University Physics I · unit UPI-10
Rotational Dynamics
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.
Torque & Lever Arms: infer an input with uncertainty
The torque of a force about a point O is τ = r×F, where r runs from O to the point where the force acts; its magnitude is the force times the perpendicular distance from O to the force's line of action. Use the supplied topic model Y_out = r*abs(sind(phi)), where Y_out is LEVER ARM r⊥ = r |sin φ| (m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|; sind/cosd/tand use degrees. phi is Angle φ from r to F, measured in °. Keep r (Distance r from O to P) = 0.3 m; f (Force magnitude F) = 80 N fixed. For this exercise use only -85.05 ° ≤ phi ≤ -9.45 ° 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 = 0.18852743 m, with a bounded reading uncertainty ±0.016553 m. 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 0.00756 °, 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.
LEVER ARM r⊥ = r |sin φ| against phi
Predictions from the supplied Torque & Lever Arms 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): LEVER ARM r⊥ = r |sin φ| / m
Horizontal axis (x): phi: Angle φ from r to F / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| phi: Angle φ from r to F / ° | LEVER ARM r⊥ = r |sin φ| / m |
|---|---|
| -85.05000000000001 | 0.2988811129478824 |
| -66.15 | 0.2743821479395841 |
| -47.25 | 0.22029675283070566 |
| -28.349999999999994 | 0.14245691696117835 |
| -9.449999999999989 | 0.049256053970658825 |
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(-85.05) = 0.298881 m and Y_out(-9.45) = 0.0492561 m. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives phi_line = -85.05 + (0.188527 − 0.298881)(-9.45 − -85.05)/(0.0492561 − 0.298881) = -51.6289 °. 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 = -38.934 °; substituting back gives Y_out = 0.18852743 m.
- The interpolation error in the input is |phi_line − phi| = 12.6949 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.17197443 m and 0.20508043 m. Sort the roots even if the response decreases: -43.1257 ° ≤ phi ≤ -34.977 °.
- The lower and upper input deviations are 4.19172 ° and 3.95696 °. 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 ≈ -38.934 °; compatible interval [-43.1257, -34.977] °. Endpoint interpolation gives -51.6289 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Moment of inertia of particle systems: infer an input with uncertainty
The moment of inertia of a system of point masses about a chosen axis is I = Σmᵢrᵢ², each mass multiplied by the square of its perpendicular distance from that axis; it measures how strongly the system resists a change in its rotation about that axis. Use the supplied topic model Y_out = m1*s^2+m2*(s-0.9)^2, where Y_out is I ABOUT THE AXIS (kg m²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Axis position s, measured in m. Keep m1 (Bead 1 mass m₁, at x = 0) = 2 kg; m2 (Bead 2 mass m₂, at x = 0.90 m) = 1 kg fixed. For this exercise use only 0.347625 m ≤ s ≤ 0.961875 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 = 1.0750562 kg m², with a bounded reading uncertainty ±0.079238 kg m². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown s by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000061425 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.
I ABOUT THE AXIS against s
Predictions from the supplied Moment of inertia of particle systems 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): I ABOUT THE AXIS / kg m²
Horizontal axis (x): s: Axis position s / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Axis position s / m | I ABOUT THE AXIS / kg m² |
|---|---|
| 0.347625 | 0.5468044218750001 |
| 0.5011875 | 0.66142923046875 |
| 0.6547499999999999 | 0.9175426874999999 |
| 0.8083125 | 1.3151447929687499 |
| 0.961875 | 1.854235546875 |
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(s) = 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(0.347625) = 0.546804 kg m² and Y_out(0.961875) = 1.85424 kg m². The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives s_line = 0.347625 + (1.07506 − 0.546804)(0.961875 − 0.347625)/(1.85424 − 0.546804) = 0.595805 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 s = 0.722317 m; substituting back gives Y_out = 1.0750562 kg m².
- The interpolation error in the input is |s_line − s| = 0.126512 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.9958182 kg m² and 1.1542942 kg m². Sort the roots even if the response decreases: 0.689794 m ≤ s ≤ 0.752509 m.
- The lower and upper input deviations are 0.0325234 m and 0.0301919 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 s ≈ 0.722317 m; compatible interval [0.689794, 0.752509] m. Endpoint interpolation gives 0.595805 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Moment of inertia of continuous bodies: infer an input with uncertainty
The moment of inertia of a continuous body about an axis is I = ∫ r⊥² dm: split the body into mass elements dm, weight each by the square of its perpendicular distance r⊥ from the axis, and integrate over the whole body. Use the supplied topic model Y_out = 100*(1-h^4), where Y_out is I LEFT, % OF SOLID DISC (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. h is Hole radius R₁/R, dimensionless. Keep n (Number of rings n) = 4 fixed. For this exercise use only 0.085 ≤ h ≤ 0.715 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 = 95.149325 %, with a bounded reading uncertainty ±0.72682 %. 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 0.000063, 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.
I LEFT, % OF SOLID DISC against h
Predictions from the supplied Moment of inertia of continuous bodies 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): I LEFT, % OF SOLID DISC / %
Horizontal axis (x): h: Hole radius R₁/R
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| h: Hole radius R₁/R | I LEFT, % OF SOLID DISC / % |
|---|---|
| 0.085 | 99.9947799375 |
| 0.2425 | 99.65418249609374 |
| 0.4 | 97.44 |
| 0.5575 | 90.33994749609376 |
| 0.715 | 73.86489993750001 |
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(0.085) = 99.9948 % and Y_out(0.715) = 73.8649 %. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives h_line = 0.085 + (95.1493 − 99.9948)(0.715 − 0.085)/(73.8649 − 99.9948) = 0.201826. 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 = 0.4693; substituting back gives Y_out = 95.149325 %.
- The interpolation error in the input is |h_line − h| = 0.267474. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 94.422505 % and 95.876145 %. Sort the roots even if the response decreases: 0.450636 ≤ h ≤ 0.48597.
- The lower and upper input deviations are 0.018664 and 0.0166704. 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 ≈ 0.4693; compatible interval [0.450636, 0.48597]. Endpoint interpolation gives 0.201826.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Parallel- & Perpendicular-Axis Theorems: infer an input with uncertainty
The parallel-axis theorem gives the moment of inertia about any axis as I_cm + Md², d being the perpendicular distance from the parallel axis through the centre of mass; the perpendicular-axis theorem gives a flat body's I_z as I_x + I_y for three perpendicular axes meeting at one point. Use the supplied topic model Y_out = 40*r^2/(1+r-p)^2, where Y_out is BALL OWN I ÷ ITS SHIFT TERM (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. p is Pivot distance p from left end, measured in m. Keep ms (Ball mass m_b) = 2 kg; r (Ball radius R) = 0.1 m fixed. For this exercise use only 0.10625 m ≤ p ≤ 0.89375 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 = 1.5177162 %, with a bounded reading uncertainty ±0.1669 %. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875 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.
BALL OWN I ÷ ITS SHIFT TERM against p
Predictions from the supplied Parallel- & Perpendicular-Axis Theorems 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): BALL OWN I ÷ ITS SHIFT TERM / %
Horizontal axis (x): p: Pivot distance p from left end / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Pivot distance p from left end / m | BALL OWN I ÷ ITS SHIFT TERM / % |
|---|---|
| 0.10625 | 0.40504726869981406 |
| 0.303125 | 0.629911572472126 |
| 0.5 | 1.111111111111111 |
| 0.6968749999999999 | 2.4613905414338064 |
| 0.8937499999999999 | 9.403122130394845 |
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(p) = 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(0.10625) = 0.405047 % and Y_out(0.89375) = 9.40312 %. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives p_line = 0.10625 + (1.51772 − 0.405047)(0.89375 − 0.10625)/(9.40312 − 0.405047) = 0.203629 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 p = 0.586625 m; substituting back gives Y_out = 1.5177162 %.
- The interpolation error in the input is |p_line − p| = 0.382996 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.3508162 % and 1.6846162 %. Sort the roots even if the response decreases: 0.555833 m ≤ p ≤ 0.612719 m.
- The lower and upper input deviations are 0.0307916 m and 0.026094 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 p ≈ 0.586625 m; compatible interval [0.555833, 0.612719] m. Endpoint interpolation gives 0.203629 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.