Classical Mechanics Practice

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1.1Model a Body as a Particle

Exam I | Problem 1.1 |

A large object is represented by a single position vector in a mechanics model. What simplifying assumption is being used?

Solution

The object is being treated as a particle.

That means we ignore its size and rotation and focus only on its translational motion.

1.2Differentiate a Position Function

Exam I | Problem 1.2 |

$$ x(t) = 2t^2 - 3t + 1, $$

what are $v(2)$ and $a(2)$?

Solution

Differentiate once to get velocity:

v(t) = \frac{dx}{dt} = 4t - 3

$$ v(2) = 4(2) - 3 = 5 $$

Differentiate again to get acceleration:

a(t) = \frac{dv}{dt} = 4

$$ a(2) = 4 $$

1.3Find the Net Force

Exam I | Problem 1.3 |

Newton's second law

A $4$ kg cart accelerates at $3\ \text{m/s}^2$ to the east. What is the net force on the cart?

Solution

Use Newton's second law:

\sum \mathbf{F} = m\mathbf{a}

So the net force is

F = ma = 4(3) = 12\ \text{N}

to the east.

1.4Compute Work from a Constant Force

Exam I | Problem 1.4 |

Work

A constant force of $15\ \text{N}$ acts at an angle of $60^\circ$ to a displacement of $4\ \text{m}$. How much work is done?

Solution

For a constant force,

W = Fd\cos\theta

Substitute the values:

W = 15(4)\cos 60^\circ

Since $\cos 60^\circ = \tfrac{1}{2}$,

W = 60 \cdot \frac{1}{2} = 30\ \text{J}

1.5Compute Linear Momentum

Exam I | Problem 1.5 |

Linear momentum

A $0.5$ kg puck moves at $8\ \text{m/s}$. What is its momentum?

Solution

Momentum is

\mathbf{p} = m\mathbf{v}

So the magnitude is

p = 0.5(8) = 4\ \text{kg}\cdot\text{m/s}

The momentum points in the same direction as the velocity.

1.6Compute Average Power

Exam I | Problem 1.6 |

Power

A machine does $180\ \text{J}$ of work in $3\ \text{s}$. What is its average power?

Solution

Power is the rate of doing work:

P = \frac{W}{t}

P = \frac{180}{3} = 60\ \text{W}

1.7Use Rotational Kinematics

Exam I | Problem 1.7 |

Rotational kinematics

A wheel starts with angular velocity $\omega_0 = 2\ \text{rad/s}$ and angular acceleration $\alpha = 4\ \text{rad/s}^2$. What is its angular velocity after $3\ \text{s}$?

Solution

Use

\omega = \omega_0 + \alpha t

Substitute:

\omega = 2 + 4(3) = 14\ \text{rad/s}

1.8Find a Torque

Exam I | Problem 1.8 |

Torque

A $12\ \text{N}$ force is applied perpendicular to a wrench $0.25\ \text{m}$ from the pivot. What is the torque?

Solution

Use

\tau = rF\sin\phi

Because the force is perpendicular, $\sin\phi = 1$. So

\tau = 0.25(12) = 3\ \text{N}\cdot\text{m}

1.9Find the Period of SHM

Exam I | Problem 1.9 |

A mass-spring system has $m = 2\ \text{kg}$ and $k = 8\ \text{N/m}$. What is its period?

Solution

For simple harmonic motion,

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{8}{2}} = 2

Then

T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi\ \text{s}

1.10Write a Lagrangian

Exam I | Problem 1.10 |

Lagrangian

For a mass-spring system with kinetic energy

T = \frac{1}{2}m\dot{x}^2

and potential energy

V = \frac{1}{2}kx^2,

what is the Lagrangian $\mathcal{L}$?

Solution

The Lagrangian is

\mathcal{L} = T - V

\mathcal{L} = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2

Difficulty

Integrated

2.1Projectile Motion to the Peak

Exam II | Problem 2.1 |

A ball is launched from level ground at $20\ \text{m/s}$ at an angle of $60^\circ$ above the horizontal. Take $g = 10\ \text{m/s}^2$. How far horizontally has it traveled when it reaches its highest point?

Solution

At the top of the flight, the vertical velocity is zero.

The initial vertical component is

v_{0y} = 20\sin 60^\circ = 10\sqrt{3}

That gives the time to the top:

t = \frac{v_{0y}}{g} = \frac{10\sqrt{3}}{10} = \sqrt{3}\ \text{s}

The horizontal speed is constant:

v_{0x} = 20\cos 60^\circ = 10

So the horizontal distance is

x = v_{0x}t = 10\sqrt{3}\ \text{m}

2.2Relative Motion on a River

Exam II | Problem 2.2 |

A swimmer moves at $3\ \text{m/s}$ due north relative to the water, while the river flows at $4\ \text{m/s}$ due east. What is the swimmer's velocity relative to the bank?

Solution

Add the velocity vectors:

\mathbf{v}_{\text{bank}} = \mathbf{v}_{\text{swimmer/water}} + \mathbf{v}_{\text{water/bank}}

So the components are $4\ \text{m/s}$ east and $3\ \text{m/s}$ north.

The speed is

\sqrt{4^2 + 3^2} = 5\ \text{m/s}

The direction is

\tan^{-1}\left(\frac{3}{4}\right) \approx 36.9^\circ

north of east.

2.3Work-Energy with Friction

Exam II | Problem 2.3 |

A $2$ kg block starts from rest. A horizontal force of $6\ \text{N}$ pulls it $5\ \text{m}$, while kinetic friction exerts a $2\ \text{N}$ force opposite the motion. What is the block's speed after moving $5\ \text{m}$?

Solution

The net work is

W_{\text{net}} = (6 - 2)(5) = 20\ \text{J}

By the work-energy theorem,

W_{\text{net}} = \Delta T = \frac{1}{2}mv^2 - 0

20 = \frac{1}{2}(2)v^2

$$ v^2 = 20 $$

v = 2\sqrt{5}\ \text{m/s}

2.4Impulse Changes Momentum

Exam II | Problem 2.4 |

A $0.5$ kg puck moves at $6\ \text{m/s}$ to the right. A force of $10\ \text{N}$ acts to the left for $0.2\ \text{s}$. What is the puck's final speed?

Solution

The impulse is

J = F\Delta t = 10(0.2) = 2\ \text{N}\cdot\text{s}

Because the force is to the left, the impulse is negative relative to the original motion.

Initial momentum:

$$ p_i = mv = 0.5(6) = 3 $$

Final momentum:

p_f = p_i - 2 = 1\ \text{kg}\cdot\text{m/s}

So the final speed is

v_f = \frac{p_f}{m} = \frac{1}{0.5} = 2\ \text{m/s}

to the right.

2.5Center of Mass of Two Masses

Exam II | Problem 2.5 |

Center of mass

Two masses lie on the $x$-axis: $2$ kg at $x = 0$ and $6$ kg at $x = 4\ \text{m}$. What is the center of mass?

Solution

Use the discrete center-of-mass formula:

x_{cm} = \frac{\sum m_i x_i}{\sum m_i}

x_{cm} = \frac{2(0) + 6(4)}{2 + 6} = \frac{24}{8} = 3\ \text{m}

2.6Rolling Without Slipping

Exam II | Problem 2.6 |

A wheel of radius $0.3\ \text{m}$ rolls without slipping at $5\ \text{m/s}$. The wheel has mass $4\ \text{kg}$ and moment of inertia $I = \frac{1}{2}mR^2$. Find its angular speed and its total kinetic energy.

Solution

The no-slip condition gives

v_{cm} = R\omega

\omega = \frac{5}{0.3} = \frac{50}{3}\ \text{rad/s}

The translational kinetic energy is

T_{trans} = \frac{1}{2}mv^2 = \frac{1}{2}(4)(25) = 50\ \text{J}

The moment of inertia is

I = \frac{1}{2}(4)(0.3^2) = 0.18\ \text{kg}\cdot\text{m}^2

The rotational kinetic energy is

T_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.18)\left(\frac{50}{3}\right)^2 = 25\ \text{J}

So the total kinetic energy is

T = 50 + 25 = 75\ \text{J}

2.7Static Support Forces

Exam II | Problem 2.7 |

A uniform $2$ m beam weighs $40\ \text{N}$ and is supported at both ends. A $60\ \text{N}$ load hangs $1.5$ m from the left end. What are the upward support forces at the left and right ends?

Solution

Let the left reaction be $R_L$ and the right reaction be $R_R$.

Force balance gives

$$ R_L + R_R = 40 + 60 = 100 $$

Take torques about the left end:

$$ 2R_R = 40(1) + 60(1.5) $$

$$ 2R_R = 40 + 90 = 130 $$

R_R = 65\ \text{N}

Then

R_L = 100 - 65 = 35\ \text{N}

2.8Derive a Simple Equation of Motion

Exam II | Problem 2.8 |

For a mass-spring oscillator with

\mathcal{L} = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2,

derive the equation of motion.

Solution

Apply the Euler-Lagrange equation:

\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right) - \frac{\partial \mathcal{L}}{\partial x} = 0

Compute the derivatives:

\frac{\partial \mathcal{L}}{\partial \dot{x}} = m\dot{x}

\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right) = m\ddot{x}

Also,

\frac{\partial \mathcal{L}}{\partial x} = -kx

Substitute:

m\ddot{x} - (-kx) = 0

so the equation of motion is

m\ddot{x} + kx = 0

Difficulty

Applied

3.1Two-Body Pulley System

Final | Problem 3.1 |

Newton's second law, Constraints

A $2$ kg block on a frictionless table is connected by a light string over a frictionless pulley to a hanging $1$ kg mass. Find the acceleration of the system and the string tension. Take $g = 10\ \text{m/s}^2$.

Solution

Let the acceleration magnitude be $a$.

For the $2$ kg block on the table:

$$ T = 2a $$

For the $1$ kg hanging mass:

$$ 10 - T = a $$

Substitute $T = 2a$:

$$ 10 - 2a = a $$

$$ 10 = 3a $$

a = \frac{10}{3}\ \text{m/s}^2

Then

T = 2a = \frac{20}{3}\ \text{N}

3.2Perfectly Inelastic Collision

Final | Problem 3.2 |

A $2$ kg cart moving at $3\ \text{m/s}$ collides with a stationary $4$ kg cart. The carts stick together. What is their final speed, and how much kinetic energy is lost?

Solution

Use conservation of momentum:

$$ 2(3) + 4(0) = (2 + 4)v_f $$

$$ 6 = 6v_f $$

v_f = 1\ \text{m/s}

Initial kinetic energy:

T_i = \frac{1}{2}(2)(3^2) = 9\ \text{J}

Final kinetic energy:

T_f = \frac{1}{2}(6)(1^2) = 3\ \text{J}

So the kinetic energy lost is

9 - 3 = 6\ \text{J}

3.3Torque, Angular Acceleration, and Rotation

Final | Problem 3.3 |

A uniform disk has mass $2$ kg and radius $0.5$ m. A tangential force of $4\ \text{N}$ is applied at the rim for $2\ \text{s}$ starting from rest. Find the angular acceleration, the angular speed after $2\ \text{s}$, and the angle turned through.

Solution

First find the moment of inertia of the disk:

I = \frac{1}{2}mR^2 = \frac{1}{2}(2)(0.5^2) = 0.25\ \text{kg}\cdot\text{m}^2

The torque is

\tau = rF = 0.5(4) = 2\ \text{N}\cdot\text{m}

So the angular acceleration is

\alpha = \frac{\tau}{I} = \frac{2}{0.25} = 8\ \text{rad/s}^2

Starting from rest,

\omega = \alpha t = 8(2) = 16\ \text{rad/s}

The angular displacement is

\theta = \frac{1}{2}\alpha t^2 = \frac{1}{2}(8)(2^2) = 16\ \text{rad}

3.4Compare Two Circular Orbits

Final | Problem 3.4 |

A satellite moves in a circular orbit of radius $r_1$ around a planet. Another satellite orbits at radius $r_2 = 4r_1$. What are the ratios $v_2/v_1$ and $T_2/T_1$?

Solution

For circular orbits,

v = \sqrt{\frac{GM}{r}}

so speed scales like $r^{-1/2}$:

\frac{v_2}{v_1} = \sqrt{\frac{r_1}{r_2}} = \sqrt{\frac{1}{4}} = \frac{1}{2}

The orbital period scales like $r^{3/2}$:

\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2} = 4^{3/2} = 8

3.5Choose the Best Mechanics Tool

Final | Problem 3.5 |

Problem-solving workflow

For each situation, name the main tool from the workflow that is most useful.

(a) A collision lasts a very short time and external impulse is negligible. (b) A rigid sign hangs motionless from two cables. (c) An object moves under only conservative forces, but the path is complicated. (d) A spinning object has no external torque acting on it.

Solution

Use the tool that matches the situation:

(a) Momentum conservation (b) Equilibrium conditions / statics (c) Energy methods (d) Angular momentum conservation

Difficulty

Challenge

4.1Rolling Disk Dropping Through a Height

Final | Problem 4.1 |

A solid disk starts from rest and rolls without slipping down a track, dropping through a vertical height $h$. Derive its speed at the bottom in terms of $g$ and $h$.

Solution

Use conservation of mechanical energy:

mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

For a solid disk,

I = \frac{1}{2}mR^2

and rolling without slipping gives

\omega = \frac{v}{R}

Substitute:

mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mR^2\right)\left(\frac{v^2}{R^2}\right)

mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2

v = \sqrt{\frac{4gh}{3}}

4.2Spin-Up from Conservation of Angular Momentum

Final | Problem 4.2 |

A skater's moment of inertia decreases from $6\ \text{kg}\cdot\text{m}^2$ to $2\ \text{kg}\cdot\text{m}^2$ while angular momentum is conserved. If the initial angular speed is $3\ \text{rad/s}$, find the final angular speed and the change in rotational kinetic energy.

Solution

Conservation of angular momentum gives

I_i\omega_i = I_f\omega_f

6(3) = 2\omega_f

\omega_f = 9\ \text{rad/s}

Initial rotational kinetic energy:

T_i = \frac{1}{2}I_i\omega_i^2 = \frac{1}{2}(6)(3^2) = 27\ \text{J}

Final rotational kinetic energy:

T_f = \frac{1}{2}(2)(9^2) = 81\ \text{J}

So the kinetic energy increases by

81 - 27 = 54\ \text{J}

4.3Beam Balance with Multiple Loads

Final | Problem 4.3 |

A $3$ m uniform beam weighs $30\ \text{N}$ and is supported at both ends. Additional loads of $20\ \text{N}$ and $50\ \text{N}$ hang at $0.5$ m and $2.5$ m from the left end, respectively. What are the support forces?

Solution

Let the left and right reactions be $R_L$ and $R_R$.

Force balance:

$$ R_L + R_R = 30 + 20 + 50 = 100 $$

Take torques about the left end:

$$ 3R_R = 30(1.5) + 20(0.5) + 50(2.5) $$

$$ 3R_R = 45 + 10 + 125 = 180 $$

R_R = 60\ \text{N}

Then

R_L = 100 - 60 = 40\ \text{N}

4.4Lagrangian with Two Springs

Final | Problem 4.4 |

A mass moves on a frictionless line between two identical springs, each with spring constant $k$. If the mass is displaced by $x$ from equilibrium, derive the equation of motion using the Lagrangian method.

Solution

Each spring contributes a potential energy of $\frac{1}{2}kx^2$, so the total potential energy is

V = \frac{1}{2}kx^2 + \frac{1}{2}kx^2 = kx^2

The kinetic energy is

T = \frac{1}{2}m\dot{x}^2

So the Lagrangian is

\mathcal{L} = \frac{1}{2}m\dot{x}^2 - kx^2

Apply the Euler-Lagrange equation:

\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right) - \frac{\partial \mathcal{L}}{\partial x} = 0

Compute the derivatives:

\frac{\partial \mathcal{L}}{\partial \dot{x}} = m\dot{x}

\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right) = m\ddot{x}

and

\frac{\partial \mathcal{L}}{\partial x} = -2kx

Substitute:

m\ddot{x} - (-2kx) = 0

Thus the equation of motion is

m\ddot{x} + 2kx = 0