University Physics I · topic UPI-10.06
Rotational kinetic energy
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.
Rotational Kinetic Energy: predict a finite change
Rotational kinetic energy is the summed ½mv² of all the particles of a rigid body turning about an axis, which collapses to K = ½Iω² with I taken about that same axis and ω in rad s⁻¹. Use the supplied topic model Y_out = 0.0675+(p-0.2)^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. p is Axis position p along the rod, measured in m. Keep w (Angular speed ω) = 4 rad/s fixed. For this exercise use only 0.221375 m ≤ p ≤ 0.625625 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. Calculate Y_out at p = 0.31031 m and 0.366905 m. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δp across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
I ABOUT THE AXIS against p
Predictions from the supplied Rotational Kinetic Energy 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): p: Axis position p along the rod / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Axis position p along the rod / m | I ABOUT THE AXIS / kg m² |
|---|---|
| 0.22137499999999996 | 0.067956890625 |
| 0.32243749999999993 | 0.08249094140624999 |
| 0.4234999999999999 | 0.11745224999999995 |
| 0.5245624999999999 | 0.17284081640624996 |
| 0.6256249999999999 | 0.2486566406249999 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of p separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At p = 0.31031 m, substitute into Y_out = 0.0675+(p-0.2)^2: Y₀ = 0.0796683 kg m².
- At p = 0.366905 m, the same fixed-parameter model gives Y₁ = 0.0953573 kg m².
- Signed change: ΔY_out = Y₁ − Y₀ = 0.015689 kg m²; the input change is Δp = 0.056595 m.
- Average sensitivity is ΔY_out/Δp = 0.277215 kg m² per m. 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 = 0.0796683 kg m², Y₁ = 0.0953573 kg m²; ΔY_out = 0.015689 kg m²; average sensitivity = 0.277215 kg m² per m.
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.
Rotational Kinetic Energy: infer an input with uncertainty
Rotational kinetic energy is the summed ½mv² of all the particles of a rigid body turning about an axis, which collapses to K = ½Iω² with I taken about that same axis and ω in rad s⁻¹. Use the supplied topic model Y_out = 0.0675+(p-0.2)^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. p is Axis position p along the rod, measured in m. Keep w (Angular speed ω) = 4 rad/s fixed. For this exercise use only 0.221375 m ≤ p ≤ 0.625625 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 = 0.13930658 kg m², with a bounded reading uncertainty ±0.010702 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 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 0.000040425 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 p
Predictions from the supplied Rotational Kinetic Energy 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): p: Axis position p along the rod / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Axis position p along the rod / m | I ABOUT THE AXIS / kg m² |
|---|---|
| 0.22137499999999996 | 0.067956890625 |
| 0.32243749999999993 | 0.08249094140624999 |
| 0.4234999999999999 | 0.11745224999999995 |
| 0.5245624999999999 | 0.17284081640624996 |
| 0.6256249999999999 | 0.2486566406249999 |
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(0.221375) = 0.0679569 kg m² and Y_out(0.625625) = 0.248657 kg m². The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives p_line = 0.221375 + (0.139307 − 0.0679569)(0.625625 − 0.221375)/(0.248657 − 0.0679569) = 0.380994 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 = 0.467967 m; substituting back gives Y_out = 0.13930658 kg m².
- The interpolation error in the input is |p_line − p| = 0.0869735 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.12860458 kg m² and 0.15000858 kg m². Sort the roots even if the response decreases: 0.447193 m ≤ p ≤ 0.487243 m.
- The lower and upper input deviations are 0.0207741 m and 0.0192756 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 ≈ 0.467967 m; compatible interval [0.447193, 0.487243] m. Endpoint interpolation gives 0.380994 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.