University Physics I · unit UPI-11
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Angular Momentum of a Particle: infer an input with uncertainty
The angular momentum of a particle about a point O is L = r×p, the cross product of its position vector measured from O with its linear momentum; its magnitude is the momentum times the perpendicular distance from O to the particle's line of motion. Use the supplied topic model Y_out = hypot(s,h), where Y_out is DISTANCE r FROM O (m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; hypot(a,b) means √(a²+b²). h is Origin offset from the line h, measured in m. Keep s (Particle position along the line s) = -3.3 m; v (Speed v (mass 2.0 kg)) = 2 m s⁻¹ fixed. For this exercise use only -1.18125 m ≤ h ≤ -0.07875 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 = 3.338982 m, with a bounded reading uncertainty ±0.0057064 m. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown h by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00011025 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.
DISTANCE r FROM O against h
Predictions from the supplied Angular Momentum of a Particle 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): DISTANCE r FROM O / m
Horizontal axis (x): h: Origin offset from the line h / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| h: Origin offset from the line h / m | DISTANCE r FROM O / m |
|---|---|
| -1.18125 | 3.505046584925798 |
| -0.9056249999999999 | 3.4220106137510733 |
| -0.6299999999999999 | 3.359598190260258 |
| -0.3543749999999999 | 3.3189729797973646 |
| -0.07874999999999988 | 3.3009394969462864 |
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(h) = 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(-1.18125) = 3.50505 m and Y_out(-0.07875) = 3.30094 m. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives h_line = -1.18125 + (3.33898 − 3.50505)(-0.07875 − -1.18125)/(3.30094 − 3.50505) = -0.284239 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 h = -0.508725 m; substituting back gives Y_out = 3.338982 m.
- The interpolation error in the input is |h_line − h| = 0.224485 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.3332756 m and 3.3446884 m. Sort the roots even if the response decreases: -0.544922 m ≤ h ≤ -0.469815 m.
- The lower and upper input deviations are 0.0361978 m and 0.0389096 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 h ≈ -0.508725 m; compatible interval [-0.544922, -0.469815] m. Endpoint interpolation gives -0.284239 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.
Torque & Angular-Momentum Change: infer an input with uncertainty
The net external torque about a point is the rate at which a system's angular momentum about that same point changes, τ⃗ₑₓₜ = dL⃗/dt, so a torque acting for a time changes L⃗ by the angular impulse ∫τ⃗ dt. Use the supplied topic model Y_out = deg(atan2(((imp*sind(phi)*1000000)/1000000), (spin+imp*cosd(phi)))), where Y_out is L⃗ TURNED BY (before display rounding) (°). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees; deg converts radians to degrees; atan2(y,x) is the angle of the vector (x,y), in radians. phi is Angle from L⃗₀ to τ⃗, φ, measured in °. Keep spin (Initial angular momentum L₀) = 4 kg m² s⁻¹; imp (Angular impulse τΔt) = 2 N m s fixed. For this exercise use only 123.075 ° ≤ phi ≤ 160.875 ° 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 = 25.474639 °, with a bounded reading uncertainty ±0.67143 °. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown phi by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00378 °, 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.
L⃗ TURNED BY (before display rounding) against phi
Predictions from the supplied Torque & Angular-Momentum Change 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): L⃗ TURNED BY (before display rounding) / °
Horizontal axis (x): phi: Angle from L⃗₀ to τ⃗, φ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| phi: Angle from L⃗₀ to τ⃗, φ / ° | L⃗ TURNED BY (before display rounding) / ° |
|---|---|
| 123.07499999999999 | 29.950848610670448 |
| 132.52499999999998 | 29.100147672077064 |
| 141.975 | 26.937264249209097 |
| 151.425 | 23.092089752378275 |
| 160.875 | 17.249195132642292 |
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(phi) = 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(123.075) = 29.9508 ° and Y_out(160.875) = 17.2492 °. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives phi_line = 123.075 + (25.4746 − 29.9508)(160.875 − 123.075)/(17.2492 − 29.9508) = 136.396 °. 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 phi = 146.133 °; substituting back gives Y_out = 25.474639 °.
- The interpolation error in the input is |phi_line − phi| = 9.73684 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 24.803209 ° and 26.146069 °. Sort the roots even if the response decreases: 144.344 ° ≤ phi ≤ 147.768 °.
- The lower and upper input deviations are 1.7892 ° and 1.63492 °. 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 phi ≈ 146.133 °; compatible interval [144.344, 147.768] °. Endpoint interpolation gives 136.396 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.