Solution to Problem 4
Problem Statement: Tape of length $L=70$ m is wound from $r_0=10$ mm to $R=25$ mm in $T=165$ s with constant angular velocity. Find the length wound in $t=110$ s.
Step 1: Relation for Radius Increase
Since the angular velocity $\omega$ is constant and the tape thickness is uniform, the rate at which the radius increases is constant.
The number of layers added per second is proportional to $\omega$.
$$ \frac{dr}{dt} = k \quad (\text{constant}) $$
Therefore, the radius $r(t)$ grows linearly with time:
$$ r(t) = r_0 + \frac{R – r_0}{T} t $$
Step 2: Relation for Length Winding
The rate at which length is added is the tangential velocity:
$$ \frac{dl}{dt} = v = \omega r(t) $$
The total length $l(t)$ wound at time $t$ is the integral of speed:
$$ l(t) = \int_0^t \omega r(\tau) d\tau = \omega \int_0^t \left( r_0 + \frac{R-r_0}{T}\tau \right) d\tau $$
$$ l(t) = \omega \left[ r_0 t + \frac{R-r_0}{2T} t^2 \right] $$
Step 3: Eliminating $\omega$
We apply the condition for the total time $T$ and total length $L$:
$$ L = \omega \left[ r_0 T + \frac{R-r_0}{2T} T^2 \right] = \omega T \left( \frac{2r_0 + R – r_0}{2} \right) = \omega T \frac{R+r_0}{2} $$
From this, $\omega = \frac{2L}{T(R+r_0)}$.
Substituting $\omega$ back into the expression for $l(t)$:
$$ l(t) = \frac{2L}{T(R+r_0)} \cdot t \cdot \left( r_0 + \frac{R-r_0}{2T} t \right) $$
$$ l(t) = \frac{L t}{T (R+r_0)} \left( 2r_0 + \frac{R-r_0}{T} t \right) $$
Step 4: Calculation
Given:
$L = 70$ m
$r_0 = 10$ mm, $R = 25$ mm (Note: Since we use ratios of radii, units can stay in mm as long as they are consistent).
$T = 165$ s, $t = 110$ s.
Ratio $\frac{t}{T} = \frac{110}{165} = \frac{2}{3}$.
Substitute into the formula:
$$ l(t) = 70 \cdot \frac{2}{3} \cdot \frac{1}{25+10} \cdot \left( 2(10) + (25-10)\frac{2}{3} \right) $$
$$ l(t) = \frac{140}{3 \cdot 35} \left( 20 + 15 \cdot \frac{2}{3} \right) $$
$$ l(t) = \frac{4}{3} \left( 20 + 10 \right) = \frac{4}{3} (30) $$
$$ l(t) = 40 \text{ m} $$
Answer: 40 m