University Physics I · topic UPI-09.04
Constant angular acceleration
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.
Constant Angular Acceleration: predict a finite change
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. Calculate Y_out at t = 1.36515 s and 1.8282 s. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δt across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
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
Substitute each value of t separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At t = 1.36515 s, substitute into Y_out = 0.5*(w+max(w+a*t,0))*min(t,w/max(-a,0.001)): Y₀ = 6.79317 rad.
- At t = 1.8282 s, the same fixed-parameter model gives Y₁ = 8.46246 rad.
- Signed change: ΔY_out = Y₁ − Y₀ = 1.66929 rad; the input change is Δt = 0.46305 s.
- Average sensitivity is ΔY_out/Δt = 3.60499 rad per s. 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 = 6.79317 rad, Y₁ = 8.46246 rad; ΔY_out = 1.66929 rad; average sensitivity = 3.60499 rad per s.
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.
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.