Dynamics Practice

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Direct

1.1Classify the Study of Motion

Exam I | Problem 1.1 |

Which branch of dynamics describes motion without considering the forces that cause it?

Solution

That is kinematics.

Kinematics describes motion itself, while kinetics connects motion to the forces and moments that produce it.

1.2Differentiate a Position Function

Exam I | Problem 1.2 |

Given

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

find $v(2)$ and $a(2)$.

Solution

Differentiate position to get velocity:

v(t) = \frac{dx}{dt} = 6t - 2

Differentiate again to get acceleration:

a(t) = \frac{dv}{dt} = 6

Evaluate at $t=2$:

$$ v(2) = 6(2) - 2 = 10 $$

$$ a(2) = 6 $$

1.3Use Constant Acceleration to Find Speed

Exam I | Problem 1.3 |

A cart starts from rest and accelerates at $4 \, \text{m/s}^2$ for $3$ s. Find its speed.

Solution

Use the constant-acceleration relation:

$$ v = v_0 + at $$

Since the cart starts from rest, $v_0 = 0$:

v = 0 + 4(3) = 12 \, \text{m/s}

1.4Find Displacement Under Constant Acceleration

Exam I | Problem 1.4 |

A particle starts at $x_0 = 5$ m with $v_0 = 2 \, \text{m/s}$ and constant acceleration $a = 1 \, \text{m/s}^2$. Find $x$ after $4$ s.

Solution

Use

x = x_0 + v_0 t + \frac{1}{2}at^2

Substitute the values:

x = 5 + 2(4) + \frac{1}{2}(1)(4^2)

x = 5 + 8 + 8 = 21 \, \text{m}

1.5Read Velocity and Acceleration from Cartesian Components

Exam I | Problem 1.5 |

For the position vector

\mathbf{r}(t) = t\,\mathbf{i} + t^2\,\mathbf{j},

find $\mathbf{v}$ and $\mathbf{a}$ at $t=3$.

Solution

Differentiate the position vector:

\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \mathbf{i} + 2t\,\mathbf{j}

Differentiate again:

\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = 2\,\mathbf{j}

At $t=3$:

\mathbf{v}(3) = \mathbf{i} + 6\,\mathbf{j}

\mathbf{a}(3) = 2\,\mathbf{j}

1.6Compute Normal Acceleration

Exam I | Problem 1.6 |

A particle moves along a curved path at a speed of $15 \, \text{m/s}$ on a path with radius of curvature $25$ m. Find its normal acceleration.

Solution

Use

a_n = \frac{v^2}{\rho}

a_n = \frac{15^2}{25} = \frac{225}{25} = 9 \, \text{m/s}^2

1.7Add Velocities in Relative Motion

Exam I | Problem 1.7 |

Relative Motion

A walkway moves east at $1.5 \, \text{m/s}$. A person walks east relative to the walkway at $0.8 \, \text{m/s}$. What is the person's speed relative to the ground?

Solution

For motion in the same direction, the velocities add:

v = 1.5 + 0.8 = 2.3 \, \text{m/s}

So the person's ground speed is $2.3 \, \text{m/s}$ east.

1.8Apply Newton's Second Law

Exam I | Problem 1.8 |

A $5$ kg particle has a net force of $20$ N to the right. Find its acceleration.

Solution

Use Newton's second law:

\sum F = ma

a = \frac{20}{5} = 4 \, \text{m/s}^2

The acceleration is $4 \, \text{m/s}^2$ to the right.

1.9Use Friction at a Contact Surface

Exam I | Problem 1.9 |

A surface has normal force $N = 40$ N and kinetic friction coefficient $\mu_k = 0.25$. Find the kinetic friction force.

Solution

Use

f_k = \mu_k N

f_k = 0.25(40) = 10 \, \text{N}

1.10Find Tangential Speed in Pure Rotation

Exam I | Problem 1.10 |

A point lies $0.4$ m from a fixed axis on a rigid body rotating at $\omega = 6 \, \text{rad/s}$. Find the point's speed.

Solution

For pure rotation,

v = \omega r

v = 6(0.4) = 2.4 \, \text{m/s}

Difficulty

Integrated

2.1Use Work-Energy with a Constant Force

Exam II | Problem 2.1 |

A $2$ kg cart moves at $3 \, \text{m/s}$. A constant force of $10$ N acts in the direction of motion over $4$ m on a level track. Find the final speed.

Solution

Initial kinetic energy:

T_1 = \frac{1}{2}mv_1^2 = \frac{1}{2}(2)(3^2) = 9 \, \text{J}

Work done by the force:

W = Fs = 10(4) = 40 \, \text{J}

T_2 = T_1 + W = 9 + 40 = 49 \, \text{J}

Then

\frac{1}{2}(2)v_2^2 = 49

$$ v_2^2 = 49 $$

v_2 = 7 \, \text{m/s}

2.2Track Energy with a Spring

Exam II | Problem 2.2 |

A $1$ kg block is released from rest by a spring with $k = 100 \, \text{N/m}$ compressed $0.20$ m on a frictionless track. Find the speed when the spring returns to its natural length.

Solution

The spring potential energy is

V_s = \frac{1}{2}kx^2 = \frac{1}{2}(100)(0.20^2) = 2 \, \text{J}

At the natural length, all of that energy becomes kinetic energy:

\frac{1}{2}mv^2 = 2

With $m=1$ kg:

$$ v^2 = 4 $$

v = 2 \, \text{m/s}

2.3Use Impulse to Find Final Velocity

Exam II | Problem 2.3 |

A $3$ kg particle starts from rest and is acted on by a $30$ N force for $0.2$ s. Find its final speed.

Solution

The impulse is

J = Ft = 30(0.2) = 6 \, \text{N}\cdot\text{s}

Impulse-momentum gives

$$ J = m(v_2 - v_1) $$

Since $v_1 = 0$,

$$ 6 = 3v_2 $$

v_2 = 2 \, \text{m/s}

2.4Solve a Friction Problem with Force Balance

Exam II | Problem 2.4 |

A $10$ kg block slides on a horizontal surface. A $50$ N horizontal force pulls it to the right, and $\mu_k = 0.2$. Take $g = 10 \, \text{m/s}^2$. Find the acceleration.

Solution

The normal force is

N = mg = 10(10) = 100 \, \text{N}

So the kinetic friction force is

f_k = \mu_k N = 0.2(100) = 20 \, \text{N}

The net horizontal force is

50 - 20 = 30 \, \text{N}

Therefore,

a = \frac{30}{10} = 3 \, \text{m/s}^2

2.5Combine Translation and Rotation

Exam II | Problem 2.5 |

Point $A$ on a rigid bar moves to the right at $1.5 \, \text{m/s}$. The bar rotates counterclockwise at $3 \, \text{rad/s}$, and point $B$ is $0.6$ m above $A$. Find $\mathbf{v}_B$.

Solution

Use

\mathbf{v}_B = \mathbf{v}_A + \boldsymbol{\omega} \times \mathbf{r}_{B/A}

Take $\mathbf{v}_A = 1.5\,\mathbf{i}$, $\boldsymbol{\omega} = 3\,\mathbf{k}$, and $\mathbf{r}_{B/A} = 0.6\,\mathbf{j}$.

Then

\boldsymbol{\omega} \times \mathbf{r}_{B/A} = 3\mathbf{k} \times 0.6\mathbf{j} = -1.8\,\mathbf{i}

\mathbf{v}_B = 1.5\,\mathbf{i} - 1.8\,\mathbf{i} = -0.3\,\mathbf{i}

Point $B$ moves left at $0.3 \, \text{m/s}$.

2.6Find Angular Acceleration from a Moment

Exam II | Problem 2.6 |

A thin rod has mass $4$ kg and length $1.5$ m. It rotates about one end. If the net moment about the end is $6$ N$\cdot$m, find its angular acceleration.

Solution

First find the mass moment of inertia about the end using the parallel-axis theorem:

I_G = \frac{1}{12}mL^2 = \frac{1}{12}(4)(1.5^2) = 0.75 \, \text{kg}\cdot\text{m}^2

The center of mass is $d = 0.75$ m from the end, so

I_O = I_G + md^2 = 0.75 + 4(0.75^2) = 3.0 \, \text{kg}\cdot\text{m}^2

Now use

\sum M_O = I_O \alpha

\alpha = \frac{6}{3.0} = 2 \, \text{rad/s}^2

2.7Choose the Best Method

Exam II | Problem 2.7 |

Choosing the Right Method

A problem asks for the speed of a block after it slides a known distance on a rough surface. Which method is usually best?

Solution

The best choice is usually work-energy.

The distance is known, so it is often faster to write the work done by gravity, friction, or other forces and relate it to the change in kinetic energy instead of solving for time first.

2.8Spot a Common Pitfall

Exam II | Problem 2.8 |

A particle has acceleration $a(t) = 3t$. Can you use $v = v_0 + at$ with $a$ taken at the final time? Why or why not?

Solution

No. The formula

$$ v = v_0 + at $$

is valid only when acceleration is constant.

Here $a(t)$ changes with time, so you must integrate

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

to find velocity.

Difficulty

Applied

3.1Recognize Variable Acceleration by Position

Final | Problem 3.1 |

A particle moves in a straight line with acceleration

a(x) = 4x \, \text{m/s}^2.

If $v = 2 \, \text{m/s}$ at $x = 0$, find the speed when $x = 3$ m.

Solution

Use the position form of the acceleration relation:

a = v\frac{dv}{dx}

v\,dv = 4x\,dx

Integrate from $(x,v) = (0,2)$ to $(3,v)$:

\int_2^v v\,dv = \int_0^3 4x\,dx

\frac{1}{2}(v^2 - 4) = 18

$$ v^2 = 40 $$

v = 2\sqrt{10} \, \text{m/s}

3.2Use Work-Energy on an Incline

Final | Problem 3.2 |

A $2$ kg crate starts from rest and slides $5$ m down a frictionless $30^\circ$ incline. Take $g = 10 \, \text{m/s}^2$. Find the speed at the bottom.

Solution

The work done by gravity is

W_g = mgs\sin 30^\circ

W_g = 2(10)(5)\left(\frac{1}{2}\right) = 50 \, \text{J}

Since the crate starts from rest,

\frac{1}{2}mv^2 = 50

With $m = 2$ kg:

$$ v^2 = 50 $$

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

3.3Use Impulse-Momentum in an Impact

Final | Problem 3.3 |

A $0.5$ kg puck moving east at $8 \, \text{m/s}$ is struck by an average $12$ N force to the west for $0.5$ s. Find its final velocity.

Solution

Take east as positive.

The impulse is

J = Ft = (-12)(0.5) = -6 \, \text{N}\cdot\text{s}

Initial momentum:

p_1 = mv_1 = 0.5(8) = 4 \, \text{kg}\cdot\text{m/s}

Impulse-momentum gives

$$ p_2 = p_1 + J = 4 - 6 = -2 $$

v_2 = \frac{p_2}{m} = \frac{-2}{0.5} = -4 \, \text{m/s}

The puck ends with speed $4 \, \text{m/s}$ to the west.

3.4Find a Point's Velocity and Acceleration on a Rotating Bar

Final | Problem 3.4 |

A rigid bar has point $A$ moving to the right at $2 \, \text{m/s}$ with acceleration $1 \, \text{m/s}^2$ to the right. The bar rotates counterclockwise with $\omega = 4 \, \text{rad/s}$ and $\alpha = 2 \, \text{rad/s}^2$. Point $B$ is $0.5$ m above $A$. Find $\mathbf{v}_B$ and $\mathbf{a}_B$.

Solution

Use the rigid-body relations:

\mathbf{v}_B = \mathbf{v}_A + \boldsymbol{\omega} \times \mathbf{r}_{B/A}

\mathbf{a}_B = \mathbf{a}_A + \boldsymbol{\alpha} \times \mathbf{r}_{B/A} + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}_{B/A})

Take

\mathbf{v}_A = 2\,\mathbf{i}, \quad \mathbf{a}_A = 1\,\mathbf{i}, \quad \boldsymbol{\omega} = 4\,\mathbf{k}, \quad \boldsymbol{\alpha} = 2\,\mathbf{k}, \quad \mathbf{r}_{B/A} = 0.5\,\mathbf{j}

Then

\boldsymbol{\omega} \times \mathbf{r}_{B/A} = 4\mathbf{k} \times 0.5\mathbf{j} = -2\,\mathbf{i}

\mathbf{v}_B = 2\,\mathbf{i} - 2\,\mathbf{i} = \mathbf{0}

For acceleration,

\boldsymbol{\alpha} \times \mathbf{r}_{B/A} = 2\mathbf{k} \times 0.5\mathbf{j} = -1\,\mathbf{i}

and

\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}_{B/A}) = 4\mathbf{k} \times (-2\,\mathbf{i}) = -8\,\mathbf{j}

Therefore,

\mathbf{a}_B = 1\,\mathbf{i} - 1\,\mathbf{i} - 8\,\mathbf{j} = -8\,\mathbf{j}

So point $B$ is instantaneously at rest, and its acceleration is $8 \, \text{m/s}^2$ downward.

3.5Choose the Right Impact Method

Final | Problem 3.5 |

A $2$ kg cart slows from $6 \, \text{m/s}$ to $2 \, \text{m/s}$ in $0.1$ s during a collision. What method should you use to find the average impact force, and what is that force?

Solution

Use impulse-momentum because the force acts over a short time and the velocity change is given.

Take the positive direction as the initial direction of motion:

\Delta p = m(v_2 - v_1) = 2(2 - 6) = -8 \, \text{kg}\cdot\text{m/s}

Average force:

F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{-8}{0.1} = -80 \, \text{N}

So the average impact force is $80$ N opposite the cart's motion.

Difficulty

Challenge

4.1Analyze Planar Motion from Position Functions

Final | Problem 4.1 |

A particle has position

x(t) = t^2, \qquad y(t) = 2t^2 - 4t.

Find $\mathbf{v}$ and $\mathbf{a}$ at $t=2$, and determine whether the particle is speeding up or slowing down at that instant.

Solution

Write the velocity vector:

\mathbf{v}(t) = \dot{x}\,\mathbf{i} + \dot{y}\,\mathbf{j}

\mathbf{v}(t) = 2t\,\mathbf{i} + (4t - 4)\,\mathbf{j}

At $t=2$:

\mathbf{v}(2) = 4\,\mathbf{i} + 4\,\mathbf{j}

Differentiate again:

\mathbf{a}(t) = 2\,\mathbf{i} + 4\,\mathbf{j}

\mathbf{a}(2) = 2\,\mathbf{i} + 4\,\mathbf{j}

The speed is

|\mathbf{v}(2)| = \sqrt{4^2 + 4^2} = 4\sqrt{2}

Since the acceleration points generally in the same direction as the velocity, the particle is speeding up at $t=2$.

4.2Velocity and Acceleration of a Rotating Point

Final | Problem 4.2 |

A wheel of radius $0.5$ m has $\omega = 2 \, \text{rad/s}$ and $\alpha = 2 \, \text{rad/s}^2$ at an instant. For a point on the rim, find the speed, tangential acceleration, normal acceleration, and total acceleration magnitude.

Solution

Use the fixed-axis rotation formulas:

v = \omega r, \qquad a_t = \alpha r, \qquad a_n = \omega^2 r

v = 2(0.5) = 1 \, \text{m/s}

a_t = 2(0.5) = 1 \, \text{m/s}^2

a_n = 2^2(0.5) = 2 \, \text{m/s}^2

The total acceleration magnitude is

a = \sqrt{a_t^2 + a_n^2} = \sqrt{1^2 + 2^2} = \sqrt{5} \, \text{m/s}^2

4.3Use Rotational Work-Energy

Final | Problem 4.3 |

A uniform rod has mass $2$ kg and length $1$ m and rotates about one end. A constant moment of $6$ N$\cdot$m acts on it as it turns through $90^\circ$ from rest. Find its angular speed at the end of the turn.

Solution

Use work-energy for rotation:

W = \Delta T

The work done by the moment is

W = M\theta = 6\left(\frac{\pi}{2}\right) = 3\pi \, \text{J}

The rod's moment of inertia about one end is

I_O = \frac{1}{3}mL^2 = \frac{1}{3}(2)(1^2) = \frac{2}{3} \, \text{kg}\cdot\text{m}^2

Since it starts from rest,

\frac{1}{2}I_O\omega^2 = 3\pi

Substitute $I_O = \frac{2}{3}$:

\frac{1}{2}\left(\frac{2}{3}\right)\omega^2 = 3\pi

\frac{1}{3}\omega^2 = 3\pi

\omega^2 = 9\pi

\omega = 3\sqrt{\pi} \, \text{rad/s}

4.4Handle a Rotating Frame with Coriolis Acceleration

Final | Problem 4.4 |

A collar slides outward in a radial slot on a disk rotating counterclockwise with constant angular velocity $\omega = 4 \, \text{rad/s}$. At the instant shown, the collar is $0.5$ m from the center and has outward relative speed $\dot{r} = 2 \, \text{m/s}$. The disk has no angular acceleration and the radial speed is constant, so $\ddot{r} = 0$. Find the Coriolis acceleration magnitude and the total acceleration vector.

Solution

For a rotating frame, the Coriolis term is

2\boldsymbol{\omega} \times \mathbf{v}_{rel}

Its magnitude is

2\omega v_{rel} = 2(4)(2) = 16 \, \text{m/s}^2

Since $\dot{r}$ is constant, $\mathbf{a}_{rel} = 0$ and $\boldsymbol{\alpha} = 0$.

The centripetal term is

\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}) = -\omega^2 r\,\mathbf{e}_r

a_r = -(4^2)(0.5) = -8 \, \text{m/s}^2

and the Coriolis term points in the tangential direction:

a_\theta = 16 \, \text{m/s}^2

Therefore,

\mathbf{a} = -8\,\mathbf{e}_r + 16\,\mathbf{e}_\theta

The Coriolis acceleration magnitude is $16 \, \text{m/s}^2$.

Dynamics Practice

Selected problems

Translation of reference frames

Direct

1.1Classify the Study of Motion

1.2Differentiate a Position Function

1.3Use Constant Acceleration to Find Speed

1.4Find Displacement Under Constant Acceleration

1.5Read Velocity and Acceleration from Cartesian Components

1.6Compute Normal Acceleration

1.7Add Velocities in Relative Motion

1.8Apply Newton's Second Law

1.9Use Friction at a Contact Surface

1.10Find Tangential Speed in Pure Rotation

Integrated

2.1Use Work-Energy with a Constant Force

2.2Track Energy with a Spring

2.3Use Impulse to Find Final Velocity

2.4Solve a Friction Problem with Force Balance

2.5Combine Translation and Rotation

2.6Find Angular Acceleration from a Moment

2.7Choose the Best Method

2.8Spot a Common Pitfall

Applied

3.1Recognize Variable Acceleration by Position

3.2Use Work-Energy on an Incline

3.3Use Impulse-Momentum in an Impact

3.4Find a Point's Velocity and Acceleration on a Rotating Bar

3.5Choose the Right Impact Method

Challenge

4.1Analyze Planar Motion from Position Functions

4.2Velocity and Acceleration of a Rotating Point

4.3Use Rotational Work-Energy

4.4Handle a Rotating Frame with Coriolis Acceleration