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Physics / Chapter 10: Rotation / Section 10-9

Work and Rotational Kinetic Energy — Question 1

Section 10-9 · Physics pages 1532–1533

Question 1 A positive torque can remove energy from clockwise motion A rotor of inertia 0.50 kg m2 starts at ω0 = −4 rad/s. Its only torque is constant +3 N m. t ω −4 2/3 1 +2 The positive torque first reduces clockwise speed, then increases counterclockwise speed; time is in seconds. (a) Find when it stops and the work done up to that instant. (b) Find its angular velocity and total work after 1 s. Explain how torque can first remove and then add energy.
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Question 1 - Solution A positive torque can remove energy from clockwise motion t ω −4 2/3 1 +2 (a) Derivation and calculation The angular acceleration and stopping time follow from the signed rate: α = τ I = 3 0.50 = 6 rad/s2, 0 = −4 + 6ts ⇒ts = 2 3 s . ∆θs = −4 (2 3 ) + 1 2(6) (2 3 )2 = −4 3 rad, Ws = 3 ( −4 3 ) = −4 J . (b) Result and interpretation At one second, ω(1) = −4 + 6 = 2 rad/s , ∆θ(1) = −4 + 3 = −1 rad, W(1) = −3 J . P = τω = 3(−4 + 6t). Power is negative before 2/3 s and positive afterward. The final energy is Kf = 4 −3 = 1 J. Physical check The positive torque does negative work while the angular displacement is negative.

Question 1

A positive torque can remove energy from clockwise motion

A rotor of inertia 0.50kgm20.50\,\mathrm{kg\,m^2} starts at ω0=−4rad/s\omega_0=-4\,\mathrm{rad/s}. Its only torque is constant +3Nm+3\,\mathrm{N\,m}.

Diagram shown in the original SVG worksheet.

  1. Find when it stops and the work done up to that instant.

  2. Find its angular velocity and total work after 1s1\,\mathrm s. Explain how torque can first remove and then add energy.

Question 1 - Solution

A positive torque can remove energy from clockwise motion

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

The angular acceleration and stopping time follow from the signed rate: α=τI=30.50=6rad/s2,0=−4+6ts⇒ts=23s.\alpha=\frac\tau I=\frac3{0.50}=6\ \mathrm{rad/s^2},\qquad 0=-4+6t_s\Rightarrow\boxed{t_s=\frac23\ \mathrm s}. Δθs=−4(23)+12(6)(23)2=−43rad,Ws=3(−43)=−4J.\Delta\theta_s=-4\left(\frac23\right)+\tfrac12(6)\left(\frac23\right)^2=-\frac43\ \mathrm{rad},\qquad \boxed{W_s=3\left(-\frac43\right)=-4\ \mathrm J}.

(b) Result and interpretation

At one second, ω(1)=−4+6=2rad/s,Δθ(1)=−4+3=−1rad,W(1)=−3J.\boxed{\omega(1)=-4+6=2\ \mathrm{rad/s}},\quad \Delta\theta(1)=-4+3=-1\ \mathrm{rad},\quad\boxed{W(1)=-3\ \mathrm J}. P=τω=3(−4+6t).P=\tau\omega=3(-4+6t). Power is negative before 2/32/3 s and positive afterward. The final energy is Kf=4−3=1K_f=4-3=1 J.

Physical check

The positive torque does negative work while the angular displacement is negative.

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