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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.

Extra challenging · 01

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.

Explore the question data

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

LEVER ARM r⊥ = r |sin φ| against phiPredictions 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. Horizontal axis: phi: Angle φ from r to F / °. Vertical axis: LEVER ARM r⊥ = r |sin φ| / m. Exact values are available in the data table below.0.039270.10670.17410.24150.3089-88.07-67.66-47.25-26.84-6.426xy

Horizontal axis (x): phi: Angle φ from r to F / °

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1phi: Angle φ from r to F / °: -85.05LEVER ARM r⊥ = r |sin φ| / m: 0.2989
Read the exact data as a table
LEVER ARM r⊥ = r |sin φ| against phi — supplied values
phi: Angle φ from r to F / °LEVER ARM r⊥ = r |sin φ| / m
-85.050000000000010.2988811129478824
-66.150.2743821479395841
-47.250.22029675283070566
-28.3499999999999940.14245691696117835
-9.4499999999999890.049256053970658825

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |phi_line − phi| = 12.6949 °. This is larger than the requested numerical tolerance.
  5. 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 °.
  6. 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.

Extra challenging · 02

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.

Explore the question data

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

I ABOUT THE AXIS against sPredictions 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. Horizontal axis: s: Axis position s / m. Vertical axis: I ABOUT THE AXIS / kg m². Exact values are available in the data table below.0.49450.84751.2011.5541.9070.32310.48890.65470.82060.9864xy

Horizontal axis (x): s: Axis position s / m

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1s: Axis position s / m: 0.3476I ABOUT THE AXIS / kg m²: 0.5468
Read the exact data as a table
I ABOUT THE AXIS against s — supplied values
s: Axis position s / mI ABOUT THE AXIS / kg m²
0.3476250.5468044218750001
0.50118750.66142923046875
0.65474999999999990.9175426874999999
0.80831251.3151447929687499
0.9618751.854235546875

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |s_line − s| = 0.126512 m. This is larger than the requested numerical tolerance.
  5. 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.
  6. 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.

Extra challenging · 03

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.

Explore the question data

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 / %

I LEFT, % OF SOLID DISC against hPredictions 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. Horizontal axis: h: Hole radius R₁/R. Vertical axis: I LEFT, % OF SOLID DISC / %. Exact values are available in the data table below.72.8279.8786.9393.981010.05980.22990.40.57010.7402xy

Horizontal axis (x): h: Hole radius R₁/R

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1h: Hole radius R₁/R: 0.085I LEFT, % OF SOLID DISC / %: 99.99
Read the exact data as a table
I LEFT, % OF SOLID DISC against h — supplied values
h: Hole radius R₁/RI LEFT, % OF SOLID DISC / %
0.08599.9947799375
0.242599.65418249609374
0.497.44
0.557590.33994749609376
0.71573.86489993750001

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
  1. 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.
  2. 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.
  3. 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 %.
  4. The interpolation error in the input is |h_line − h| = 0.267474. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 94.422505 % and 95.876145 %. Sort the roots even if the response decreases: 0.450636 ≤ h ≤ 0.48597.
  6. 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.

Extra challenging · 04

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.

Explore the question data

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 / %

BALL OWN I ÷ ITS SHIFT TERM against pPredictions 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. Horizontal axis: p: Pivot distance p from left end / m. Vertical axis: BALL OWN I ÷ ITS SHIFT TERM / %. Exact values are available in the data table below.0.045122.4754.9047.3349.7630.074750.28740.50.71260.9252xy

Horizontal axis (x): p: Pivot distance p from left end / m

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1p: Pivot distance p from left end / m: 0.1062BALL OWN I ÷ ITS SHIFT TERM / %: 0.405
Read the exact data as a table
BALL OWN I ÷ ITS SHIFT TERM against p — supplied values
p: Pivot distance p from left end / mBALL OWN I ÷ ITS SHIFT TERM / %
0.106250.40504726869981406
0.3031250.629911572472126
0.51.111111111111111
0.69687499999999992.4613905414338064
0.89374999999999999.403122130394845

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
  1. 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.
  2. 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.
  3. 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 %.
  4. The interpolation error in the input is |p_line − p| = 0.382996 m. This is larger than the requested numerical tolerance.
  5. 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.
  6. 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.

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