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Difficulty
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.
Given
find $v(2)$ and $a(2)$.
Solution
Differentiate position to get velocity:
Differentiate again to get acceleration:
Evaluate at $t=2$:
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:
Since the cart starts from rest, $v_0 = 0$:
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
Substitute the values:
For the position vector
find $\mathbf{v}$ and $\mathbf{a}$ at $t=3$.
Solution
Differentiate the position vector:
Differentiate again:
At $t=3$:
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
So
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:
So the person's ground speed is $2.3 \, \text{m/s}$ east.
A $5$ kg particle has a net force of $20$ N to the right. Find its acceleration.
Solution
Use Newton's second law:
So
The acceleration is $4 \, \text{m/s}^2$ to the right.
A surface has normal force $N = 40$ N and kinetic friction coefficient $\mu_k = 0.25$. Find the kinetic friction force.
Solution
Use
So
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,
So
Difficulty
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:
Work done by the force:
So
Then
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
At the natural length, all of that energy becomes kinetic energy:
With $m=1$ kg:
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
Impulse-momentum gives
Since $v_1 = 0$,
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
So the kinetic friction force is
The net horizontal force is
Therefore,
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
Take $\mathbf{v}_A = 1.5\,\mathbf{i}$, $\boldsymbol{\omega} = 3\,\mathbf{k}$, and $\mathbf{r}_{B/A} = 0.6\,\mathbf{j}$.
Then
So
Point $B$ moves left at $0.3 \, \text{m/s}$.
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:
The center of mass is $d = 0.75$ m from the end, so
Now use
So
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.
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
is valid only when acceleration is constant.
Here $a(t)$ changes with time, so you must integrate
to find velocity.
Difficulty
A particle moves in a straight line with acceleration
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:
So
Integrate from $(x,v) = (0,2)$ to $(3,v)$:
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
So
Since the crate starts from rest,
With $m = 2$ kg:
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
Initial momentum:
Impulse-momentum gives
So
The puck ends with speed $4 \, \text{m/s}$ to the west.
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:
Take
Then
so
For acceleration,
and
Therefore,
So point $B$ is instantaneously at rest, and its acceleration is $8 \, \text{m/s}^2$ downward.
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:
Average force:
So the average impact force is $80$ N opposite the cart's motion.
Difficulty
A particle has position
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:
so
At $t=2$:
Differentiate again:
so
The speed is
Since the acceleration points generally in the same direction as the velocity, the particle is speeding up at $t=2$.
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:
So
The total acceleration magnitude is
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:
The work done by the moment is
The rod's moment of inertia about one end is
Since it starts from rest,
Substitute $I_O = \frac{2}{3}$:
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
Its magnitude is
Since $\dot{r}$ is constant, $\mathbf{a}_{rel} = 0$ and $\boldsymbol{\alpha} = 0$.
The centripetal term is
so
and the Coriolis term points in the tangential direction:
Therefore,
The Coriolis acceleration magnitude is $16 \, \text{m/s}^2$.
