1 University of California, Berkeley Physics 7A, Section 1, Spring 2009

University of California, Berkeley Physics 7A, Section 1, Spring 2009 (Yury Kolomensky)

SOLUTIONS TO THE SECOND MIDTERM Maximum score: 100 points

1. (10 points) Blitz (a) hoop, disk, sphere This is a set of simple questions to warm you up. The (b) disk, hoop, sphere problem consists of ﬁve questions, 2 points each. (c) sphere, hoop, disk 1. Circle correct answer. Imagine a basketball bouncing on a court. After a dozen or so (d) sphere, disk, hoop bounces, the ball stops. From this, you can con- (e) hoop, sphere, disk clude that Answer: (d). The object with highest moment of (a) Energy is not conserved. inertia has the largest kinetic energy at the bot- (b) Mechanical energy is conserved, but only tom, and would reach highest elevation. The ob- for a single bounce. ject with the smallest moment of inertia (sphere) (c) Total energy in conserved, but mechanical will therefore reach lowest altitude, and the ob- energy is transformed into other types of ject with the largest moment of inertia (hoop) energy (e.g. heat). will be highest.

(d) Potential energy is transformed into kinetic 4. A particle attached to the end of a massless rod energy of length R is rotating counter-clockwise in the (e) None of the above. horizontal plane with a constant angular velocity ω as shown in the picture below. Answer: (c) 2. Circle correct answer. In an elastic collision ω (a) momentum is not conserved but kinetic energy is conserved; (b) momentum is conserved but kinetic energy R is not conserved; (c) both momentum and kinetic energy are conserved;

(d) neither kinetic energy nor momentum is Which direction does the vector ~ω point ? conserved; (e) none of the above. (a) Tangential to the circle, in the counter- clockwise direction Answer: (c) (b) Along the radius, towards the center of the 3. Circle correct answer. A hoop, a uniform disk, circle and a uniform sphere, all with the same mass (c) Along the radius, away from the center of and outer radius, start with the same speed and the circle roll without sliding up identical inclines. Rank the objects according to how high they go, least (d) Normal to the page, towards you to greatest. (e) Normal to the page, away from you 2

Answer: (d), according to the right-hand rule 50% efficient, i.e. half of the energy from burning diesel fuel goes to locomotive motion, and 5. Clean-and-jerk is an Olympic weightlifting half goes to waste. Burning one gallon of diesel event. In the first phase (clean), a competitor fuel produces about 150 MJ of energy. How stands up with a heavy barbell from the squat many gallons of diesel is spent on the 8 km trip position, lifting the bar by about 1 m in less than up Grapevine Grade ? 1 second. The Olympic record for this event is 263.5 kg (580.3 pounds), set by Hossein Reza Zadeh of Iran in 2004 Olympics in Athens. Es- 2. Solution timate (to 1 digit) the power delivered to the (a) There are four forces acting on the truck and the barbell by Zadeh during this lift in horsepower trailer: weight mg, directed downward, normal force (1 hp = 750 W). from the surface, air drag (pointing along the surface in the direction opposite to the velocity of the truck), Answer: 3 hp. 4 hp would also be an allowed and the propulsion force Fpr. The propulsion force is answer (e.g., if you used g = 10m/s2, or ac- the force of static friction between the truck’s tires and counted for the elevation of the weightlifter’s ground, and it prevents the wheels from slipping. Since body). the velocity of the truck is constant, the forces are bal- anced. The balance of forces along the slope leads to 2. (15 points) Grapevine Grade the equation: Grapevine grade is a famous place on interstate I-5 between Bakersfield and Los Angeles, where a 5 mile (8 Fdrag + mg sin θ = Fpr = 7.3 kN km) stretch of the road has an average incline of 6% ◦ (3.4 ). Imagine a heavy duty pickup truck towing a The power provided by the force Fpr (which ultimately trailer up Grapevine Grade at the constant speed of 65 comes from the truck’s drivetrain) is mph (29 m/s). At this speed, the air drag force acting on P = F v = 2.1 · 105 J = 280hp the truck is about Fdrag = 1.5 kN. The total mass of the pr truck and the trailer is m = 10 tons (104 kg). (b) The useful work done by the propulsion force is (a) (10 points) Estimate the power output of the truck’s engine (assume that all power goes into W = FprL = 58 MJ locomotive motion of the truck). Express you answer in Horsepower (1 hp = 750 W) One gallon of diesel fuel produces 0.5∗150 MJ = 75 MJ of useful work. So the entire trip would use up 60/75 = (b) (5 points) The modern diesel engines are about 0.8 gal of diesel.

3. (20 points) Dangerous Demo

You may remember this demonstration from earlier in the semester. A 200 g wooden block is placed on top of a muzzle of a 0.22” rifle, mounted vertically next to a meter stick (see picture below). The rifle fires a 3 g bullet into the block. The bullet stops inside the block, and we observe that the block (with the bullet inside) lifts up to the elevation h = 240 cm above the muzzle (this is not what happened in class as we used a low-velocity lighter bullet). This is a convenient way to measure the muzzle velocity (velocity with which it exits the rifle) of the bullet. 3

(a) (10 points) What velocity does the bullet have just before it hits the block ? That is the muzzle velocity.

(b) (10 points) Obviously, the rifles are not normally used this way. Usually, you would hold the rifle in your hands, press the butt end against your shoulder, and fire. When the bullet is fired, you feel the rifle recoil, i.e. the butt of the rifle strike you in the shoulder. Why does the rifle recoil ? What average force does the rifle exert on the shoulder while the bullet accelerates ? You will need to know the length of the rifle barrel (68.5 cm).

3. Solution Energy conservation requires (a) This part should be familiar, since it has been worked out in class when we discussed inelastic col- Uinitial + Kinitial = Ufinal + Kfinal lisions. Plugging in the values, we can find V : First, momentum conservation relates the velocity v of the bullet just before the collision to the velocity V V = 2gh of the block-bullet combination just after the collision: p (which is the standard answer from kinematics as well), pbefore = mv = pafter = (m + M)V and plug it into Eq. (1): where m = 3 g is the mass of the bullet, and M = 200 g m + M v = 2gh = 464 m/s . is the mass of the block. The collision is fully inelastic: m p the bullet is embedded in the block. So if we know the (b) The recoil of the rifle is another demonstration of velocity V of the block after the collision, we can find momentum conservation. Initially, the rifle and the bul- out the velocity of the bullet: let are at rest, and their total momentum is zero. Af- m + M v = V (1) ter the bullet is fired and before it and the rifle have m a chance to interact with the surrounding bodies (e.g. Since we know the elevation of the block, we can re- your shoulder), the momentum of the system has to re- late it to the velocity V using the conservation of energy. main the same. So if the bullet is traveling fast away At the muzzle, the potential energy of the block-bullet from you, the rifle has to recoil towards you, to keep the combo is zero, and the kinetic energy is total momentum zero. Ignoring the powder gas (which m + M 2 escapes through the holes on the side of the muzzle and Kinitial = V 2 carries no forward momentum – I did not tell you about The block rises to the elevation h = 2.4 m, where its it in the problem, and it’s safe to ignore this), the mo- kinetic energy is zero, and its potential energy is mentum of the bullet is equal in magnitude and opposite

Uﬁnal = (m + M)gh in direction to the momentum the riﬂe picks up. 4

You can also think about it in terms of the 3rd New- up by the bullet, if we ignore the powder gases), and ton’s law. Since the bullet is accelerated, there must be a ∆t is the time it takes for the bullet to fully accelerate force applied to it. The source of this force, ultimately, (i.e. leave the barrel). The change in momentum is (by is the rifle, or more precisely, the gun powder gas that magnitude) pushes against the rifle and accelerates the bullet. By 3rd Newton’s law, the force that accelerates the bullet ∆p = mv has to have a pair – the force acting on the rifle in the opposite direction. To find the time interval ∆t, we can use kinematics To compute the force acting on the shoulder, we will equations for the bullet: put a 3rd Newton’s law to good use. The force that accelerates the bullet is equal and opposite to the force ∆t2 l = hai that is applied to the rifle and, through the buttstock, 2 transfered to the shoulder. So if we find the force that v = hai∆t accelerates the bullet, we get the force acting on the rifle and the shoulder. where hai is the average acceleration. These can be The easiest way to find the average force is to invoke solved to express energy conservation (recall problem #6 on the Problem of the Week page). The kinetic energy the bullet picks 2l ∆t = up is v

mv2 Then, the average force is K = 2 mv mv2 and it must be equal to the work done by external forces hF i = = = 470 N – which provide acceleration: ∆t 2l which is the same result. K = Wext = hF il 4. (30 points) Daredevil Gymnast where l = 68.5 cm is the distance over which the bullet is accelerated, i.e. the length of the barrel. The average A gymnast of mass m = 60 kg is rotating around a force is hence horizontal bar, holding on to it with his arms (this skill is called the giants). When his body is vertical and is 2 mv −1 hF i = = 470 N below the bar, his angular velocity is ω0 = 7 sec . 2l In the following, treat the gymnast’s body as a uniform This is a significant force (a bit over 100 lbs !) The ri- solid rod of length L = 2 m (arms to toes), rotating fle recoil can be quite tiring after a few rounds, espe- around its end. The moment of inertia of such a rod is cially for higher power weapons. Have you noticed that I = mL2/3. Ignore friction and air resistance. in (realistic) movies, a sharpshooter never puts his eye right up against the scope of a rifle (like you would be inclined to do looking through binoculars) ? He would not want to get a black eye, or, worse yet, end up with a broken eye socket ! The other way to find the force is to use the impulse

theorem: )

∆p = hF i∆t where ∆p is the change in momentum picked up by the riﬂe (it is equal and opposite to the momentum picked (a) 5

which can be solved for N:

2 L ) Nbot = m(g + acp,bot)= m g + ωbot = 3.5 kN 2 (6 times his weight !) By 3rd Newton’s law, this is

) ) the force the gymnast is applying to the bar, and he is

) clearly pulling himself toward the bar to stay on a cir- 90o cular orbit. (b) This is a two-part question. First, assuming that the

gymnast has angular velocity ωtop 6= ωbot at the top of the swing, the normal force of the bar Ntop and, con- (b) (c) sequently, the force which which the gymnast pulls the bar, can be found from the 2nd Newton’s law: (a) (5 points) What force does the gymnast apply to the bar in the bottom position ? Is he pushing Ntop + mg = macp,top (3)

himself away from the bar, or pulling toward it ? 2 where acp,top = ωtopL/2 is the centripetal acceleration (b) (10 points) What force does the gymnast apply to of the center of mass in the top position (directed down- the bar in the top position ? Is he pushing himself ward). Notice the difference in sign of the normal force away from the bar, or pulling toward it ? with respect to Eq. (2): we assume that both the weight of the gymnast and the normal force are directed down- (c) (15 points) Suppose that after a few giants ward. We will be able to test this assumption: if the nor- around the bar the gymnast releases the bar when mal force happens to be directed upward, it will come ◦ he is at 90 with the vertical. His center of mass out negative when we solve Eq. (3). continues moving vertically while he performs a Now we need to ﬁnd ωtop. Since there is no friction layout somersault by rotating around his center involved, the total mechanical energy of the gymnast is of mass in a prone position (see picture). How conserved. On the bottom of the swing (when the gym- high over the bar would his center of mass be at nast is vertical and is below the bar), his kinetic energy the highest point ? How many revolutions would is the gymnast complete before falling back to the Iω2 bar elevation ? K = bot bot 2 and his potential energy (taking the bar elevation to be 4. Solution zero) is (a) The acceleration of the center of mass of the gym- L nast rotating with angular velocity ωbot = 7 rad/sec is U = −mg bot 2 2 acp,bot = ωbotRcm When the gymnast is at his highest point over the bar, his potential energy is where Rcm = L/2 is the distance between the center of L mass of the gymnast (approximated to be a uniformly Utop =+mg dense solid rod) and the axis of rotation. There are two 2 forces acting on the gymnast: his weight mg, directed and his kinetic energy is downward, and the normal force of the bar Nbot. We Iω2 K = top will assume that it is directed toward the bar, i.e. up- top 2 ward. The 2nd Newton’s law relates these forces to the Energy conservation then implies the equation centripetal acceleration: Iω2 L Iω2 L bot − mg = top + mg (4) Nbot − mg = macp,bot (2) 2 2 2 2 6

2 2 which can be solved for ωtop: Now we can plug ωside, I = mL /3, and I0 = mL /12 into Eq. (5) and solve for H: 2mgL 6g − 2 2 1 2 2 ωtop = ωbot − = ωbot − = 4.4 sec ω L s I r L H = side = 1.7m 8mg We can now plug that into Eq. (3) and solve for Ntop: That’s quite impressive ! L N = m(a − g)= m ω2 − 4g = 540 N Finally, to find the number of revolutions, we need top cp,top bot 2 to figure out the amount of time it takes the gymnast to So indeed, the gymnast is pulling himself toward the fly up to elevation H and come back down. The motion bar, with the force that is slightly less than his own of center of mass is a free fall. Therefore, it takes weight. 2H (c) This is another application of the energy conserva- tfall = = 0.6s s g tion. As the gymnast lets go of the bar, the velocity of his center of mass is directed upward. Since the only to fall from elevation H above the bar to the bar level, force acting on him now is gravity, the center of mass and therefore the total flight time.is will continue moving upward to some elevation H. At t = 2t = 1.2s the same time, his body will be rotating around its cen- total fall ter of mass. Since gravity exerts no torque relative to The number of revolutions is the center of mass, the angular velocity relative to the ωsidettotal center of mass remains constant and equal to what it N = = 1 2π was when the gymnast let go, i.e., ω2 . At the top of side So the gymnast makes (almost) exactly one revolution. the upward trajectory (elevation H above the bar), the Indeed, if he adds a half-twist, he would be facing the gymnast has potential energy bar again and would be able to catch it. That is one of a U(H)= mgH gymnastics release moves. and kinetic energy of 5. (25 points) Another Lecture Demo This was one of the most recent lecture demonstrations. I ω2 K(H)= 0 side A spool consists of an inner hollow cylinder of radius 2 r = 5 cm and mass mc = 4 kg and two solid disks of 2 2 where I0 = I − mL /4 = mL /12 (according to the radius R = 10 cm and mass md = 1 kg each, attached parallel axis theorem). Equaling it to the total mechan- coaxially to the ends of the cylinder. A rope is wound 2 tightly around the inner cylinder, and a force N ical energy Eside = Kside = Iωside/2, we get an equa- F = 16 tion for H: is applied to the end of the rope in the horizontal di- 2 2 rection, as shown on the picture below. The spool rolls Iω I0ω side = mgH + side without slipping on a horizontal table under the influ- 2 2 2 ence of this force. ωside H = (I − I0) (5) 2mg F We still need to find ωside, and that is straightforwardly r done using energy conservation, similar to Eq. (4): R Iω2 L Iω2 bot − mg = side 2 2 2 which is solved for (a) (5 points) Which direction does the spool roll ?

(b) (10 points) Find acceleration of the center of 2 mgL 2 3g −1 ωside = ωbot − = ωbot − = 5.9 sec s I r L mass of the spool, and its angular acceleration. 7

(c) (10 points) Find the smallest coefﬁcient of static 0 = N − Mg (Y ) (7) friction between the spool and the table for the roll to proceed without slipping. Third equation comes from the balance of torques:

5. Solution Iα = F r + F R (8) (a) We worked out this problem in some detail in lec- fr ture, though there is one important difference in the setup above: the rope goes over the top of the spool. where α = a/R is the angular acceleration, and I is the There are two forces acting on the spool: F, which moment of inertia of the spool. Eq. (6) and Eq. (8) can points to the right, and the friction force Ffr, which be solved for a: points to the left. Both produce torque that would rotate the spool in the clockwise direction. Therefore, the F 1+ r/R a = (9) spool would rotate in the clockwise direction, and roll M 1+ I/MR2 to the right, in the direction of F. (b) Now we can be numeric. We start with the 2nd New- The ﬁnal step is to ﬁnd the moment of inertia I. The ton’s law: spool consists of a hollow cylinder with moment of inertia I = m r2 and two solid disks, each with moments F + F + Mg + N = Ma c c fr 2 of inertia Id = mdR /2. Therefore, the total moment where a is the linear acceleration of the center of mass of inertia is (points to the right), N is the normal force, and M = mc + 2md = 6 kg is the total mass of the spool. We can 2 2 I = Ic + 2Id = mcr + mdR project it onto the horizontal X and vertical Y axes:

Ma = F − Ffr (X) (6) Finally, the linear and angular accelerations are

F 1+ r/R 2 a = = 2 2 =3m/s (10) [mc + 2md] [1 + mcr /((mc + 2md)R )+ md/(mc + 2md)] a α = = 30 rad/s2 (11) R

(c) The sign of the friction force is very curious. What it The spool does not slip as long as the friction force means is that the friction force actually points in the

Ffr has not reached its maximal value Ffr max = µsN, direction opposite to what we assumed, i.e., it points where N is the magnitude of the normal force. Hence, to the right ! That means that if the spool slips, it will we have the inequality for the coefﬁcient of static fric- spin rather than slide. You can explicitly check that our tion: expression for acceleration is still valid. The condition F of no slipping is that the coefﬁcient of friction should µ ≥ fr s N indeed be positive and larger than Given acceleration a from Eq. (11), Ffr and N can be F − Ma Ma − F computed from Eq. (6) and Eq. (7): µs ≥ = = 0.03 Mg Mg

N = Mg = 59 N

Ffr = F − Ma = −2 N