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Physics / Chapter 11: Rolling, Torque, and Angular Momentum / Section 11-1

Rolling as Translation and Rotation Combined — Question 5

Section 11-1 · Physics pages 1560–1561

Question 5 Acceleration at the front is not purely tangential A wheel of radius 0.25 m rolls right without slip. At an instant vC = 2 m/s and aC = 3 m/s2 rightward. C vC Side view; arrows and labels specify the directions. (a) Find angular acceleration and the acceleration of the rightmost rim point. (b) Find the acceleration of the bottom material point.
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Question 5 - Solution Acceleration at the front is not purely tangential C vC (a) Derivation and calculation Differentiating the rolling constraint gives ω = −V/R = −8 rad/s, α = −aC/R = −12 rad/s2 . Use the complete acceleration field:⃗ aP =⃗aC +⃗α ×⃗r −ω2⃗r. At⃗r = .25ˆx m,⃗ ar = 3ˆx −3ˆy −16ˆx = (−13ˆx −3ˆy) m/s2 . (b) Result and interpretation At the bottom,⃗r = −.25ˆy m:⃗ α ×⃗r = −3ˆx, −ω2⃗r = 16ˆy. Hence⃗ ab = 3ˆx −3ˆx + 16ˆy = 16ˆy m/s2 . The tangential constraint cancels horizontal acceleration; the upward normal acceleration remains. Physical check The bottom material point has zero instantaneous velocity while accelerating away from the surface.

Question 5

Acceleration at the front is not purely tangential

A wheel of radius 0.25m0.25\,\mathrm m rolls right without slip. At an instant vC=2m/sv_C=2\,\mathrm{m/s} and aC=3m/s2a_C=3\,\mathrm{m/s^2} rightward.

Diagram shown in the original SVG worksheet.

  1. Find angular acceleration and the acceleration of the rightmost rim point.

  2. Find the acceleration of the bottom material point.

Question 5 - Solution

Acceleration at the front is not purely tangential

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Differentiating the rolling constraint gives ω=−V/R=−8rad/s,α=−aC/R=−12rad/s2.\omega=-V/R=-8\ \mathrm{rad/s},\qquad\boxed{\alpha=-a_C/R=-12\ \mathrm{rad/s^2}}. Use the complete acceleration field: a→P=a→C+α→×r→−ω2r→.\vec a_P=\vec a_C+\vec\alpha\times\vec r-\omega^2\vec r. At r→=.25x̂\vec r=.25\hat x m, a→r=3x̂−3ŷ−16x̂=(−13x̂−3ŷ)m/s2.\vec a_r=3\hat x-3\hat y-16\hat x =\boxed{(-13\hat x-3\hat y)\ \mathrm{m/s^2}}.

(b) Result and interpretation

At the bottom, r→=−.25ŷ\vec r=-.25\hat y m: α→×r→=−3x̂,−ω2r→=16ŷ.\vec\alpha\times\vec r=-3\hat x,\qquad-\omega^2\vec r=16\hat y. Hence a→b=3x̂−3x̂+16ŷ=16ŷm/s2.\boxed{\vec a_b=3\hat x-3\hat x+16\hat y=16\hat y\ \mathrm{m/s^2}}. The tangential constraint cancels horizontal acceleration; the upward normal acceleration remains.

Physical check

The bottom material point has zero instantaneous velocity while accelerating away from the surface.

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