Pomodoro
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Difficulty
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.
If
what are $v(2)$ and $a(2)$?
Solution
Differentiate once to get velocity:
So
Differentiate again to get acceleration:
So
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:
So the net force is
to the east.
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,
Substitute the values:
Since $\cos 60^\circ = \tfrac{1}{2}$,
A $0.5$ kg puck moves at $8\ \text{m/s}$. What is its momentum?
Solution
Momentum is
So the magnitude is
The momentum points in the same direction as the velocity.
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:
So
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
Substitute:
A $12\ \text{N}$ force is applied perpendicular to a wrench $0.25\ \text{m}$ from the pivot. What is the torque?
Solution
Use
Because the force is perpendicular, $\sin\phi = 1$. So
A mass-spring system has $m = 2\ \text{kg}$ and $k = 8\ \text{N/m}$. What is its period?
Solution
For simple harmonic motion,
Then
For a mass-spring system with kinetic energy
and potential energy
what is the Lagrangian $\mathcal{L}$?
Solution
The Lagrangian is
So
Difficulty
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
That gives the time to the top:
The horizontal speed is constant:
So the horizontal distance is
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:
So the components are $4\ \text{m/s}$ east and $3\ \text{m/s}$ north.
The speed is
The direction is
north of east.
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
By the work-energy theorem,
So
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
Because the force is to the left, the impulse is negative relative to the original motion.
Initial momentum:
Final momentum:
So the final speed is
to the right.
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:
So
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
so
The translational kinetic energy is
The moment of inertia is
The rotational kinetic energy is
So the total kinetic energy is
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
Take torques about the left end:
Then
For a mass-spring oscillator with
derive the equation of motion.
Solution
Apply the Euler-Lagrange equation:
Compute the derivatives:
so
Also,
Substitute:
so the equation of motion is
Difficulty
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:
For the $1$ kg hanging mass:
Substitute $T = 2a$:
Then
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:
So
Initial kinetic energy:
Final kinetic energy:
So the kinetic energy lost is
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:
The torque is
So the angular acceleration is
Starting from rest,
The angular displacement is
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,
so speed scales like $r^{-1/2}$:
The orbital period scales like $r^{3/2}$:
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
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:
For a solid disk,
and rolling without slipping gives
Substitute:
So
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
So
Initial rotational kinetic energy:
Final rotational kinetic energy:
So the kinetic energy increases by
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:
Take torques about the left end:
Then
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
The kinetic energy is
So the Lagrangian is
Apply the Euler-Lagrange equation:
Compute the derivatives:
so
and
Substitute:
Thus the equation of motion is
