Rotation with Constant Angular Acceleration — Question 3

Section 10-3 · Physics pages 1416–1417

Question 3 Equal-time angle increments reveal a constant acceleration Over three consecutive 1.0 s intervals a rotor advances by 3, 5 and 7 rad, with no reversal. Assume constant angular acceleration throughout. t ∆θn 3 1 5 2 7 3 9 4 Equal one-second intervals give an arithmetic progression of angular increments; the fourth is a prediction. (a) Infer the initial angular velocity and acceleration. (b) Predict the next interval displacement and state what a measured 8 rad would imply.
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Question 3 - Solution Equal-time angle increments reveal a constant acceleration t ∆θn 3 1 5 2 7 3 9 4 (a) Derivation and calculation The increment in interval n of duration h is ∆θn = ω0h + αh2(n −1 2), ∆θn+1 −∆θn = αh2. α = 5 −3 12 = 2 rad/s2 , ω0 = 3 1 −2(1) 2 = 2 rad/s . (b) Result and interpretation The next predicted displacement is ∆θ4 = 2 + 2(4 −1 2) = 9 rad . An exact reading of 8 gives a residual 8 −9 = −1 rad and violates the common constant-acceleration model; it cannot fit all the stated exact increments. Physical check The observed successive differences are both 2 rad, providing two consistent estimates of acceleration.

Question 3

Equal-time angle increments reveal a constant acceleration

Over three consecutive 1.0s1.0\,\mathrm s intervals a rotor advances by 33, 55 and 7rad7\,\mathrm{rad}, with no reversal. Assume constant angular acceleration throughout.

Diagram shown in the original SVG worksheet.

  1. Infer the initial angular velocity and acceleration.

  2. Predict the next interval displacement and state what a measured 8rad8\,\mathrm{rad} would imply.

Question 3 - Solution

Equal-time angle increments reveal a constant acceleration

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

The increment in interval nn of duration hh is Δθn=ω0h+αh2(n−12),Δθn+1−Δθn=αh2.\Delta\theta_n=\omega_0h+\alpha h^2(n-\tfrac12),\qquad \Delta\theta_{n+1}-\Delta\theta_n=\alpha h^2. α=5−312=2rad/s2,ω0=31−2(1)2=2rad/s.\boxed{\alpha=\frac{5-3}{1^2}=2\ \mathrm{rad/s^2}},\qquad \boxed{\omega_0=\frac31-\frac{2(1)}2=2\ \mathrm{rad/s}}.

(b) Result and interpretation

The next predicted displacement is Δθ4=2+2(4−12)=9rad.\boxed{\Delta\theta_4=2+2(4-\tfrac12)=9\ \mathrm{rad}}. An exact reading of 8 gives a residual 8−9=−18-9=-1 rad and violates the common constant-acceleration model; it cannot fit all the stated exact increments.

Physical check

The observed successive differences are both 2 rad, providing two consistent estimates of acceleration.