Stem-Master
Physics / Chapter 11: Rolling, Torque, and Angular Momentum / Section 11-5

Torque Revisited — Question 2

Section 11-5 · Physics pages 1634–1635

Question 2 An oblique axle responds to the projected moment A body is constrained to rotate about a fixed axis through the origin with unit vector ˆn = (ˆx + ˆy)/ √ 2. Applied torque is (2, 4, 6) N m. Its allowed angular velocity is 3ˆn rad/s. axis ˆn⃗ τ τnˆn⃗ τ⊥⃗ ω = 3ˆn Torque resolves into components parallel and perpendicular to the permitted rotation axis. (a) Find torque about the axle and applied mechanical power. (b) Find the torque component perpendicular to the axle and explain its role.
Show solutionHide solution
Question 2 - Solution An oblique axle responds to the projected moment axis ˆn⃗ τ τnˆn⃗ τ⊥⃗ ω = 3ˆn (a) Derivation and calculation Project torque onto the unit axle direction: τn = ˆn ·⃗τ = 2 + 4 √ 2 = 3 √ 2 N m . The allowed angular velocity is⃗ω = 3ˆn, so P =⃗τ ·⃗ω = 3τn = 9 √ 2 = 12.73 W . (b) Result and interpretation The axial vector component is⃗ τ∥= τnˆn = (3, 3, 0) N m. Hence⃗ τ⊥=⃗τ −⃗τ∥= (−1, 1, 6) N m . It contributes zero power because⃗τ⊥·⃗ω = 0. Its effect must be included with the bearing torque in the transverse constraint balance. Physical check The transverse component is orthogonal to the axle: (−1 + 1)/ √ 2 = 0.

Question 2

An oblique axle responds to the projected moment

A body is constrained to rotate about a fixed axis through the origin with unit vector n̂=(x̂+ŷ)/2\hat n=(\hat x+\hat y)/\sqrt2. Applied torque is (2,4,6)Nm(2,4,6)\,\mathrm{N\,m}. Its allowed angular velocity is 3n̂rad/s3\hat n\,\mathrm{rad/s}.

Diagram shown in the original SVG worksheet.

  1. Find torque about the axle and applied mechanical power.

  2. Find the torque component perpendicular to the axle and explain its role.

Question 2 - Solution

An oblique axle responds to the projected moment

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Project torque onto the unit axle direction: τn=n̂⋅τ→=2+42=32Nm.\boxed{\tau_n=\hat n\cdot\vec\tau=\frac{2+4}{\sqrt2}=3\sqrt2\ \mathrm{N\,m}}. The allowed angular velocity is ω→=3n̂\vec\omega=3\hat n, so 𝒫=τ→⋅ω→=3τn=92=12.73W.\boxed{\mathcal P=\vec\tau\cdot\vec\omega=3\tau_n=9\sqrt2=12.73\ \mathrm W}.

(b) Result and interpretation

The axial vector component is τ→∥=τnn̂=(3,3,0)Nm.\vec\tau_\parallel=\tau_n\hat n=(3,3,0)\ \mathrm{N\,m}. Hence τ→⟂=τ→−τ→∥=(−1,1,6)Nm.\boxed{\vec\tau_\perp=\vec\tau-\vec\tau_\parallel=(-1,1,6)\ \mathrm{N\,m}}. It contributes zero power because τ→⟂⋅ω→=0\vec\tau_\perp\cdot\vec\omega=0. Its effect must be included with the bearing torque in the transverse constraint balance.

Physical check

The transverse component is orthogonal to the axle: (−1+1)/2=0(-1+1)/\sqrt2=0.

← Section questionsChapter contents