University Physics I · topic UPI-09.05
Linear and angular relations
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.
Linear and angular relations: predict a finite change
The linear–angular relations turn a rigid body's shared θ, ω and α into the motion of one point: multiply each by that point's perpendicular distance r from the axis to get its arc length, speed and tangential acceleration. Use the supplied topic model Y_out = w*w*r, where Y_out is RADIAL aᵣ = ω²r (m s⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. w is Angular velocity ω, measured in rad s⁻¹. Keep r (Distance of P from axis r) = 0.25 m; al (Angular acceleration α) = 3 rad s⁻² fixed. For this exercise use only 0.7125 rad s⁻¹ ≤ w ≤ 2.2875 rad s⁻¹ 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 w = 1.059 rad s⁻¹ and 1.2795 rad s⁻¹. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δw across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
RADIAL aᵣ = ω²r against w
Predictions from the supplied Linear and angular relations 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): RADIAL aᵣ = ω²r / m s⁻²
Horizontal axis (x): w: Angular velocity ω / rad s⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Angular velocity ω / rad s⁻¹ | RADIAL aᵣ = ω²r / m s⁻² |
|---|---|
| 0.7125 | 0.1269140625 |
| 1.10625 | 0.30594726562499996 |
| 1.5 | 0.5625 |
| 1.8937499999999998 | 0.8965722656249998 |
| 2.2874999999999996 | 1.3081640624999995 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of w separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At w = 1.059 rad s⁻¹, substitute into Y_out = w*w*r: Y₀ = 0.28037 m s⁻².
- At w = 1.2795 rad s⁻¹, the same fixed-parameter model gives Y₁ = 0.40928 m s⁻².
- Signed change: ΔY_out = Y₁ − Y₀ = 0.12891 m s⁻²; the input change is Δw = 0.2205 rad s⁻¹.
- Average sensitivity is ΔY_out/Δw = 0.584625 m s⁻² per rad s⁻¹. 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 = 0.28037 m s⁻², Y₁ = 0.40928 m s⁻²; ΔY_out = 0.12891 m s⁻²; average sensitivity = 0.584625 m s⁻² per rad s⁻¹.
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.
Linear and angular relations: infer an input with uncertainty
The linear–angular relations turn a rigid body's shared θ, ω and α into the motion of one point: multiply each by that point's perpendicular distance r from the axis to get its arc length, speed and tangential acceleration. Use the supplied topic model Y_out = w*w*r, where Y_out is RADIAL aᵣ = ω²r (m s⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. w is Angular velocity ω, measured in rad s⁻¹. Keep r (Distance of P from axis r) = 0.25 m; al (Angular acceleration α) = 3 rad s⁻² fixed. For this exercise use only 0.7125 rad s⁻¹ ≤ w ≤ 2.2875 rad s⁻¹ 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 = 0.69994139 m s⁻², with a bounded reading uncertainty ±0.085954 m s⁻². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown w by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575 rad s⁻¹, 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.
RADIAL aᵣ = ω²r against w
Predictions from the supplied Linear and angular relations 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): RADIAL aᵣ = ω²r / m s⁻²
Horizontal axis (x): w: Angular velocity ω / rad s⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Angular velocity ω / rad s⁻¹ | RADIAL aᵣ = ω²r / m s⁻² |
|---|---|
| 0.7125 | 0.1269140625 |
| 1.10625 | 0.30594726562499996 |
| 1.5 | 0.5625 |
| 1.8937499999999998 | 0.8965722656249998 |
| 2.2874999999999996 | 1.3081640624999995 |
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(w) = 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(0.7125) = 0.126914 m s⁻² and Y_out(2.2875) = 1.30816 m s⁻². The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives w_line = 0.7125 + (0.699941 − 0.126914)(2.2875 − 0.7125)/(1.30816 − 0.126914) = 1.47654 rad s⁻¹. 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 w = 1.67325 rad s⁻¹; substituting back gives Y_out = 0.69994139 m s⁻².
- The interpolation error in the input is |w_line − w| = 0.196714 rad s⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.61398739 m s⁻² and 0.78589539 m s⁻². Sort the roots even if the response decreases: 1.56715 rad s⁻¹ ≤ w ≤ 1.77301 rad s⁻¹.
- The lower and upper input deviations are 0.106103 rad s⁻¹ and 0.0997648 rad s⁻¹. 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 w ≈ 1.67325 rad s⁻¹; compatible interval [1.56715, 1.77301] rad s⁻¹. Endpoint interpolation gives 1.47654 rad s⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.