Chapter 21, Problem 79GP

To determine

To find the terminal velocity of the bar as its slides frictionless down the rails.

Answer to Problem 79GP

Solution:

The bar slides with a velocity of $\text{1}\text{.2 m/s}$.

Explanation of Solution

Given info:

$\text{Ω}$ is constant, R is resistance, $\text{v}$ is the velocity of component, $\text{I}$ is amount of current and $\text{ε}$ is the value of voltage.

Formula used:

$\text{I=}\frac{\text{ε}}{\text{R}}$

Calculation:

${\text{F}}_{\text{net}}{\text{=F}}_{\text{B}}\text{cos}\text{θ-mgsinθ}\mathrm{.............................}\left(\text{1}\right)$

The net power following up on the bar.

${\text{F}}_{\text{B}}\text{=BIl}\mathrm{.....................................}\left(\text{2}\right)$

The expression for the induced emf is,

$\text{ε=Blv}\text{cos}\text{θ}$

$\text{v}\text{cos}\text{θ}$ is the component of velocity of the bar which can be determine from ohms law.

$\begin{array}{l}\text{I=}\frac{\text{ε}}{\text{R}}\\ \text{=}\frac{\text{Blv}\text{cos}\text{θ}}{\text{R}}\\ {\text{F}}_{\text{B}}\text{=B}\left(\frac{\text{Blv}\text{cos}\text{θ}}{\text{R}}\right)\\ \text{=}\frac{{\text{B}}^{\text{2}}{\text{l}}^{\text{2}}\text{vcos}\text{θ}}{\text{R}}\\ \text{0=}\left(\frac{{\text{B}}^{\text{2}}{\text{l}}^{\text{2}}\text{v}\text{cos}\text{θ}}{\text{R}}\right)\text{cos}\text{θ-mg}\text{sin}\text{θ}\\ \frac{{\text{B}}^{\text{2}}{\text{l}}^{\text{2}}\text{v}\text{cos}\text{θ}}{\text{R}}\text{=mg}\text{sin}\text{θ}\\ \text{v=}\frac{\text{Rmg}\text{sin}\text{θ}}{{\text{B}}^{\text{2}}{\text{l}}^{\text{2}}\text{cos}\text{θ}}\mathrm{................................}\left(\text{4}\right)\end{array}$

$\begin{array}{l}\\ \text{l=32}\text{.0 cm}\left(\frac{\text{1}\text{.00m}}{\text{100cm}}\right)\\ \text{=32}{\text{.0×10}}^{\text{-2}}\text{ m}\\ \\ \text{v=}\frac{\text{Rmg}\text{sin}\text{θ}}{{\text{B}}^{\text{2}}{\text{l}}^{\text{2}}\text{cos θ}}\\ \\ \text{=}\frac{\left(\text{0}\text{.60}\text{Ω}\right)\left(\text{0}\text{.40 kg}\right)\left(\text{9}{\text{.80 m/s}}^{\text{2}}\right)\text{sin 6}\text{.0°}}{{\left(\text{0}\text{.45}\text{T}\right)}^{\text{2}}{\left(\text{32}{\text{.0×10}}^{\text{-2}}\text{m}\right)}^{\text{2}}{\text{cos}}^{\text{2}}\text{6}\text{.0°}}\\ \text{=1}\text{.198}\text{m/s}\\ \text{=1}\text{.2}\text{m/s}\end{array}$