Induction and Energy Transfers — Question 6

Section 30-4 · Physics pages 4822–4823

Question 6 Increasing circuit resistance with travel changes the braking distance A coasting rod of mass m starts at x=0 with speed v0. Its circuit resistance grows as R(x) = R0 + αx, with alpha positive. Let C = B2ℓ2. Ignore inductance and friction. R(x) rod v ⊗B ℓ Schematic; labels, not drawing lengths, specify the data. (a) Find speed as a function of position. (b) Find the limiting distance and recover the constant-resistance limit as alpha tends to zero.
Show solutionHide solution
Question 6 - Solution Increasing circuit resistance with travel changes the braking distance R(x) rod v ⊗B ℓ (a) Method and calculation mv dv dx = − Cv R0 + αx ⇒mdv dx = − C R0 + αx. v(x) = v0 −C mα ln(1 + αx/R0) . (b) Interpretation and further calculation v(x∞) = 0 ⇒x∞= R0 α ( emαv0/C −1 ) . α →0 : emαv0/C −1 ≃mαv0/C ⇒x∞→mR0v0/C. Physical check The cancellation of v in the differential equation is used while v is positive; the final zero-speed position is approached asymptotically.

Question 6

Increasing circuit resistance with travel changes the braking distance

A coasting rod of mass m starts at x=0 with speed v0. Its circuit resistance grows as R(x)=R0+αxR(x)=R_0+\alpha x, with alpha positive. Let C=B2ℓ2C=B^2\ell^2. Ignore inductance and friction.

Diagram shown in the original SVG worksheet.

  1. Find speed as a function of position.

  2. Find the limiting distance and recover the constant-resistance limit as alpha tends to zero.

Question 6 - Solution

Increasing circuit resistance with travel changes the braking distance

Diagram shown in the original SVG worksheet.

(a) Method and calculation

mvdvdx=−CvR0+αx⇒mdvdx=−CR0+αx.mv\frac{dv}{dx}=-\frac{Cv}{R_0+\alpha x}\Rightarrow m\frac{dv}{dx}=-\frac C{R_0+\alpha x}.v(x)=v0−Cmαln⁡(1+αx/R0).\boxed{v(x)=v_0-\frac{C}{m\alpha}\ln(1+\alpha x/R_0)}.

(b) Interpretation and further calculation

v(x∞)=0⇒x∞=R0α(emαv0/C−1).v(x_\infty)=0\Rightarrow\boxed{x_\infty=\frac{R_0}{\alpha}\left(e^{m\alpha v_0/C}-1\right)}.α→0:emαv0/C−1≃mαv0/C⇒x∞→mR0v0/C.\alpha\to0:\quad e^{m\alpha v_0/C}-1\simeq m\alpha v_0/C\Rightarrow x_\infty\to mR_0v_0/C.

Physical check

The cancellation of v in the differential equation is used while v is positive; the final zero-speed position is approached asymptotically.