University Physics I · topic UPI-11.07
Angular momentum problem studio
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 momentum problem studio: predict a finite change
Angular momentum about a point O is conserved over an event when the net external torque about O, integrated over that event, is zero — a condition that depends on the system you choose and on where you put O. Use the supplied topic model Y_out = m*v*d/(2/3+m*d^2), where Y_out is ω JUST AFTER (rad s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Impact distance d below pivot, measured in m. Keep m (Putty mass m) = 0.5 kg; v (Putty speed v) = 6 m s⁻¹ fixed. For this exercise use only 0.285 m ≤ d ≤ 0.915 m 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 = 0.4236 m and 0.5118 m. 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.
ω JUST AFTER against d
Predictions from the supplied Angular momentum problem studio 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): ω JUST AFTER / rad s⁻¹
Horizontal axis (x): d: Impact distance d below pivot / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Impact distance d below pivot / m | ω JUST AFTER / rad s⁻¹ |
|---|---|
| 0.28500000000000003 | 1.208857888504657 |
| 0.4425 | 1.736270533401818 |
| 0.6000000000000001 | 2.125984251968504 |
| 0.7575000000000001 | 2.3831501583414076 |
| 0.915 | 2.529303136289818 |
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 = 0.4236 m, substitute into Y_out = m*v*d/(2/3+m*d^2): Y₀ = 1.6801 rad s⁻¹.
- At d = 0.5118 m, the same fixed-parameter model gives Y₁ = 1.92494 rad s⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.244841 rad s⁻¹; the input change is Δd = 0.0882 m.
- Average sensitivity is ΔY_out/Δd = 2.77598 rad s⁻¹ per m. 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 = 1.6801 rad s⁻¹, Y₁ = 1.92494 rad s⁻¹; ΔY_out = 0.244841 rad s⁻¹; average sensitivity = 2.77598 rad s⁻¹ per m.
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 momentum problem studio: infer an input with uncertainty
Angular momentum about a point O is conserved over an event when the net external torque about O, integrated over that event, is zero — a condition that depends on the system you choose and on where you put O. Use the supplied topic model Y_out = m*v*d/(2/3+m*d^2), where Y_out is ω JUST AFTER (rad s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Impact distance d below pivot, measured in m. Keep m (Putty mass m) = 0.5 kg; v (Putty speed v) = 6 m s⁻¹ fixed. For this exercise use only 0.285 m ≤ d ≤ 0.915 m 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 = 2.2544262 rad s⁻¹, with a bounded reading uncertainty ±0.041232 rad s⁻¹. 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 0.000063 m, 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.
ω JUST AFTER against d
Predictions from the supplied Angular momentum problem studio 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): ω JUST AFTER / rad s⁻¹
Horizontal axis (x): d: Impact distance d below pivot / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Impact distance d below pivot / m | ω JUST AFTER / rad s⁻¹ |
|---|---|
| 0.28500000000000003 | 1.208857888504657 |
| 0.4425 | 1.736270533401818 |
| 0.6000000000000001 | 2.125984251968504 |
| 0.7575000000000001 | 2.3831501583414076 |
| 0.915 | 2.529303136289818 |
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(0.285) = 1.20886 rad s⁻¹ and Y_out(0.915) = 2.5293 rad s⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives d_line = 0.285 + (2.25443 − 1.20886)(0.915 − 0.285)/(2.5293 − 1.20886) = 0.783853 m. 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 = 0.6693 m; substituting back gives Y_out = 2.2544262 rad s⁻¹.
- The interpolation error in the input is |d_line − d| = 0.114553 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.2131942 rad s⁻¹ and 2.2956582 rad s⁻¹. Sort the roots even if the response decreases: 0.645531 m ≤ d ≤ 0.694907 m.
- The lower and upper input deviations are 0.0237689 m and 0.0256067 m. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer d ≈ 0.6693 m; compatible interval [0.645531, 0.694907] m. Endpoint interpolation gives 0.783853 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.