University Physics I · topic UPI-11.02
Angular momentum of a rigid body
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 of a Rigid Body: predict a finite change
A rigid body's angular momentum is the sum of r × mv over all of its mass; for rotation about a fixed axis, its component along that axis is L_z = Iω, where I = ∫R² dm. Use the supplied topic model Y_out = 2*0.5*a^2*w*sind(theta)*cosd(theta), where Y_out is ACROSS THE AXLE L_⊥ (kg m² s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. theta is Tilt θ between rod and axle, measured in °. Keep w (Spin rate ω) = 6 rad s⁻¹; a (Half-length a of the rod) = 0.25 m fixed. For this exercise use only 27.4375 ° ≤ theta ≤ 43.975 ° 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 theta = 31.0757 ° and 33.391 °. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δtheta across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
ACROSS THE AXLE L_⊥ against theta
Predictions from the supplied Angular Momentum of a Rigid Body 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): ACROSS THE AXLE L_⊥ / kg m² s⁻¹
Horizontal axis (x): theta: Tilt θ between rod and axle / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| theta: Tilt θ between rod and axle / ° | ACROSS THE AXLE L_⊥ / kg m² s⁻¹ |
|---|---|
| 27.4375 | 0.15335601490861786 |
| 31.571875 | 0.1672767638218028 |
| 35.70625 | 0.17771962004632624 |
| 39.840625 | 0.18446746356813368 |
| 43.975 | 0.18737999829879923 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of theta separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At theta = 31.0757 °, substitute into Y_out = 2*0.5*a^2*w*sind(theta)*cosd(theta): Y₀ = 0.165785 kg m² s⁻¹.
- At theta = 33.391 °, the same fixed-parameter model gives Y₁ = 0.172315 kg m² s⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.00652981 kg m² s⁻¹; the input change is Δtheta = 2.31525 °.
- Average sensitivity is ΔY_out/Δtheta = 0.00282035 kg m² s⁻¹ per °. 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 = 0.165785 kg m² s⁻¹, Y₁ = 0.172315 kg m² s⁻¹; ΔY_out = 0.00652981 kg m² s⁻¹; average sensitivity = 0.00282035 kg m² s⁻¹ per °.
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 of a Rigid Body: infer an input with uncertainty
A rigid body's angular momentum is the sum of r × mv over all of its mass; for rotation about a fixed axis, its component along that axis is L_z = Iω, where I = ∫R² dm. Use the supplied topic model Y_out = 2*0.5*a^2*w*sind(theta)*cosd(theta), where Y_out is ACROSS THE AXLE L_⊥ (kg m² s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. theta is Tilt θ between rod and axle, measured in °. Keep w (Spin rate ω) = 6 rad s⁻¹; a (Half-length a of the rod) = 0.25 m fixed. For this exercise use only 27.4375 ° ≤ theta ≤ 43.975 ° 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 = 0.18115401 kg m² s⁻¹, with a bounded reading uncertainty ±0.0009339 kg m² 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 theta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00165375 °, 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.
ACROSS THE AXLE L_⊥ against theta
Predictions from the supplied Angular Momentum of a Rigid Body 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): ACROSS THE AXLE L_⊥ / kg m² s⁻¹
Horizontal axis (x): theta: Tilt θ between rod and axle / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| theta: Tilt θ between rod and axle / ° | ACROSS THE AXLE L_⊥ / kg m² s⁻¹ |
|---|---|
| 27.4375 | 0.15335601490861786 |
| 31.571875 | 0.1672767638218028 |
| 35.70625 | 0.17771962004632624 |
| 39.840625 | 0.18446746356813368 |
| 43.975 | 0.18737999829879923 |
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(theta) = 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(27.4375) = 0.153356 kg m² s⁻¹ and Y_out(43.975) = 0.18738 kg m² s⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives theta_line = 27.4375 + (0.181154 − 0.153356)(43.975 − 27.4375)/(0.18738 − 0.153356) = 40.9488 °. 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 theta = 37.5254 °; substituting back gives Y_out = 0.18115401 kg m² s⁻¹.
- The interpolation error in the input is |theta_line − theta| = 3.42345 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.18022011 kg m² s⁻¹ and 0.18208791 kg m² s⁻¹. Sort the roots even if the response decreases: 36.9909 ° ≤ theta ≤ 38.1001 °.
- The lower and upper input deviations are 0.534495 ° and 0.574771 °. 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 theta ≈ 37.5254 °; compatible interval [36.9909, 38.1001] °. Endpoint interpolation gives 40.9488 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.