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Physics / Chapter 11: Rolling, Torque, and Angular Momentum / Section 11-3

The Forces of Rolling — Question 10

Section 11-3 · Physics pages 1610–1611

Question 10 Braking torque needs backward traction without contact dissipation A hoop of mass 2 kg and radius 0.30 m rolls right at 3 m/s. A constant counterclockwise brake torque 1.2 N m acts while it moves. Assume no slip and sufficient friction. C vC Side view; arrows and labels specify the directions. (a) Find acceleration, friction and stopping distance. (b) Account for the initial energy using brake work.
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Question 10 - Solution Braking torque needs backward traction without contact dissipation C vC (a) Derivation and calculation For a hoop with counterclockwise brake torque τb, f = Ma, τb + Rf = Iα = −MRa. Hence a = − τb 2MR = − 1.2 2(2)(.30) = −1.00 m/s2 , f = Ma = −2.00 N , s = V 2 0 2|a| = 4.50 m . (b) Result and interpretation The signed rotation to rest is ∆θ = −s/R = −4.5/.30 = −15.0 rad. Brake work is Wb = τb∆θ = 1.2(−15) = −18.0 J . This equals ∆K = −MV 2 0 = −2(3)2 = −18 J; stationary-ground static friction contributes no work. Physical check The braking torque and angular velocity have opposite signs, so the brake removes energy.

Question 10

Braking torque needs backward traction without contact dissipation

A hoop of mass 2kg2\,\mathrm{kg} and radius 0.30m0.30\,\mathrm m rolls right at 3m/s3\,\mathrm{m/s}. A constant counterclockwise brake torque 1.2Nm1.2\,\mathrm{N\,m} acts while it moves. Assume no slip and sufficient friction.

Diagram shown in the original SVG worksheet.

  1. Find acceleration, friction and stopping distance.

  2. Account for the initial energy using brake work.

Question 10 - Solution

Braking torque needs backward traction without contact dissipation

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

For a hoop with counterclockwise brake torque τb\tau_b, f=Ma,τb+Rf=Iα=−MRa.f=Ma,\qquad\tau_b+Rf=I\alpha=-MRa. Hence a=−τb2MR=−1.22(2)(.30)=−1.00m/s2,\boxed{a=-\frac{\tau_b}{2MR}=-\frac{1.2}{2(2)(.30)}=-1.00\ \mathrm{m/s^2}}, f=Ma=−2.00N,s=V022|a|=4.50m.\boxed{f=Ma=-2.00\ \mathrm N},\qquad \boxed{s=\frac{V_0^2}{2|a|}=4.50\ \mathrm m}.

(b) Result and interpretation

The signed rotation to rest is Δθ=−s/R=−4.5/.30=−15.0rad.\Delta\theta=-s/R=-4.5/.30=-15.0\ \mathrm{rad}. Brake work is Wb=τbΔθ=1.2(−15)=−18.0J.\boxed{W_b=\tau_b\Delta\theta=1.2(-15)=-18.0\ \mathrm J}. This equals ΔK=−MV02=−2(3)2=−18\Delta K=-MV_0^2=-2(3)^2=-18 J; stationary-ground static friction contributes no work.

Physical check

The braking torque and angular velocity have opposite signs, so the brake removes energy.

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