University Physics I · topic UPI-11.04
Conservation of angular momentum
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.
Conservation of angular momentum: predict a finite change
Angular momentum is conserved about a point when the net external torque about that point is zero: the system's total L⃗ then keeps its size and direction, however its parts rearrange, collide or grip one another. Use the supplied topic model Y_out = 3*m*d/(0.324+m*d^2), where Y_out is ω JUST AFTER IMPACT (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 from pivot, measured in m. Keep m (Puck mass m) = 0.4 kg fixed. For this exercise use only 0.185 m ≤ d ≤ 0.815 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 = 0.3236 m and 0.4118 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 IMPACT against d
Predictions from the supplied Conservation of angular momentum 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 IMPACT / rad s⁻¹
Horizontal axis (x): d: Impact distance d from 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 from pivot / m | ω JUST AFTER IMPACT / rad s⁻¹ |
|---|---|
| 0.185 | 0.6574076816014689 |
| 0.34249999999999997 | 1.1080481771798691 |
| 0.49999999999999994 | 1.4150943396226414 |
| 0.6575 | 1.587772741222223 |
| 0.815 | 1.6584985331275754 |
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 = 0.3236 m, substitute into Y_out = 3*m*d/(0.324+m*d^2): Y₀ = 1.06131 rad s⁻¹.
- At d = 0.4118 m, the same fixed-parameter model gives Y₁ = 1.26115 rad s⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.199842 rad s⁻¹; the input change is Δd = 0.0882 m.
- Average sensitivity is ΔY_out/Δd = 2.26578 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 = 1.06131 rad s⁻¹, Y₁ = 1.26115 rad s⁻¹; ΔY_out = 0.199842 rad s⁻¹; average sensitivity = 2.26578 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.
Conservation of angular momentum: infer an input with uncertainty
Angular momentum is conserved about a point when the net external torque about that point is zero: the system's total L⃗ then keeps its size and direction, however its parts rearrange, collide or grip one another. Use the supplied topic model Y_out = 3*m*d/(0.324+m*d^2), where Y_out is ω JUST AFTER IMPACT (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 from pivot, measured in m. Keep m (Puck mass m) = 0.4 kg fixed. For this exercise use only 0.185 m ≤ d ≤ 0.815 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 = 1.5059485 rad s⁻¹, with a bounded reading uncertainty ±0.022883 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 IMPACT against d
Predictions from the supplied Conservation of angular momentum 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 IMPACT / rad s⁻¹
Horizontal axis (x): d: Impact distance d from 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 from pivot / m | ω JUST AFTER IMPACT / rad s⁻¹ |
|---|---|
| 0.185 | 0.6574076816014689 |
| 0.34249999999999997 | 1.1080481771798691 |
| 0.49999999999999994 | 1.4150943396226414 |
| 0.6575 | 1.587772741222223 |
| 0.815 | 1.6584985331275754 |
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(0.185) = 0.657408 rad s⁻¹ and Y_out(0.815) = 1.6585 rad s⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives d_line = 0.185 + (1.50595 − 0.657408)(0.815 − 0.185)/(1.6585 − 0.657408) = 0.718998 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 = 0.5693 m; substituting back gives Y_out = 1.5059485 rad s⁻¹.
- The interpolation error in the input is |d_line − d| = 0.149698 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.4830655 rad s⁻¹ and 1.5288315 rad s⁻¹. Sort the roots even if the response decreases: 0.549935 m ≤ d ≤ 0.590452 m.
- The lower and upper input deviations are 0.0193654 m and 0.0211516 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 ≈ 0.5693 m; compatible interval [0.549935, 0.590452] m. Endpoint interpolation gives 0.718998 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.