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The Rotational Variables — Question 10

Section 10-1 · Physics pages 1390–1391

Question 10 One completed revolution need not take one local period A rotor starts with angular velocity 2 rad/s and follows ω(t) = 2 + 2t in SI units. Its angle starts at zero. t ω 2 T area = 2π The first revolution is completed when the accumulated area reaches 2π rad. (a) Find the time for its first complete revolution. (b) Compare with 2π/ω(0) and explain why that expression is not the elapsed turn time here.
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Question 10 - Solution One completed revolution need not take one local period t ω 2 T area = 2π (a) Derivation and calculation Integrate the rate from the specified initial angle: θ(T) −θ(0) = ∫T 0 (2 + 2t) dt = 2T + T 2. T 2 + 2T −2π = 0 =⇒ T = −1 + √ 1 + 2π = 1.699 s . The second root is negative and outside the elapsed-time domain. (b) Result and interpretation The initial-rate estimate assumes that the rate never increases: T0 = 2π ω(0) = π = 3.142 s > T. Indeed, the mean rate during the actual turn is ¯ω = 2 + ω(T) 2 = 2 + T > 2, T = 2π ¯ω . Physical check Positive angular acceleration makes the first revolution shorter than its constant-initial-speed estimate.

Question 10

One completed revolution need not take one local period

A rotor starts with angular velocity 2rad/s2\,\mathrm{rad/s} and follows ω(t)=2+2t\omega(t)=2+2t in SI units. Its angle starts at zero.

Diagram shown in the original SVG worksheet.

  1. Find the time for its first complete revolution.

  2. Compare with 2π/ω(0)2\pi/\omega(0) and explain why that expression is not the elapsed turn time here.

Question 10 - Solution

One completed revolution need not take one local period

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Integrate the rate from the specified initial angle: θ(T)−θ(0)=∫0T(2+2t)dt=2T+T2.\theta(T)-\theta(0)=\int_0^T(2+2t)\,dt=2T+T^2. T2+2T−2π=0⇒T=−1+1+2π=1.699s.T^2+2T-2\pi=0\quad\Longrightarrow\quad \boxed{T=-1+\sqrt{1+2\pi}=1.699\ \mathrm s}. The second root is negative and outside the elapsed-time domain.

(b) Result and interpretation

The initial-rate estimate assumes that the rate never increases: T0=2πω(0)=π=3.142s>T.T_0=\frac{2\pi}{\omega(0)}=\pi=3.142\ \mathrm s> T. Indeed, the mean rate during the actual turn is ω‾=2+ω(T)2=2+T>2,T=2πω‾.\bar\omega=\frac{2+\omega(T)}2=2+T>2,\qquad T=\frac{2\pi}{\bar\omega}.

Physical check

Positive angular acceleration makes the first revolution shorter than its constant-initial-speed estimate.

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