^2, which changes its direction.

Torque is what changes rotation. Its size is τ=rFsin⁡θ\tau = rF\sin\theta, where rr is the distance from the axis to where the force acts and θ\theta is the angle between rr and FF. Pushing perpendicular to a door at the handle gives the most torque, and pushing straight toward the hinge gives zero.

Moment of inertia II is the rotational version of mass. For a point mass at distance rr from an axis, I=mr2I = mr^2. For a solid disk or thin ring about an axis through its center and perpendicular to its plane, I=12MR2I = \tfrac{1}{2}MR^2 or I=MR2I = MR^2, respectively. For rotation about a fixed axis, Newton's second law becomes ∑τ=Iα\sum\tau = I\alpha.

Worked example. A string pulls with 10 N tangent to the rim of a 4.0 kg solid disk of radius 0.20 m that starts at rest. The disk rotates about a fixed, frictionless axle through its center, perpendicular to its face. The torque is τ=(0.20)(10)sin⁡90°=2.0\tau = (0.20)(10)\sin 90° = 2.0 N·m. The moment of inertia is I=12(4.0)(0.20)2=0.080I = \tfrac{1}{2}(4.0)(0.20)^2 = 0.080 kg·m², so α=τ/I=25\alpha = \tau/I = 25 rad/s². After 2.0 s, ω=0+(25)(2.0)=50\omega = 0 + (25)(2.0) = 50 rad/s.

  • Angular acceleration ($\alpha$): The rate at which angular velocity changes, measured in rad/s².
  • Torque ($\tau$): The turning effect of a force, τ=rFsin⁡θ\tau = rF\sin\theta, measured in N·m.
  • Moment of inertia ($I$): A measure of how hard it is to change an object's rotation, which grows when mass sits farther from the axis.
  • Tangential acceleration: The part of acceleration along the circle that changes speed, equal to rαr\alpha.

Check yourself. A solid disk and a thin ring have the same mass and radius. Each rotates about a fixed axis through its center and perpendicular to its plane, and the same net torque acts on each. Which spins up faster? (Answer: The solid disk. Its I=12MR2I = \tfrac{1}{2}MR^2 is half the ring's MR2MR^2, so with α=τ/I\alpha = \tau/I it gets twice the angular acceleration.)

LinearRotationalConnection
Displacement $x$ (m)Angle $\theta$ (rad)$s = r\theta$
Velocity $v$ (m/s)Angular velocity $\omega$ (rad/s)$v = r\omega$
Acceleration $a$ (m/s²)Angular acceleration $\alpha$ (rad/s²)$a_t = r\alpha$
Force $F$ (N)Torque $\tau$ (N·m)$\tau = rF\sin\theta$
Mass $m$ (kg)Moment of inertia $I$ (kg·m²)$I = \sum mr^2$
$\sum F = ma$$\sum\tau = I\alpha$Same form, angular variables

Key terms

TermDefinition
RadianThe unit of angle in physics, with $2\pi$ rad in one full revolution.
Angular velocityThe rate at which an object's angle changes, in rad/s. Average angular velocity is $\omega_{\text{avg}} = \Delta\theta/\Delta t$.
PeriodThe time for one complete revolution. At constant angular speed, $T = 2\pi/|\omega|$.
Centripetal accelerationThe acceleration toward the center of a circular path, with magnitude $v^2/r$.
Centripetal forceThe net inward force, $mv^2/r$, that real forces must supply for an object to move in a circle.
Tangential accelerationThe component of acceleration along the path that changes an object's speed, equal to $r\alpha$.
Angular accelerationThe rate of change of angular velocity, in rad/s². Average angular acceleration is $\alpha_{\text{avg}} = \Delta\omega/\Delta t$.
TorqueThe turning effect of a force about an axis, $\tau = rF\sin\theta$, in N·m.
Moment of inertiaAn object's resistance to changes in rotation, set by its mass and how far that mass is from the axis.

Mistakes to avoid in Circular and Rotational Motion

Drawing centripetal force as an extra arrow on the free-body diagram

The name makes it sound like its own force, so it is easy to add it next to tension or friction and count it twice. Draw only real forces, then set the sum of their inward components equal to mv2/rmv^2/r.

Thinking an object flies straight outward when released

Riding in a turning car feels like being pushed outward, but that feeling is your body trying to keep going straight. A ball whose string snaps initially moves along the tangent to the circle, in the direction its velocity was already pointing. Its later path can curve if another force, such as gravity, acts on it.

Plugging degrees or rpm into $v = r\omega$ or $s = r\theta$

These formulas only work with radians, because the radian is defined by arc length divided by radius. Convert first, using 2π2\pi rad per revolution and 60 s per minute, so 30 rpm becomes π\pi rad/s.

Using the full distance in torque without the angle

Torque is not simply rFrF unless the force is perpendicular to the line from the axis. Include sin⁡θ\sin\theta, or use only the force component perpendicular to rr, and a force aimed straight at the axis gives zero torque.

Study unit 6 other ways

Ready to review Unit 6?Start with the notes, check the topic cards, and use the practice or resource links when they are available for this course.