Question

A magnetic disc is moving on an air-cushioned table above a metal sheet. Due to the generated eddy currents there is a retarding force exerted on the disc, which is proportional to the speed of the disc. Moving above a sheet made of aluminum the disc stops after covering a distance of 30 cm, but when the sheet is made of copper the disc stops after covering only 20 cm. How much distance will the disc cover if first it travels above a piece of copper sheet of width 15 cm, and then continues its motion above an aluminum sheet? (The initial speeds of the disc are the same in all the three cases.)

Answer 1

22.5 cm

Answer 2

Let $m$ be the mass of the disc and $v$ be its speed. The retarding force due to eddy currents is given as $F = -kv$, where $k$ is a constant that depends on the material of the metal sheet.

The equation of motion is $m \frac{dv}{dt} = -kv$.

We can write $\frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$.

So the equation becomes $m v \frac{dv}{dx} = -kv$.

For $v \neq 0$, we can divide by $v$: $m \frac{dv}{dx} = -k$.

This implies $\frac{dv}{dx} = -\frac{k}{m}$.

Integrating with respect to $x$ from initial position $x=0$ (where speed is $v_{initial}$) to position $x$ (where speed is $v(x)$):

$\int_{v_{initial}}^{v(x)} dv = \int_{0}^{x} -\frac{k}{m} dx$

$v(x) - v_{initial} = -\frac{k}{m} x$

$v(x) = v_{initial} - \frac{k}{m} x$.

The disc stops when $v(x) = 0$. Let the total distance covered until it stops be $D$.

$0 = v_{initial} - \frac{k}{m} D$.

$D = \frac{m v_{initial}}{k}$.

Let $v_0$ be the initial speed in all cases.

For the aluminum sheet, the distance covered is $D_A = 30$ cm. Let the constant be $k_A$.

$30 = \frac{m v_0}{k_A}$. This gives $\frac{k_A}{m} = \frac{v_0}{30}$.

For the copper sheet, the distance covered is $D_C = 20$ cm. Let the constant be $k_C$.

$20 = \frac{m v_0}{k_C}$. This gives $\frac{k_C}{m} = \frac{v_0}{20}$.

Now, the disc first travels above a copper sheet of width 15 cm, and then above an aluminum sheet. The initial speed is $v_0$.

Let $v_1$ be the speed of the disc after covering 15 cm over the copper sheet.

Using the equation $v(x) = v_{initial} - \frac{k}{m} x$:

For the motion over the copper sheet, $v_{initial} = v_0$, $x = 15$ cm, and $k = k_C$.

$v_1 = v_0 - \frac{k_C}{m} (15)$.

Substitute $\frac{k_C}{m} = \frac{v_0}{20}$:

$v_1 = v_0 - \frac{v_0}{20} (15) = v_0 - \frac{15}{20} v_0 = v_0 - \frac{3}{4} v_0 = \frac{1}{4} v_0$.

Now the disc moves over the aluminum sheet, starting with speed $v_1 = \frac{1}{4} v_0$. Let $x_A$ be the distance covered on the aluminum sheet until the disc stops (speed becomes 0).

Using the equation $v(x) = v_{initial} - \frac{k}{m} x$ again:

For the motion over the aluminum sheet, $v_{initial} = v_1 = \frac{1}{4} v_0$, the final speed is 0, and $k = k_A$. Let the distance covered be $x_A$.

$0 = v_1 - \frac{k_A}{m} x_A$.

$x_A = \frac{v_1}{k_A/m}$.

Substitute $v_1 = \frac{1}{4} v_0$ and $\frac{k_A}{m} = \frac{v_0}{30}$:

$x_A = \frac{\frac{1}{4} v_0}{v_0/30} = \frac{1/4}{1/30} = \frac{30}{4} = 7.5$ cm.

The total distance covered by the disc is the sum of the distance covered on the copper sheet and the distance covered on the aluminum sheet.

Total distance = (distance on copper) + (distance on aluminum)

Total distance = 15 cm + 7.5 cm = 22.5 cm.