University Physics I · topic UPI-09.01
Angular position and displacement
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Angular position and displacement: predict a finite change
Angular position θ is the signed angle, in radians, from a chosen reference line to a line fixed in the rotating body; angular displacement Δθ = θ − θ₀ is the change in that angle, with every full turn counted. Use the supplied topic model Y_out = d/(2*pi), where Y_out is REVOLUTIONS Δθ/2π (rev). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Angular displacement Δθ, measured in rad. Keep t0 (Starting angle θ₀) = 0.5 rad; r (Radius of the mark r) = 0.4 m fixed. For this exercise use only -7.48125 rad ≤ d ≤ 7.48125 rad on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at d = -4.1895 rad and -2.09475 rad. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δd across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
REVOLUTIONS Δθ/2π against d
Predictions from the supplied Angular position and displacement model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): REVOLUTIONS Δθ/2π / rev
Horizontal axis (x): d: Angular displacement Δθ / rad
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Angular displacement Δθ / rad | REVOLUTIONS Δθ/2π / rev |
|---|---|
| -7.48125 | -1.1906779180062421 |
| -3.7406250000000005 | -0.5953389590031211 |
| -8.881784197001252e-16 | -1.4135798584282297e-16 |
| 3.7406249999999988 | 0.5953389590031208 |
| 7.481249999999998 | 1.1906779180062417 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of d separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At d = -4.1895 rad, substitute into Y_out = d/(2*pi): Y₀ = -0.66678 rev.
- At d = -2.09475 rad, the same fixed-parameter model gives Y₁ = -0.33339 rev.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.33339 rev; the input change is Δd = 2.09475 rad.
- Average sensitivity is ΔY_out/Δd = 0.159155 rev per rad. Its positive sign means that this response increases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = -0.66678 rev, Y₁ = -0.33339 rev; ΔY_out = 0.33339 rev; average sensitivity = 0.159155 rev per rad.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Angular position and displacement: infer an input with uncertainty
Angular position θ is the signed angle, in radians, from a chosen reference line to a line fixed in the rotating body; angular displacement Δθ = θ − θ₀ is the change in that angle, with every full turn counted. Use the supplied topic model Y_out = d/(2*pi), where Y_out is REVOLUTIONS Δθ/2π (rev). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Angular displacement Δθ, measured in rad. Keep t0 (Starting angle θ₀) = 0.5 rad; r (Radius of the mark r) = 0.4 m fixed. For this exercise use only -7.48125 rad ≤ d ≤ 7.48125 rad on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.26194914 rev, with a bounded reading uncertainty ±0.13931 rev. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00149625 rad, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
REVOLUTIONS Δθ/2π against d
Predictions from the supplied Angular position and displacement model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): REVOLUTIONS Δθ/2π / rev
Horizontal axis (x): d: Angular displacement Δθ / rad
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Angular displacement Δθ / rad | REVOLUTIONS Δθ/2π / rev |
|---|---|
| -7.48125 | -1.1906779180062421 |
| -3.7406250000000005 | -0.5953389590031211 |
| -8.881784197001252e-16 | -1.4135798584282297e-16 |
| 3.7406249999999988 | 0.5953389590031208 |
| 7.481249999999998 | 1.1906779180062417 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(d) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(-7.48125) = -1.19068 rev and Y_out(7.48125) = 1.19068 rev. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives d_line = -7.48125 + (0.261949 − -1.19068)(7.48125 − -7.48125)/(1.19068 − -1.19068) = 1.64587 rad. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives d = 1.64587 rad; substituting back gives Y_out = 0.26194914 rev.
- The interpolation error in the input is |d_line − d| = 0 rad. This is within the requested numerical tolerance.
- Invert both limits Y_out = 0.12263914 rev and 0.40125914 rev. Sort the roots even if the response decreases: 0.770564 rad ≤ d ≤ 2.52119 rad.
- The lower and upper input deviations are 0.875311 rad and 0.875311 rad. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer d ≈ 1.64587 rad; compatible interval [0.770564, 2.52119] rad. Endpoint interpolation gives 1.64587 rad.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.