Solution to Question 11
1. Kinematics of the Bead and Cord
The bead slides down the cord with a constant velocity $v$ relative to the ceiling. The cord itself stretches as the bead moves down. We need to distinguish between the position of the bead ($z$) and the amount of cord material passed ($y$).
Let $y$ be the unstretched length of the cord above the bead. The stiffness of this segment of length $y$ is given by:
$$k_{seg} = \frac{k l}{y}$$
The tension in the cord above the bead supports the weight of the bead (since it moves at constant speed, $f = mg$). Thus, the extension $\delta$ of the segment $y$ is:
$$\delta = \frac{\text{Tension}}{k_{seg}} = \frac{mg}{kl/y} = \frac{mg y}{kl}$$
The actual position of the bead $z$ (distance from ceiling) is the sum of the unstretched length and the extension:
$$z = y + \delta = y + \frac{mg}{kl} y = y \left( 1 + \frac{mg}{kl} \right)$$
2. Velocity Relations
Differentiating the position $z$ with respect to time gives the velocity $v$:
$$v = \frac{dz}{dt} = \frac{dy}{dt} \left( 1 + \frac{mg}{kl} \right)$$
Here, $\dot{y}$ represents the rate at which the bead moves along the unstretched material of the cord. We can solve for $\dot{y}$:
$$\dot{y} = v \left( \frac{kl}{kl + mg} \right)$$
3. Energy Balance and Power Dissipation
We can calculate the thermal power dissipated using the energy conservation principle. The rate of work done by gravity is balanced by the rate of increase in elastic potential energy and the thermal power dissipated.
$$P_{gravity} = P_{elastic} + P_{thermal}$$
Power from Gravity: $P_{gravity} = F_g \cdot v = mgv$.
Rate of Elastic Energy Storage: The elastic energy stored in the segment $y$ is:
$$U = \frac{1}{2} k_{seg} \delta^2 = \frac{1}{2} \left( \frac{kl}{y} \right) \left( \frac{mgy}{kl} \right)^2 = \frac{1}{2} \frac{(mg)^2}{kl} y$$
The rate of change of elastic energy is:
$$P_{elastic} = \frac{dU}{dt} = \frac{(mg)^2}{2kl} \frac{dy}{dt}$$
Substituting $\dot{y}$:
$$P_{elastic} = \frac{(mg)^2}{2kl} \cdot v \left( \frac{kl}{kl + mg} \right) = \frac{1}{2} \frac{(mg)^2 v}{kl + mg}$$
4. Calculating Thermal Power
Now, solve for $P_{thermal}$:
$$P_{thermal} = mgv – \frac{1}{2} \frac{(mg)^2 v}{kl + mg}$$
Factor out $mgv$:
$$P_{thermal} = mgv \left[ 1 – \frac{mg}{2(kl + mg)} \right]$$
Find a common denominator:
$$P_{thermal} = mgv \left[ \frac{2(kl + mg) – mg}{2(kl + mg)} \right]$$
$$P_{thermal} = mgv \left[ \frac{mg + 2kl}{2(mg + kl)} \right]$$