The Rotational Variables — Question 7

Section 10-1 · Physics pages 1384–1385

Question 7 A shared angle origin does not alter angular rates Two encoders observe the same fixed-axis rotation. Encoder A uses θA = 0.4 + 3t −t2 in SI units. Encoder B has a zero shifted by π/2 but keeps the same positive sense. x y + B zero A zero The two encoders use reference directions separated by a right angle. (a) Write θB and compare both angular velocities and accelerations. (b) If B also reverses its positive sense, how do the three signed quantities transform?
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Question 7 - Solution A shared angle origin does not alter angular rates x y + B zero A zero (a) Derivation and calculation With B’s zero at A angle π/2 and the same positive sense, θB(t) = 0.4 + 3t −t2 −π 2 . ωB = dθB dt = ωA = 3 −2t , αB = αA = −2 . Rates are in rad/s and rad/s2, respectively. (b) Result and interpretation Reversing B’s positive sense also multiplies its coordinate by −1: θB = −(θA −π/2), ωB = −ωA = −3 + 2t, αB = −αA = +2 . A constant origin shift disappears under differentiation; a sense reversal changes every signed rate. Physical check Both conventions predict the same stopping time t = 1.5 s.

Question 7

A shared angle origin does not alter angular rates

Two encoders observe the same fixed-axis rotation. Encoder A uses θA=0.4+3t−t2\theta_A=0.4+3t-t^2 in SI units. Encoder B has a zero shifted by π/2\pi/2 but keeps the same positive sense.

Diagram shown in the original SVG worksheet.

  1. Write θB\theta_B and compare both angular velocities and accelerations.

  2. If B also reverses its positive sense, how do the three signed quantities transform?

Question 7 - Solution

A shared angle origin does not alter angular rates

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

With B’s zero at A angle π/2\pi/2 and the same positive sense, θB(t)=0.4+3t−t2−π2.\boxed{\theta_B(t)=0.4+3t-t^2-\frac\pi2}. ωB=dθBdt=ωA=3−2t,αB=αA=−2.\omega_B=\frac{d\theta_B}{dt}=\omega_A=\boxed{3-2t},\qquad \alpha_B=\alpha_A=\boxed{-2}. Rates are in rad/s and rad/s2^2, respectively.

(b) Result and interpretation

Reversing B’s positive sense also multiplies its coordinate by −1-1: θB=−(θA−π/2),ωB=−ωA=−3+2t,αB=−αA=+2.\boxed{\theta_B=-(\theta_A-\pi/2),\quad \omega_B=-\omega_A=-3+2t,\quad\alpha_B=-\alpha_A=+2}. A constant origin shift disappears under differentiation; a sense reversal changes every signed rate.

Physical check

Both conventions predict the same stopping time t=1.5t=1.5 s.