Radian vs Degree: What Is the Difference?

A degree is one 360th of a turn, chosen by convention. A radian is an arc length divided by a radius, so it is a ratio of two lengths and carries no dimensions at all. That definition is what makes every equation linking an angle to a length true in radians and false in degrees.

AP Physics: Unit 5 (topics 5.1 Rotational Kinematics, 5.2 Connecting Linear and Rotational Motion, 7.2 Frequency and Period of SHM). The radian enters AP Physics 1 at Unit 5, Torque and Rotational Dynamics, weighted at 10 to 15 percent of the multiple-choice section. EK 5.1.A.1 puts the unit inside the definition: angular displacement is the measurement of the angle, in radians, through which a point on a rigid system rotates about a specified axis. Topic 5.2 is where radians become compulsory rather than conventional, at EK 5.2.A.1, delta s = r delta theta, and EK 5.2.A.2, which gives s = r theta, v = r omega and a_T = r alpha. Suggested skills are 1.B, 2.A, 2.D, 3.A and 3.C for Topic 5.1 and 1.C, 2.A, 2.C and 3.B for Topic 5.2. The small-angle approximation is named in AP Physics C: Mechanics, Unit 7, Topic 7.5, Simple and Physical Pendulums, at EK 7.5.A.2.ii, with derived equations sin theta approx theta and tau = -mgd theta = I alpha, and EK 7.5.A.2.iii giving the second-order differential equation for SHM. AP Physics 1 restricts the simple pendulum period to a small angle at EK 7.2.A.1.ii without naming the approximation. AP Physics 2 lists it as an exam convention on its Table of Information, for single-slit and double-slit diffraction. On the sheets, verified from rendered images of all four appendices: all four print s = r theta in the geometry table under Circle, with an arc figure; all four print a table of trigonometric values in degrees for 0, 30, 37, 45, 53, 60 and 90 degrees; all four print v = r omega, a_T = r alpha, delta x_cm = r delta theta and the three constant-angular-acceleration equations; all four print T = 1/f, with the two Physics C sheets adding T = 2 pi / omega = 1/f. The radian appears in none of the four unit-symbols boxes, which list 8 symbols on AP Physics 1, 18 on AP Physics 2, 7 on AP Physics C: Mechanics and 15 on AP Physics C: Electricity and Magnetism; neither does the degree, which is consistent with both being dimensionless.

One of them is defined by geometry, the other by agreement

A degree is one three hundred and sixtieth of a full turn. There is no reason for the number beyond history. Nothing in mathematics or physics prefers it, and nothing breaks if you pick a different divisor.

A radian is not like that. The radian is the angle for which the arc length equals the radius, so it is defined by the circle rather than imposed on it:

θ=srequivalentlys=rθ\theta = \frac{s}{r} \qquad\text{equivalently}\qquad s = r\theta

That second form is printed in the geometry table of all four AP equation sheets, under "Circle", beside A=πr2A = \pi r^2 and C=2πrC = 2\pi r, with a diagram of an arc labelled ss and a radius labelled rr. It is on the sheet, and it holds only in radians.

Two consequences follow, and the whole page is those two consequences worked out.

First: the radian is dimensionless. It is a length divided by a length. That is not a technicality, it is the reason s=rθs = r\theta works dimensionally at all: metres on the left, metres times nothing on the right. Push it through and you get m×rad/s=m/s\text{m} \times \text{rad/s} = \text{m/s}, which is why v=rωv = r\omega balances, and m×rad/s2=m/s2\text{m} \times \text{rad/s}^2 = \text{m/s}^2, which is why aT=rαa_T = r\alpha balances. A degree is also dimensionless, and this is where it goes wrong: it is dimensionless and it is not equal to one. One radian equals one. One degree equals π180\frac{\pi}{180}, which is 0.01745330.0174533. So substituting degrees into s=rθs = r\theta scales the answer by 57.295857.2958 while the units still appear to work.

Second: no unit check will ever catch a radian error. Both angular units are dimensionless, so both cancel out of a dimensional analysis identically. The equation looks right, the units come out right, and the number is wrong by a factor of nearly sixty. That is the mechanism behind most silent errors in rotational physics, and it is why the habit has to be conscious.

The CED builds the requirement into its definition rather than leaving it as a note. EK 5.1.A.1: angular displacement is the measurement of the angle, in radians, through which a point on a rigid system rotates about a specified axis. The unit is in the definition of the quantity.

Radian vs degree, side by side

PropertyRadianDegree
DefinitionArc length divided by radiusOne 360th of a full turn
Symbolrad^\circ
Full turn2π6.28322\pi \approx 6.2832360360
Right angleπ21.5708\frac{\pi}{2} \approx 1.57089090
Value of one, as a pure number11π180=0.0174533\frac{\pi}{180} = 0.0174533
DimensionsNoneNone
Conversion1 rad=180π=57.29581\ \text{rad} = \frac{180^\circ}{\pi} = 57.2958^\circ1=0.0174533 rad1^\circ = 0.0174533\ \text{rad}
Required in s=rθs = r\theta, v=rωv = r\omega, aT=rαa_T = r\alphaYesNever
Required for sinθθ\sin\theta \approx \thetaYesNever
Valid in the constant-angular-acceleration equationsYesYes, if used throughout
Valid inside sin\sin, cos\cos, tan\tanYes, with the calculator in radian modeYes, with the calculator in degree mode
Used by the AP trigonometric values tableNoYes: 00^\circ, 3030^\circ, 3737^\circ, 4545^\circ, 5353^\circ, 6060^\circ, 9090^\circ
Listed in the AP unit symbols boxesNoNo
Appears in the CED's definition of angular displacementYes, EK 5.1.A.1No

The row about the unit symbols is worth a sentence, because it is easy to misread as an oversight. The four AP sheets each print a box of unit symbols: eight entries on AP Physics 1, eighteen on AP Physics 2, seven on AP Physics C: Mechanics and fifteen on AP Physics C: Electricity and Magnetism. The radian is in none of them, and neither is the degree. That is consistent rather than careless: the boxes list units of dimensional quantities, and an angle is not one.

The row that produces the most wrong answers is the second to last: both units are correct inside a trigonometric function, provided the calculator agrees with you. That is the only place on this page where degrees are as good as radians, and it is also where the most common silent error lives.

The case that separates them: the same theta doing two different jobs

Look at what θ\theta is asked to do on a single AP equation sheet. The AP Physics 1 symbol key lists θ\theta twice, as "angle" in the translational column and as "angle or angular position" in the rotational column. Those are two different jobs and they take different units.

Equation, and where it is printedWhat θ\theta means thereUnit required
W=Fd=FdcosθW = F_{\parallel}d = Fd\cos\thetaThe geometric angle between force and displacementEither, with the calculator set to match
τ=rF=rFsinθ\tau = r_{\perp}F = rF\sin\thetaThe geometric angle between the radius and the forceEither, with the calculator set to match
L=rmvsinθL = rmv\sin\theta, AP Physics 1 and 2The geometric angle between the radial direction and the velocityEither, with the calculator set to match
n1sinθ1=n2sinθ2n_1\sin\theta_1 = n_2\sin\theta_2, AP Physics 2The geometric angle from the normalEither, with the calculator set to match
s=rθs = r\thetaAn angular position, multiplying a lengthRadians only
θ=θ0+ω0t+12αt2\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2An angular positionEither, used consistently
Δxcm=rΔθ\Delta x_{\text{cm}} = r\Delta\thetaAn angular displacement, multiplying a lengthRadians only

The dividing line is not "rotational versus translational". It is whether the angle sits inside a trigonometric function or multiplies a length.

Inside sin\sin or cos\cos, the function itself absorbs the unit: sin30\sin 30^\circ and sinπ6\sin\frac{\pi}{6} are both 0.50.5, and the only requirement is that your calculator knows which convention you meant. That is why the AP sheets can print a table of trigonometric values in degrees, for 00^\circ, 3030^\circ, 3737^\circ, 4545^\circ, 5353^\circ, 6060^\circ and 9090^\circ, without contradicting the rotational block three lines away. All four sheets carry that table and all four print it in degrees.

Multiplying a length, there is no function to absorb anything. The angle's numerical value goes straight into the answer, and only the radian's value makes s=rθs = r\theta an identity.

The third case, the constant-angular-acceleration equations, is the one people over-correct. Those contain no length at all, so every term carries the same power of the angular unit and a change of unit multiplies both sides by the same factor. Degrees work in all three of them. You can run an entire angular kinematics problem in degrees per second and degrees per second squared and every answer converts back correctly, right up until you multiply by a radius. Linear vs angular velocity works a turntable problem both ways to show it.

The practical rule that follows: convert to radians before the first equation containing an rr, and simply converting at the start is always safe.

Why omega and f differ by 2 pi and not by a conversion

Here is the dimensionlessness doing something visible. Angular frequency and ordinary frequency have the same dimensions and are not the same number.

ω=2πf\omega = 2\pi f
  • ff is in hertz, which is s1\text{s}^{-1}: cycles per second.
  • ω\omega is in radians per second, which is also s1\text{s}^{-1}, because the radian contributes nothing.

So ω\omega and ff are quantities with identical dimensions that differ by a factor of 6.28326.2832. Nothing in the units warns you, and no dimensional check will separate them. The 2π2\pi is not a unit conversion; it is the number of radians in the cycle that ff counts as one.

A turntable at 331333\frac{1}{3} revolutions per minute makes the point in four numbers, all describing the same motion:

DescriptionValue
Revolutions per minute33.33 rev/min33.33\ \text{rev/min}
Frequency ff0.55560.5556 Hz
Angular velocity ω\omega3.4907 rad/s3.4907\ \text{rad/s}
In degrees per second200 deg/s200\ \text{deg/s}
Period TT1.81.8 s

Two independent conversions are in play and mixing them up is a standard error: dividing by 6060 turns per-minute into per-second, and multiplying by 2π2\pi turns revolutions into radians. Doing only one leaves the answer wrong by 6060 or by 6.28326.2832.

The period check catches it. All four sheets print T=1fT = \frac{1}{f}, and the two Physics C sheets also print T=2πω=1fT = \frac{2\pi}{\omega} = \frac{1}{f}. Both routes here give 1.81.8 s, so the two numbers describe the same rotation.

The same collapse of units happens elsewhere and is worth recognising. rad/s2\text{rad/s}^2 and s2\text{s}^{-2} are the same dimensions. kgm2/s\text{kg}\cdot\text{m}^2/\text{s} serves for angular momentum whether you got there through IωI\omega or through rmvsinθrmv\sin\theta. In every one of these the radian is written down for clarity and dropped whenever the algebra needs it to disappear, which is exactly what a dimensionless unit is entitled to do, and exactly what makes it easy to lose track of.

The small-angle approximation only exists in radians

sinθθ\sin\theta \approx \theta

That statement is meaningless in degrees. sin10\sin 10^\circ is 0.17360.1736 and 1010 is 1010; they are not close. In radians, 1010^\circ is 0.17450.1745, and 0.17450.1745 against 0.17360.1736 is within half a percent. The approximation is not a fact about small angles. It is a fact about the radian, and it comes from the same definition as everything else on this page: for a small angle the arc and the chord are nearly the same length, and the radian is the arc.

How good it is, in numbers:

AngleIn radianssinθ\sin\thetaError in using θ\theta for sinθ\sin\theta
11^\circ0.0174530.0174530.0174520.0174520.005%0.005\%
55^\circ0.0872660.0872660.0871560.0871560.13%0.13\%
1010^\circ0.1745330.1745330.1736480.1736480.51%0.51\%
1515^\circ0.2617990.2617990.2588190.2588191.15%1.15\%
2020^\circ0.3490660.3490660.3420200.3420202.06%2.06\%
3030^\circ0.5235990.5235990.5000000.5000004.72%4.72\%

The approximation always overestimates, because the arc is longer than the chord.

The AP courses use it in two places and name it in one of them.

AP Physics C: Mechanics prints it as a derived equation. Topic 7.5, Simple and Physical Pendulums, EK 7.5.A.2.ii: for small amplitudes of motion, the small-angle approximation can be applied to the restoring torque, with the derived equations sinθθ\sin\theta \approx \theta and τ=mgdθ=Iα\tau = -mgd\theta = I\alpha. EK 7.5.A.2.iii then says the small-angle approximation and Newton's second law in rotational form yield a second-order differential equation that describes simple harmonic motion, d2θdt2=ω2θ\frac{d^2\theta}{dt^2} = -\omega^2\theta.

AP Physics 1 uses the restriction without naming the approximation. EK 7.2.A.1.ii gives the period of a simple pendulum displaced by a small angle as Tp=2πgT_p = 2\pi\sqrt{\frac{\ell}{g}}, and leaves the reason implicit. So the same physical move appears in both courses with only one of them showing its working.

AP Physics 2 puts it in the exam conventions. The conventions box on the AP Physics 2 Table of Information includes the bullet: the small angle approximation is valid for single- and double-slit diffraction. That is what licenses the small-angle forms on its sheet, a(yminL)mλa\left(\frac{y_{\text{min}}}{L}\right) \approx m\lambda and d(ymaxL)mλd\left(\frac{y_{\text{max}}}{L}\right) \approx m\lambda, which replace sinθ\sin\theta with tanθ\tan\theta and then with yL\frac{y}{L}. Both replacements are the same approximation used twice, and both need radians to be stated as sinθtanθθ\sin\theta \approx \tan\theta \approx \theta.

Which is worth noticing: the small-angle approximation reaches further into the AP courses through optics than through mechanics, and in that setting it is a printed exam convention rather than something you choose to apply.

Calculator mode, and how to catch it in one keystroke

A calculator left in the wrong angle mode is the most self-inflicted error available in this material, because everything about the working is right.

Feed a degree value to a calculator set to radians and it will quietly interpret the number as radians. It returns a real answer, in range, with no complaint. Two examples, both with the same wrong setting:

CalculationCorrect, degree modeWrong, radian modeWould you notice?
τ=rFsinθ\tau = rF\sin\theta with r=0.35r = 0.35 m, F=60F = 60 N, θ=40\theta = 40^\circ13.5 Nm13.5\ \text{N}\cdot\text{m}15.6 Nm15.6\ \text{N}\cdot\text{m}Probably not
W=FdcosθW = Fd\cos\theta with F=80F = 80 N, d=4.0d = 4.0 m, θ=40\theta = 40^\circ245245 J213-213 JYes, the sign flipped
Range =v02sin2θg= \frac{v_0^2\sin 2\theta}{g} with v0=25v_0 = 25 m/s, θ=30\theta = 30^\circ55.255.2 m19.4-19.4 mYes, a negative range

The torque row is the dangerous one. 15.615.6 against 13.513.5 is a sixteen percent error, the sign is right, the magnitude is plausible, and nothing in the answer looks wrong. The other two rows are loud because sin40\sin 40 and cos40\cos 40 in radians happen to land somewhere unhelpful. You cannot rely on the error being loud.

The check takes one keystroke: compute sin30\sin 30. If it returns 0.50.5, you are in degree mode. If it returns 0.988-0.988, you are in radians. The AP sheets even hand you the reference value: the trigonometric values table prints sin30=12\sin 30^\circ = \frac{1}{2} on all four sheets, so the answer is on the page in front of you.

A second check, for the other direction: sin3.14159\sin 3.14159 should be about zero in radian mode and about 0.05480.0548 in degree mode.

Which mode you want depends on the equation, not on the topic:

  • Degree mode for geometric angles inside sin\sin, cos\cos and tan\tan: W=FdcosθW = Fd\cos\theta, τ=rFsinθ\tau = rF\sin\theta, L=rmvsinθL = rmv\sin\theta, n1sinθ1=n2sinθ2n_1\sin\theta_1 = n_2\sin\theta_2, projectile launch angles, incline angles, vector components. This covers most of AP Physics 1 and 2, so degree mode is the sensible default for the algebra-based exams.
  • Radian mode whenever an angle goes into a trigonometric function as part of an oscillation or wave: x=Acos(2πft)x = A\cos(2\pi f t), x=xmaxcos(ωt+ϕ)x = x_{\text{max}}\cos(\omega t + \phi), and anything with ωt\omega t in it. The argument ωt\omega t is in radians because ω\omega is, so a calculator in degree mode will return the wrong displacement for a perfectly correct equation.

And separately from the mode entirely: any angle that multiplies a radius must be a radian value regardless of what the calculator is set to, because no trigonometric function is involved and the mode setting never comes into it. That is s=rθs = r\theta, v=rωv = r\omega, aT=rαa_T = r\alpha and Δxcm=rΔθ\Delta x_{\text{cm}} = r\Delta\theta.

Where the confusion costs a mark

  • Using degrees in s=rθs = r\theta, v=rωv = r\omega, aT=rαa_T = r\alpha or Δxcm=rΔθ\Delta x_{\text{cm}} = r\Delta\theta. Wrong by 57.295857.2958, and no unit check will find it.
  • Using degrees per second in K=12Iω2K = \frac{1}{2}I\omega^2. Wrong by 57.295857.2958 squared, which is about 32833283.
  • Assuming degrees are forbidden in the angular kinematics equations. They are not. ω=ω0+αt\omega = \omega_0 + \alpha t, θ=θ0+ω0t+12αt2\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2 and ω2=ω02+2α(θθ0)\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0) contain no length and work in any consistent angular unit.
  • Applying sinθθ\sin\theta \approx \theta with θ\theta in degrees. The approximation is a statement about radians. In degrees it is not even approximately true.
  • Leaving the calculator in the wrong angle mode. Compute sin30\sin 30 before you start; degree mode gives 0.50.5.
  • Working an oscillation in degree mode. x=Acos(2πft)x = A\cos(2\pi f t) and x=xmaxcos(ωt+ϕ)x = x_{\text{max}}\cos(\omega t + \phi) have arguments in radians, since ω\omega is in radians per second.
  • Converting revolutions per minute in one step. Divide by 6060 and multiply by 2π2\pi. Doing one and not the other is wrong by a factor of 6060 or of 6.28326.2832.
  • Confusing ω\omega with ff because both are per second. They differ by 2π2\pi, and they have identical dimensions, so nothing but attention separates them.
  • Reading the sheet's degree-based trigonometry table as permission to use degrees everywhere. That table is for geometric angles inside trigonometric functions, which is the one place either unit works.
  • Writing an angular answer with no unit at all because the radian is dimensionless. Write "rad", even though it cancels. The reader cannot otherwise tell whether you meant radians, degrees or revolutions.

When it makes no difference, and why that lulls you

Four situations let you use either unit and never find out which you were in.

Any angle that only ever appears inside a trigonometric function. Almost the whole of AP Physics 1 Unit 2 is like this. Incline angles, projectile launch angles, the angle in W=FdcosθW = Fd\cos\theta, the angle in a vector decomposition: put the calculator in degree mode and the radian never comes up. This is the single largest reason the distinction is learned late, because it holds for the first three or four units of the course.

Any angular kinematics problem with no radius in it. Degrees work throughout, and every answer converts back correctly.

Any ratio or proportional-reasoning question. "If the angular velocity doubles, what happens to the arc length?" needs no unit at all, since the factor cancels.

Any problem stated in revolutions. Counting turns dodges both units, and the conversion only becomes necessary when you multiply by a radius.

Then the distinction arrives, and it arrives all at once, in Unit 5.

  • Every equation in the rotational block that contains an rr. Four of them on the AP Physics 1 sheet.
  • Every energy or momentum expression built on v=rωv = r\omega, which includes K=12Iω2K = \frac{1}{2}I\omega^2 and L=IωL = I\omega even though no rr is visible in either.
  • Every oscillation written with ωt\omega t, which is Unit 7.
  • Every small-angle approximation, which is the pendulum in Unit 7 and the diffraction pattern in AP Physics 2.

That is why the safe habit is a conversion rather than a judgement: turn every angular quantity into radians the moment you write it down, and set the calculator to degrees for the geometric angles. Those two rules never conflict, because the geometric angles never multiply a radius and the angular quantities never sit inside a sin\sin.

What the CED asks, and what the sheets print

The radian enters the AP Physics 1 course at Unit 5, Torque and Rotational Dynamics, weighted at 10 to 15 percent of the multiple-choice section. Topic 5.1, Rotational Kinematics, learning objective 5.1.A, describe the rotation of a system with respect to time using angular displacement, angular velocity, and angular acceleration. EK 5.1.A.1 puts the unit inside the definition: angular displacement is the measurement of the angle, in radians, through which a point on a rigid system rotates about a specified axis, with the relevant equation Δθ=θθ0\Delta\theta = \theta - \theta_0. EK 5.1.A.2 and EK 5.1.A.3 define the average angular velocity and average angular acceleration, and EK 5.1.A.4.i gives the three constant-angular-acceleration equations. Suggested skills for Topic 5.1 are 1.B, 2.A, 2.D, 3.A and 3.C.

Topic 5.2, Connecting Linear and Rotational Motion, is where the unit becomes compulsory rather than conventional. EK 5.2.A.1 gives Δs=rΔθ\Delta s = r\Delta\theta and EK 5.2.A.2 gives s=rθs = r\theta, v=rωv = r\omega and aT=rαa_T = r\alpha. Suggested skills for Topic 5.2 are 1.C, 2.A, 2.C and 3.B.

The small-angle approximation is named in AP Physics C: Mechanics, Unit 7, Oscillations, at Topic 7.5, Simple and Physical Pendulums, EK 7.5.A.2.ii, with the derived equations sinθθ\sin\theta \approx \theta and τ=mgdθ=Iα\tau = -mgd\theta = I\alpha, and EK 7.5.A.2.iii giving d2θdt2=ω2θ\frac{d^2\theta}{dt^2} = -\omega^2\theta. Suggested skills for Topic 7.5 are 1.B, 2.A, 2.B, 3.A and 3.B. AP Physics 1 restricts the simple pendulum to a small angle at EK 7.2.A.1.ii without naming the approximation. AP Physics 2 lists it as an exam convention on its Table of Information, for single-slit and double-slit diffraction.

On the sheets, read off rendered images of all four appendices rather than a transcription:

  • All four print s=rθs = r\theta, in the geometry table under "Circle", with a figure showing the arc ss, the radius rr and the angle θ\theta, and a symbol key giving ss as arc length and θ\theta as angle. This line is easy to miss because it sits in the geometry table rather than the mechanics block, and a transcription that covers only the physics equations will not have it.
  • All four print a table of trigonometric values in degrees, headed "Values of Trigonometric Functions for Common Angles", for 00^\circ, 3030^\circ, 3737^\circ, 4545^\circ, 5353^\circ, 6060^\circ and 9090^\circ.
  • All four print v=rωv = r\omega and aT=rαa_T = r\alpha in the rotational block, along with Δxcm=rΔθ\Delta x_{\text{cm}} = r\Delta\theta and the three constant-angular-acceleration equations.
  • All four print T=1fT = \frac{1}{f}, and the two Physics C sheets also print T=2πω=1fT = \frac{2\pi}{\omega} = \frac{1}{f}, which is the ω\omega against ff relationship in disguise.
  • The radian appears in no unit symbols box on any of the four. Neither does the degree.

The symbol keys are worth reading for this too. The AP Physics 1 and AP Physics 2 rotational keys give θ\theta as "angle or angular position" and ω\omega as "angular speed"; the two Physics C keys give θ\theta as "angular position", ω\omega as "angular frequency or angular speed" and add ϕ\phi as "phase angle". In every case the key names the quantity and not the unit, which is one more reason the radian requirement has to be carried in your head.

Related material: Topic 5.1 and Topic 5.2 carry the CED framing, rotational kinematics works the three angular equations, simple harmonic motion is where ωt\omega t starts to matter, and the definitions live at radian, arc length, angular frequency and the small-angle approximation. The AP Physics 1 equation sheet shows the geometry table and the rotational block on adjacent pages.

The same three calculations in the wrong calculator mode

A calculator is left in radian mode while a student works three problems whose angles are stated in degrees: a torque with r=0.35r = 0.35 m, F=60F = 60 N and θ=40\theta = 40^\circ; the work done by an 8080 N force over 4.04.0 m at 4040^\circ to the displacement; and the range of a projectile launched at 2525 m/s at 3030^\circ above the horizontal, using range=v02sin2θg\text{range} = \frac{v_0^2\sin 2\theta}{g} with g=9.8 m/s2g = 9.8\ \text{m/s}^2. Give the correct and incorrect answers for each and say which errors are detectable from the answer alone.

  1. Reference values. In degree mode, sin40=0.6428\sin 40^\circ = 0.6428 and cos40=0.7660\cos 40^\circ = 0.7660. In radian mode the calculator reads the same keystroke as 4040 radians, giving sin40=0.7451\sin 40 = 0.7451 and cos40=0.6669\cos 40 = -0.6669.

  2. Torque, correct. τ=rFsinθ=(0.35)(60)(0.6428)=(21)(0.6428)=13.50 Nm\tau = rF\sin\theta = (0.35)(60)(0.6428) = (21)(0.6428) = 13.50\ \text{N}\cdot\text{m}.

  3. Torque, radian mode. τ=(21)(0.7451)=15.65 Nm\tau = (21)(0.7451) = 15.65\ \text{N}\cdot\text{m}. That is 15.9%15.9\% too large, positive, and entirely plausible for the numbers given. Nothing about this answer announces itself as wrong.

  4. Work, correct. W=Fdcosθ=(80)(4.0)(0.7660)=(320)(0.7660)=245W = Fd\cos\theta = (80)(4.0)(0.7660) = (320)(0.7660) = 245 J.

  5. Work, radian mode. W=(320)(0.6669)=213W = (320)(-0.6669) = -213 J. The sign flipped, so a force with a component along the motion appears to be taking energy out. This one is catchable, if you were watching the sign.

  6. Range, correct. 2θ=602\theta = 60^\circ, sin60=0.8660\sin 60^\circ = 0.8660, so range =(25)2(0.8660)9.8=625×0.86609.8=541.39.8=55.2= \frac{(25)^2(0.8660)}{9.8} = \frac{625 \times 0.8660}{9.8} = \frac{541.3}{9.8} = 55.2 m.

  7. Range, radian mode. sin60=0.3048\sin 60 = -0.3048, so range =625×(0.3048)9.8=19.4= \frac{625 \times (-0.3048)}{9.8} = -19.4 m. A negative range. Obviously wrong.

  8. The lesson is in the spread. One error announced itself with a negative range, one with a flipped sign, and one produced a wrong number that looks entirely reasonable. You cannot rely on a degree-mode error being visible, so it has to be prevented rather than detected.

  9. The one-keystroke check. Compute sin30\sin 30. Degree mode returns 0.50.5; radian mode returns 0.988-0.988. The correct value is printed on all four AP equation sheets in the trigonometric values table, at sin30=12\sin 30^\circ = \frac{1}{2}, so the reference is on the page.

  10. For completeness, none of these three errors would be caught by a units check either. The angle is dimensionless in both conventions, so Nm\text{N}\cdot\text{m}, joules and meters all come out correctly whichever mode the calculator is in.

Torque: 13.50 Nm13.50\ \text{N}\cdot\text{m} correct against 15.65 Nm15.65\ \text{N}\cdot\text{m} in radian mode, an undetectable 15.9%15.9\% error. Work: 245245 J against 213-213 J, detectable by the sign. Range: 55.255.2 m against 19.4-19.4 m, obviously wrong. Check the mode by computing sin30\sin 30, which is 0.50.5 in degrees.

One turntable, five ways of saying how fast it turns

A turntable runs at 331333\frac{1}{3} revolutions per minute. Express its rate as a frequency in hertz, an angular velocity in radians per second, a rate in degrees per second, and a period in seconds. Then find the linear speed and the arc length in one second for a point 0.150.15 m from the center, and show what using degrees per second in v=rωv = r\omega would give.

  1. Revolutions per second: 33.33360=0.5556 rev/s\frac{33.333}{60} = 0.5556\ \text{rev/s}. Since one revolution is one cycle, this is the frequency: f=0.5556f = 0.5556 Hz.

  2. Angular velocity: one revolution is 2π2\pi radians, so ω=2πf=2π(0.55556)=3.4907 rad/s\omega = 2\pi f = 2\pi(0.55556) = 3.4907\ \text{rad/s}. Note the two separate conversions: divide by 6060 for the time unit, multiply by 2π2\pi for the angle unit. They are independent and both are required.

  3. Degrees per second: one revolution is 360360^\circ, so (0.55556)(360)=200 deg/s(0.55556)(360) = 200\ \text{deg/s} exactly.

  4. Period, from T=1fT = \frac{1}{f} as printed on all four sheets: T=10.55556=1.80T = \frac{1}{0.55556} = 1.80 s. Cross-check through the Physics C form T=2πω=6.283193.49066=1.80T = \frac{2\pi}{\omega} = \frac{6.28319}{3.49066} = 1.80 s. The two agree.

  5. Stop and note the trap. f=0.5556 s1f = 0.5556\ \text{s}^{-1} and ω=3.4907 s1\omega = 3.4907\ \text{s}^{-1} have the same dimensions, because the radian and the cycle are both dimensionless. They differ by a factor of 2π2\pi and no unit check separates them.

  6. Linear speed of a point at r=0.15r = 0.15 m, using EK 5.2.A.2 with radians: v=rω=(0.15)(3.4907)=0.524v = r\omega = (0.15)(3.4907) = 0.524 m/s.

  7. Arc length in one second, from EK 5.2.A.1: Δs=rΔθ=(0.15)(3.4907)=0.524\Delta s = r\Delta\theta = (0.15)(3.4907) = 0.524 m, which is the same number because the interval is one second. Check it another way: the point traces 0.55560.5556 of a circle of circumference 2π(0.15)=0.94252\pi(0.15) = 0.9425 m, giving (0.55556)(0.9425)=0.524(0.55556)(0.9425) = 0.524 m. Agreement.

  8. Now the degrees test. Substituting 200 deg/s200\ \text{deg/s} into v=rωv = r\omega gives (0.15)(200)=30(0.15)(200) = 30 m/s, for a record player. Too large by exactly 2003.4907=57.2958\frac{200}{3.4907} = 57.2958, which is the number of degrees in a radian.

  9. And the frequency test, a subtler version of the same error. Substituting f=0.5556f = 0.5556 Hz into v=rωv = r\omega gives (0.15)(0.5556)=0.083(0.15)(0.5556) = 0.083 m/s, too small by 2π2\pi. Both wrong answers came out in metres per second and neither would fail a dimensional check.

f=0.5556f = 0.5556 Hz, ω=3.4907 rad/s\omega = 3.4907\ \text{rad/s}, 200 deg/s200\ \text{deg/s}, T=1.80T = 1.80 s. A point at 0.150.15 m moves at 0.5240.524 m/s and covers 0.5240.524 m of arc per second. Using 200 deg/s200\ \text{deg/s} in v=rωv = r\omega gives 3030 m/s, too large by 57.295857.2958; using 0.55560.5556 Hz gives 0.0830.083 m/s, too small by 2π2\pi.

How small is a small angle, and what the approximation costs

A pendulum bob of mass 0.250.25 kg hangs 0.800.80 m below a pivot. Find the restoring torque about the pivot at a displacement of 1515^\circ, first exactly from τ=mgdsinθ\tau = -mgd\sin\theta and then with the small-angle approximation sinθθ\sin\theta \approx \theta. Give the percentage error. Then tabulate the error at 55^\circ, 1010^\circ, 2020^\circ and 3030^\circ, and find the pendulum's period. Use g=9.8 m/s2g = 9.8\ \text{m/s}^2.

  1. Convert first: 15=15π180=0.26179915^\circ = \frac{15\pi}{180} = 0.261799 rad. Both the exact and the approximate calculation need this number, the exact one because the calculator needs a mode and the approximate one because the approximation is a statement about radians.

  2. Exact restoring torque, from the derived equation the AP Physics C: Mechanics CED gives at EK 7.5.A.2.i: τ=mgdsinθ\tau = -mgd\sin\theta, so τ=(0.25)(9.8)(0.80)sin15=(1.96)(0.258819)=0.5073 Nm\lvert\tau\rvert = (0.25)(9.8)(0.80)\sin 15^\circ = (1.96)(0.258819) = 0.5073\ \text{N}\cdot\text{m}.

  3. Small-angle version, from EK 7.5.A.2.ii: τmgdθ=(1.96)(0.261799)=0.5131 Nm\lvert\tau\rvert \approx mgd\,\theta = (1.96)(0.261799) = 0.5131\ \text{N}\cdot\text{m}.

  4. Percentage error: 0.513130.507290.50729=0.01151\frac{0.51313 - 0.50729}{0.50729} = 0.01151, that is 1.15%1.15\% too large. The approximation always overestimates, because θ>sinθ\theta > \sin\theta for any positive angle: the arc is longer than the chord it subtends.

  5. The error at other amplitudes, computed the same way: 55^\circ gives 0.13%0.13\%, 1010^\circ gives 0.51%0.51\%, 1515^\circ gives 1.15%1.15\%, 2020^\circ gives 2.06%2.06\%, 3030^\circ gives 4.72%4.72\%. The error grows roughly with the square of the angle, so halving the amplitude quarters it.

  6. Now do the same thing in degrees to see why the approximation is a claim about radians. At 1515^\circ, sinθ=0.2588\sin\theta = 0.2588 and the degree measure of the angle is 1515. Those differ by a factor of 5858, so sinθθ\sin\theta \approx \theta is not slightly wrong in degrees, it is not a statement at all.

  7. Period of the pendulum, from EK 7.2.A.1.ii in AP Physics 1: Tp=2πg=2π0.809.8=2π0.081633=2π(0.285714)=1.795T_p = 2\pi\sqrt{\frac{\ell}{g}} = 2\pi\sqrt{\frac{0.80}{9.8}} = 2\pi\sqrt{0.081633} = 2\pi(0.285714) = 1.795 s.

  8. Note what the small-angle approximation bought. The exact restoring torque, proportional to sinθ\sin\theta, is not proportional to the displacement, so the motion is not simple harmonic. Replacing sinθ\sin\theta by θ\theta makes the torque proportional to θ\theta, which is what makes it simple harmonic and what makes the period formula exist at all. The CED spells this out at EK 7.5.A.2.iii: the small-angle approximation and Newton's second law in rotational form yield a second-order differential equation that describes simple harmonic motion.

  9. So the 1.15%1.15\% error at 1515^\circ is the price of having a period formula. That is why the CED's AP Physics 1 statement, at EK 7.2.A.1.ii, restricts the formula to a pendulum displaced by a small angle rather than to any pendulum.

At 1515^\circ: 0.5073 Nm0.5073\ \text{N}\cdot\text{m} exactly, 0.5131 Nm0.5131\ \text{N}\cdot\text{m} with the approximation, 1.15%1.15\% too large. The error is 0.13%0.13\% at 55^\circ, 0.51%0.51\% at 1010^\circ, 2.06%2.06\% at 2020^\circ and 4.72%4.72\% at 3030^\circ, always an overestimate. The period is 1.7951.795 s. In degrees the approximation is not merely inaccurate, it is meaningless.

Frequently asked questions

What is the difference between a radian and a degree?

A degree is one 360th of a full turn, a number fixed by convention. A radian is defined by the circle itself: it is the angle whose arc length equals the radius, so an angle in radians is an arc length divided by a radius. That makes the radian a ratio of two lengths, and therefore dimensionless with a numerical value of exactly one. A degree is also dimensionless but its numerical value is pi over 180, about 0.0175. One radian is 180 over pi degrees, about 57.2958 degrees, and a full turn is 2 pi radians.

Why do physics equations use radians instead of degrees?

Because any equation that links an angle to a length is only true in radians. Arc length equals r theta is the definition of the radian rearranged, and every equation descended from it inherits the requirement: v equals r omega, tangential acceleration equals r alpha, and the center-of-mass displacement of a rolling object equals r delta theta. Substituting degrees makes each of them wrong by 57.2958. The AP Physics 1 CED writes the unit into the definition at essential knowledge 5.1.A.1, which defines angular displacement as the angle, in radians, through which a point on a rigid system rotates.

When can you use degrees in physics?

Whenever the angle sits inside a trigonometric function and your calculator is set to match. That covers the angle in work equals F d cosine theta, the angle in torque equals r F sine theta, refraction angles, projectile launch angles and vector components, so degrees are the sensible default for most of AP Physics 1 and 2. All four AP equation sheets print a table of trigonometric values in degrees for exactly this reason. Degrees also work throughout the three constant-angular-acceleration equations, since those contain no length. What degrees never work in is any equation where an angle multiplies a radius.

Does the small-angle approximation work in degrees?

No. Sine theta approximately equals theta is a statement about the radian and nothing else. The sine of 10 degrees is 0.1736, and in radians 10 degrees is 0.1745, which agrees to within half a percent; the number 10 is nowhere near either. The approximation always slightly overestimates, because the arc is longer than the chord: the error is 0.13 percent at 5 degrees, 0.51 percent at 10 degrees, 1.15 percent at 15 degrees and 4.72 percent at 30 degrees. The AP Physics C: Mechanics CED prints sine theta approximately equals theta as a derived equation at essential knowledge 7.5.A.2.ii.

How do you tell if your calculator is in degree or radian mode?

Compute the sine of 30. Degree mode returns 0.5 and radian mode returns about negative 0.988. The correct value is printed on all four AP equation sheets in the table of trigonometric values for common angles, so the reference is on the page in front of you. This check is worth making before every calculation involving an angle, because a mode error is not reliably visible in the answer. A torque of 0.35 meters times 60 newtons times the sine of 40 degrees is 13.5 newton meters correctly and 15.6 in radian mode, and nothing about the second number looks wrong.

Why is angular velocity 2 pi times frequency and not a unit conversion?

Because the 2 pi counts radians per cycle rather than converting between units of time. Frequency counts cycles per second and angular velocity counts radians per second, and a cycle is 2 pi radians, so omega equals 2 pi f. Both quantities have dimensions of inverse seconds, because the radian is a length divided by a length and so carries no dimensions at all, which is exactly why they are so easy to confuse and why no dimensional check will separate them. A turntable at 33 and a third revolutions per minute has a frequency of 0.5556 hertz and an angular velocity of 3.4907 radians per second, and the two describe the same motion.

Is s equals r theta on the AP equation sheet?

Yes, on all four of them, but not where you would look for it. It is printed in the geometry and trigonometry table under the heading Circle, next to the area and circumference of a circle, with a diagram showing an arc labelled s, a radius labelled r and an angle theta, and a symbol key defining s as arc length. It also appears in the AP Physics 1 CED as one of the derived relationships at essential knowledge 5.2.A.2. Because it sits in the geometry table rather than the mechanics block, it is easy to miss when scanning the sheet for rotational equations.