University Physics I · unit UPI-09
Rotational Kinematics
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 position and displacement: infer an input with uncertainty
Angular position θ is the signed angle, in radians, from a chosen reference line to a line fixed in the rotating body; angular displacement Δθ = θ − θ₀ is the change in that angle, with every full turn counted. Use the supplied topic model Y_out = d/(2*pi), where Y_out is REVOLUTIONS Δθ/2π (rev). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Angular displacement Δθ, measured in rad. Keep t0 (Starting angle θ₀) = 0.5 rad; r (Radius of the mark r) = 0.4 m fixed. For this exercise use only -7.48125 rad ≤ d ≤ 7.48125 rad 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 = 0.26194914 rev, with a bounded reading uncertainty ±0.13931 rev. 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 0.00149625 rad, 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.
REVOLUTIONS Δθ/2π against d
Predictions from the supplied Angular position and displacement 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): REVOLUTIONS Δθ/2π / rev
Horizontal axis (x): d: Angular displacement Δθ / rad
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Angular displacement Δθ / rad | REVOLUTIONS Δθ/2π / rev |
|---|---|
| -7.48125 | -1.1906779180062421 |
| -3.7406250000000005 | -0.5953389590031211 |
| -8.881784197001252e-16 | -1.4135798584282297e-16 |
| 3.7406249999999988 | 0.5953389590031208 |
| 7.481249999999998 | 1.1906779180062417 |
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(-7.48125) = -1.19068 rev and Y_out(7.48125) = 1.19068 rev. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives d_line = -7.48125 + (0.261949 − -1.19068)(7.48125 − -7.48125)/(1.19068 − -1.19068) = 1.64587 rad. 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 = 1.64587 rad; substituting back gives Y_out = 0.26194914 rev.
- The interpolation error in the input is |d_line − d| = 0 rad. This is within the requested numerical tolerance.
- Invert both limits Y_out = 0.12263914 rev and 0.40125914 rev. Sort the roots even if the response decreases: 0.770564 rad ≤ d ≤ 2.52119 rad.
- The lower and upper input deviations are 0.875311 rad and 0.875311 rad. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer d ≈ 1.64587 rad; compatible interval [0.770564, 2.52119] rad. Endpoint interpolation gives 1.64587 rad.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Angular Velocity, Period & Frequency: infer an input with uncertainty
Angular velocity is the rate at which a rotating body's angle changes, ω = dθ/dt in rad s⁻¹; its sign, or its direction along the axle by the right-hand rule, records which way the body turns. Use the supplied topic model Y_out = (pi/2*d+k*(sin(pi*(t+d)/2)-sin(pi*t/2)))/max(d,0.1), where Y_out is ω_avg OVER Δt (SECANT) (rad s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value; unmarked trigonometric functions use radians. t is Instant t, measured in s. Keep d (Averaging interval Δt) = 2 s; k (Push unevenness k) = 0.5 fixed. For this exercise use only 1.0525 s ≤ t ≤ 1.7875 s 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 = 1.2177153 rad s⁻¹, with a bounded reading uncertainty ±0.021783 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 t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000735 s, 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_avg OVER Δt (SECANT) against t
Predictions from the supplied Angular Velocity, Period & Frequency 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_avg OVER Δt (SECANT) / rad s⁻¹
Horizontal axis (x): t: Instant t / s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Instant t / s | ω_avg OVER Δt (SECANT) / rad s⁻¹ |
|---|---|
| 1.0525 | 1.0724955570312307 |
| 1.23625 | 1.104831939100146 |
| 1.42 | 1.1757188206070512 |
| 1.6037499999999998 | 1.2792915373032208 |
| 1.7874999999999999 | 1.4069812370140495 |
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(t) = 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(1.0525) = 1.0725 rad s⁻¹ and Y_out(1.7875) = 1.40698 rad s⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 1.0525 + (1.21772 − 1.0725)(1.7875 − 1.0525)/(1.40698 − 1.0725) = 1.37161 s. 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 t = 1.50085 s; substituting back gives Y_out = 1.2177153 rad s⁻¹.
- The interpolation error in the input is |t_line − t| = 0.129244 s. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.1959323 rad s⁻¹ and 1.2394983 rad s⁻¹. Sort the roots even if the response decreases: 1.46037 s ≤ t ≤ 1.53891 s.
- The lower and upper input deviations are 0.0404814 s and 0.0380593 s. 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 t ≈ 1.50085 s; compatible interval [1.46037, 1.53891] s. Endpoint interpolation gives 1.37161 s.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Angular Acceleration: infer an input with uncertainty
Angular acceleration α is the rate at which angular velocity changes, α = dω/dt, measured in rad s⁻² and signed by the same positive sense of rotation chosen for θ and ω. Use the supplied topic model Y_out = wf+(w0-wf)*exp(-t/1.5), where Y_out is ω AT t (rad s⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. t is Cursor time t, measured in s. Keep w0 (Initial angular velocity ω₀) = -6 rad s⁻¹; wf (Late-time angular velocity ω∞) = 5 rad s⁻¹ fixed. For this exercise use only 0.860938 s ≤ t ≤ 5.38906 s 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 = 4.0173483 rad s⁻¹, with a bounded reading uncertainty ±0.10198 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 t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000452812 s, 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 AT t against t
Predictions from the supplied Angular Acceleration 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 AT t / rad s⁻¹
Horizontal axis (x): t: Cursor time t / s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Cursor time t / s | ω AT t / rad s⁻¹ |
|---|---|
| 0.8609375 | -1.196204576154149 |
| 1.9929687499999997 | 2.086807847848962 |
| 3.1249999999999996 | 3.6303408141146467 |
| 4.25703125 | 4.356044439397876 |
| 5.3890625 | 4.697239453213067 |
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(t) = 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(0.860938) = -1.1962 rad s⁻¹ and Y_out(5.38906) = 4.69724 rad s⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 0.860938 + (4.01735 − -1.1962)(5.38906 − 0.860938)/(4.69724 − -1.1962) = 4.86668 s. 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 t = 3.62309 s; substituting back gives Y_out = 4.0173483 rad s⁻¹.
- The interpolation error in the input is |t_line − t| = 1.24359 s. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.9153683 rad s⁻¹ and 4.1193283 rad s⁻¹. Sort the roots even if the response decreases: 3.47498 s ≤ t ≤ 3.78745 s.
- The lower and upper input deviations are 0.148112 s and 0.164355 s. 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 t ≈ 3.62309 s; compatible interval [3.47498, 3.78745] s. Endpoint interpolation gives 4.86668 s.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Constant Angular Acceleration: infer an input with uncertainty
Constant angular acceleration means ω changes by the same amount in every second, so ω is linear in time and the angle swept is quadratic in time: ω = ω₀ + αt and θ = θ₀ + ω₀t + ½αt². Use the supplied topic model Y_out = 0.5*(w+max(w+a*t,0))*min(t,w/max(-a,0.001)), where Y_out is BRAKED WHEEL Δθ (rad). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value; max(a,b) selects the larger value. t is Elapsed time t, measured in s. Keep w (Initial angular velocity ω₀) = 6 rad s⁻¹; a (Angular acceleration α) = -1.5 rad s⁻² fixed. For this exercise use only 0.6375 s ≤ t ≤ 3.945 s 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 = 10.643383 rad, with a bounded reading uncertainty ±0.20315 rad. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00033075 s, 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.
BRAKED WHEEL Δθ against t
Predictions from the supplied Constant Angular Acceleration 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): BRAKED WHEEL Δθ / rad
Horizontal axis (x): t: Elapsed time t / s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Elapsed time t / s | BRAKED WHEEL Δθ / rad |
|---|---|
| 0.6375 | 3.5201953124999994 |
| 1.464375 | 7.177954394531251 |
| 2.29125 | 9.810130078124999 |
| 3.118125 | 11.416722363281249 |
| 3.9450000000000003 | 11.99773125 |
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(t) = 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(0.6375) = 3.5202 rad and Y_out(3.945) = 11.9977 rad. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 0.6375 + (10.6434 − 3.5202)(3.945 − 0.6375)/(11.9977 − 3.5202) = 3.4166 s. 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 t = 2.65508 s; substituting back gives Y_out = 10.643383 rad.
- The interpolation error in the input is |t_line − t| = 0.761528 s. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 10.440233 rad and 10.846533 rad. Sort the roots even if the response decreases: 2.55789 s ≤ t ≤ 2.75986 s.
- The lower and upper input deviations are 0.097188 s and 0.104781 s. 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 t ≈ 2.65508 s; compatible interval [2.55789, 2.75986] s. Endpoint interpolation gives 3.4166 s.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.