In this section we present some physics exercises which you can use to test your understanding. Don't be discouraged if you find the exercises difficult—these are meant to be hard.

When solving the exercises, I recommend that you use the following approach:

1. Figure out what type of problem you are dealing with (kinematics? angular motion? energy?).
2. Draw a diagram which describes the physical situation. Label things clearly.
3. Copy over from your formula sheet all the equations which you plan to use.
4. Substitute the known quantities into the equations and analyze which is the

unknown you are looking for. Visualize the steps (plug what into what?) which

  you will use to solve for the unknowns. 
- Solve for the unknowns.

Make sure you attempt to do each of the exercises on your own before looking at the solutions.

A ball is thrown from the ground upwards with an initial velocity of 20[m/s]. How long will it stay in the air before it comes back to the ground?

Sol: This is a kinematics question. Using the equation $v(t) = at+v_i$ we can find $t_{top}$ (we know that $v(t_{top})= 0$). Ans: $t_{flight} = 2t_{top}=4.1$[s].

A disk is rotating with an angular velocity of $\omega=5$[rad/s]. A slug is sliding along the surface of the disk in the radial direction. Imagine the slug starting from the centre of the disk an moving outwards. If the coefficient of friction between the slug and the disk is $\mu_k = 0.4$, how far will the slug be able to slide to before it flies off the surface?

Ans: The normal force of the between the slug and the turntable is $N=mg$. The friction force available is $F_f=0.4mg$. The centripetal acceleration required to keep the slug on the disk when it is at a radius $R$ is $F_r=ma_r=m\frac{(R\omega)^2}{R}$. The slug will fly off when the friction force is not sufficient to keep it turning, which happens at distance $R=\frac{0.4g}{\omega^2}$ from the centre of the disk.

You are moving your fridge and you have it loaded on the elevator. Because of static friction, a force is required to start the fridge sliding across the floor of the elevator. Rank the forces required, from smallest to largest, in three situations: (a) a stationary elevator, (b) when the elevator is accelerating upwards, © when the elevator is accelerating downwards.

Ans: $F_{fs}$ (upwards) $> F_{fs}$ (static) $> F_fs$ downwards. The equation for F_fs is $F_{fs} = \mu_s N$, where $N$ is the normal force (contact force between the box and the floor of the elevator). In the $y$ direction the force diagram on the match box reads $\sum F_y = N - mg = ma_y$. When the elevator is static we have a_y = 0 so N = mg. If $a_y > 0$ (accelerating upwards) then we must have $N > mg$ hence the friction force will be larger than when static. When $a_y < 0$ (accelerating downwards) $N$ must be smaller than $mg$ and consequently there will be less $F_{fs}$.

Three coins are placed on a rotating turntable. One coin is placed at $5$[cm] from the centre, another is placed at $10$[cm] from the centre and the third is at $15$[cm] from the centre. Initially, due to static friction, the pennies are moving together with the turntable as it starts to rotate. The angular speed $\omega$ is then increased slowly. Assuming each penny has the same coefficients of friction with the turntable surface, which penny starts to slide first?

Sol: This is a circular motion question. The penny that is furthest from the centre will fly off the first. That is because the centripetal force required to keep it turning is the largest. Recall that $F_r = ma_r$, that $a_r = v^2/R$ and that $v =\omega R$. If the turn table is turning with angular velocity $\omega$, then the centripetal acceleration required to keep a coin turning in a circle of radius $R$ is $F_r = m \omega^2 R$. This centripetal force must be supplied by the static force of friction $F_{fs}$ between the coin and the turn table. Large $R$ requires more $F_{fs}$ hence the coin furthest will fly off first.

Three identical balls (a, b, and c) are thrown upwards with identical speeds but different angles. Ball is (a) is thrown directly upwards, ball (b) is thrown at an angle of $30^\circ$ with the vertical while ball (c ) is thrown at an angle of $45^\circ$. Rank the balls in order of their speed when they have reached a height of one meter. Assume all of the balls have enough energy to get to this height.

Sol: This is a projectile question. The vertically-thrown ball (a) will have the largest $v_iy$ and zero $v_ix$. The angled shots trade decreased $v_iy$ for ininitial $v_ix$. For the $y$ direction we have $v_{fy}^2 = v_{iy}^2 + 2(-9.81)y$. For the $x$ direction we have $v_{fx} = v_{ix}$. Any investment you put into $v_ix$ is conserved, while investments in $v_iy$ get transformed by the function $f(a) = \sqrt{ a^2 - \textrm{ const} } < |a|$ so you are better off investing in $v_ix$ than in $v_iy$. To maximize the speed of the ball when it reaches a height of one meter you should choose the ball with the largest horizontal velocity (c ).

Two identical (same moment of inertia) pulleys have strings wound around them. The first pulley has the string wound around the outer radius $R$, while the sting on the second pulley is wound on a smaller radius $r < R$. The same force F is used to rotate the pulleys. After a fixed time $t$, which pulley has the larger speed? Which puck has the larger rotational kinetic energy?

Sol: This is an angular motion question. The two pucks have identical rotational resistance: they have the same moment of inertia $I$. The torque produced by the string is larger for the pulley which has the string wound around the larger radius $R$. Higher torque will produce a more angular acceleration and hence a bigger angular velocity (and KE).

These following exercises will require you to mix techniques from different sections.

The disk brake pads on your new bicycle squeeze the brake disks with a force of 5000[N] (from each side) and you have one on each tire. The coefficient of friction between brake pads and brake disk is $\mu_k=0.3$. The brake disks have radius $r=6$[cm] while the tire has a radius $R=20$[cm]. 1. What is the total force of friction in each brake?
2. What is the torque exerted by each brake?
3. How many turns of wheels will it take before to stop if you are moving with 10[m/s] and apply broth brakes. Assume that the combined mass of you and your bicycle is 100[kg]? 4. What will be the braking distance?

Sol: 1. The friction force is proportional to the normal force so we have $F_f=0.3\times 5000=1500$[N] of friction on each side of the disk for a total force of $F_f=3000$[N] per wheel. 2. This friction force of the brakes acts with a leverage of $0.06$[m] so the torque produced by each brake is $\mathcal{T} = 0.06 \times 3000 = 180$[N$\:$m]. 3. The kinetic energy of a 100kg object moving at 10[m/s] is equal to $K_i=\frac{1}{2}100(10)^2=5000$[J]. We will use $K_i - W = 0$ where $W$ is the work done by the brakes. Denote $\theta_{stop}$ the angle rotation of the tires. The work done by each brake is $180\theta_{stop}$ so to it will take a total of $\theta_{stop} = \frac{5000}{360}=13.\overline{8}$[rad] to stop the bike. This is 2.21 turns of the wheels. 4. Your stopping distance is $13.\overline{8}\times 0.20=2.\overline{7}$[m]. Yey for disk brakes!

A half-naked dude swings from a long rope attached to the ceiling. The rope has length 6.00[m], the dude swings from an initial angle of $-50.0^\circ$ ($50^\circ$ to the left of the vertical line) all the way to the angle $+10.0^\circ$ at which point he lets go of the rope. How far will Tarzan fall (as measured from the centre position of the swing), i.e., I am asking you to find $x_f = 6\sin(10) + d$ where $d$ is the distance travelled by the “projectile”.

Sol: This is an energy problem followed by a projectile motion problem. The energy equations is $\sum E_i = \sum E_f$ which in this case is $U_i = U_f + K_f$ or $mg(6-6\cos50^\circ)=mg(6-6\cos10^\circ)+\frac{1}{2}mv^2$ which can be simplified to $v^2 = 12g(\cos10^\circ - \cos50^\circ)$, and solving for $v$ we find $v=4.48$[m/s]. Now for the projectile motion part. The initial velocity is $4.48$[m/s] at an angle of $10^\circ$ with respect to the ground so $v_i=(4.42,0.778)$[m/s]. The initial position is $(x_i,y_i) = ( 6\sin(10), 6[1-\cos(10)] )= (1.04, 0.0911)$[m]. To find the total time of flight we solve for $t$ in $0=-4.9 t^2+ 0.778t+ 0.0911$ and we find $t=0.237$[s]. The final position will be $x_f = 6\sin(10)+ 4.42\times 0.237=2.08$[m].

Two disgruntled airport employees decide to vandalize a long moving sidewalk by suspending a leaking-paint-bucket pendulum on top of the moving sidewalk and letting it run. The oscillations of the pendulum are small and transverse to the direction of the motion of the sidewalk. The pendulum is composed of a long cable (considered massless) and a paint bucket with a hole in the bottom. Find the equation $y(x)$ of the the resulting pattern of pain on the moving sidewalk in terms of the pendulum's maximum (angular) displacement $\theta_{max}$, its length $\ell$, and the speed of the sidewalk $v$. Assume that $x$ measures distances along the sidewalk and $y$ denotes the transversal displacement of the pendulum. Does the loss of pendulum mass affect the waveform you see on the moving sidewalk?

Sol: This is a simple harmonic motion question involving a pendulum. We can therefore begin by writing down the general equation of motion for a pendulum $\theta(t) = \theta_{max} \cos( \omega t )$, where $\omega=\sqrt{ g/\ell }$. Enter sidewalk. The sidewalk is moving to the left at velocity $v$. If we choose the $x=0$ coordinate at a time when $\theta(t) = \theta_{max}$, then the pattern on the sidewalk will be described by the equation $y(x) = \ell \sin(\theta_{max})\cos( kx )$ where $k=2\pi/\lambda$ and $\lambda$ is how long (as a distance in the $x$ direction) it takes to complete one cycle. One full swing of the bucket takes $T = 2\pi/\omega$[s]. In that time, the moving sidewalk will have moved a distance of $vT$ meters. So one cycle in space (one wavelength) is $\lambda=vT = v 2\pi/\omega$. We conclude that the equation of the paint on the moving sidewalk is $y(x) = \ell \sin(\theta_max) \cos( (\omega/v) x )$. Observe that the the angular frequency parameters $\omega=\sqrt{ g/\ell }$ does not depend on the mass of the pendulum, thus the change in mass (as the paint leaks out) will not affect the motion.

A ball of radius $r$ is rolling back and forth inside a ramp. The ramp opens upwards and is circular in shape with an radius $R$. The ball rolls to one side, slows down, stops, then rolls back to the centre of the ramp and continues rolling up the other side and so on and so forth. What is the period of this oscillation?

Sol: http://www.chaostoy.com/cd/html/pendul_e.htm

[ Physics exercises. ]
http://en.wikibooks.org/wiki/Physics_Exercises
http://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)

NOINDENT [ Word problems ]
http://www.physics.umd.edu/perg/abp/think/mech/mechdyn.htm

NOINDENT [ Lots of worked examples. ]
http://farside.ph.utexas.edu/teaching/301/lectures/lectures.html