Physics 7A: Classical Mechanics - Problem Set Solutions and Concepts

Problem 1: Circular Motion and Energy Conservation

a) Centripetal Motion and Tension at the Top of a Trajectory

When a block and ball system moves in a circular path, the tension at the top of the trajectory must provide the necessary centripetal force. The minimum velocity condition ensures the tension is just enough to maintain circular motion.

• Centripetal Force Condition: The tension $T$ at the top is given by $T = 0$, so only gravity provides the centripetal force.

• Equation: $(m_1 + m_2) \frac{v_{\text{min}}^2}{L} = (m_1 + m_2)g \Rightarrow v_{\text{min}} = \sqrt{gL}$

• Key Concept: Minimum velocity at the top of a vertical circle is $\sqrt{gL}$.

b) Energy Conservation in Vertical Motion

Energy conservation relates kinetic and potential energy at different points in the trajectory.

• At the bottom: All energy is kinetic: $E = K = \frac{1}{2}(m_1 + m_2)v^2$

• At the top: $K = \frac{1}{2}(m_1 + m_2)v_{\text{min}}^2$, $U = (m_1 + m_2)g(2L)$

• Energy Conservation Equation: $\frac{1}{2}(m_1 + m_2)v^2 = \frac{1}{2}(m_1 + m_2)gL + (m_1 + m_2)g(2L)$

• Result: $v = \sqrt{5gL}$

c) Collision and Drag Work

After a collision, the block and ball move together, and drag force does negative work.

• Post-collision velocity: $K = \frac{1}{2}(m_1 + m_2)v_f^2$

• Work by Drag: $W = -d(m_1 + m_2)gL$

• Energy Conservation: $\frac{1}{2}(m_1 + m_2)v^2 = \frac{1}{2}(m_1 + m_2)v_f^2 + (m_1 + m_2)g(2L) + W$

• Result: $v_f = \sqrt{3gL}$

Problem 2: Escape Velocity and Satellite Orbits

a) Escape Velocity from Earth

Escape velocity is the minimum speed needed to break free from a planet's gravitational field.

• Formula: $v_{\text{esc}} = \sqrt{\frac{2GM}{d}}$

• Application: For $d = 11R/10$, $v = \sqrt{\frac{20GM}{11R}}$

b) Energy After Launch

Calculating the energy right after launch involves both kinetic and gravitational potential energy.

• Equation: $\frac{1}{2}mv^2 - \frac{GMm}{R} = \frac{5GMm}{22R} - \frac{GMm}{R} = \frac{17GMm}{22R}$

c) Energy Conservation

By energy conservation, the change in energy is equated to the work done.

• Equation: $\frac{17GMm}{22R} = \frac{1}{2}mv^2 - \frac{10GMm}{11R}$

• Result: $v = \sqrt{\frac{3GM}{11R}}$

d) Work-Energy Theorem for Circular Orbits

The work required to move a satellite into a stable orbit is the change in kinetic energy.

• Gravitational Force: $\frac{GMm}{R^2}$

• Kinetic Energy Change: $\Delta K = \frac{1}{2}m\left(\frac{10GM}{11R}\right) - \frac{17GMm}{22R} = W$

e) Mechanical Energy Conservation

Mechanical energy is conserved before the satellite is positioned; after, work is needed to achieve a stable orbit.

• Key Point: Energy is traded between kinetic and potential forms; additional work is required for orbit insertion.

Problem 3: Rain Falling on a Moving Boat

a) Mass Accumulation Due to Rain

Rain falls at a constant rate $\beta$, increasing the mass of the boat over time.

• Equation: $m(t) = m_0 + \beta t$

b) Buoyant Force and Drag

The buoyant force counteracts the weight of the boat, and drag force slows it down.

• Free Body Diagram: Shows forces acting on the boat: drag ($F_D$) and buoyancy.

c) Drag Force Equation

The drag force is the only external horizontal force, proportional to velocity.

• Equation: $F_D = m(t)\frac{dv}{dt} = -bv^2$

• Rewritten: $\frac{dv}{v^2} = -\frac{b}{m(t)}dt$

d) Terminal Speed

Terminal speed is reached when the drag force balances the driving force; here, it goes to zero as $t \to \infty$.

• Key Point: The boat eventually stops due to drag.

e) Integrating the Drag Equation

Solving the differential equation for velocity as a function of time.

• Integrated Form: $\frac{1}{v_0} - \frac{1}{v(t)} = \frac{b}{\beta} \log\left(\frac{m_0 + \beta t}{m_0}\right)$

• Rewriting: $v(t) = \left(\frac{1}{v_0} + \frac{b}{\beta} \log\left(\frac{m_0 + \beta t}{m_0}\right)\right)^{-1}$

Problem 4: Block and Pulley System - Kinematics and Rotational Dynamics

a) Kinematics of the Falling Block

The block falls under gravity, with tension and gravity as the only forces.

• Equation: $ma = mg - T_1$

• Acceleration: $a = \frac{2h}{t_1^2}$

• Tension: $T_1 = mg - \frac{2mh}{t_1^2}$

c) Rotational Acceleration of the Pulley

The pulley rotates due to the tension in the rope, with angular acceleration $\alpha$.

• Rotational Acceleration: $\alpha = \frac{a}{R}$

• Moment of Inertia: $I_p = \frac{1}{2}MR^2$

• Torque Equation: $I_{\text{tot}} \alpha = T_2 R - T_1 R$

d) Torque and Moment of Inertia

The tension $T_2$ induces a torque, causing the entire system to rotate.

• Angular Acceleration: $T_2 R = I_{\text{tot}} \alpha$

• Moment of Inertia: $I_{\text{tot}} = I + I_0$

• Equations for $I$:

  • $I = I_{ ext{tot}} - I_0$

  • $I = r_0^2 m_0$

  • $I = r_0^2 m_0 + \left(m \frac{r^2}{2h}\right)$

Additional info: Some equations and steps have been expanded for clarity and completeness, and notation has been standardized for readability.