many 11-1
How degrees in 1 radian? A) 1 rad = 2p degrees B) 1 rad = 180 o C) 1 rad = 10 o D) 1 rad = 57.3 o E) Radian is not a measure of angle, so the question makes no sense.
Answer D: 2 pi radians is 360 degrees (right?), so dividing both sides by 2 pi, you get 1 rad = 360 deg/2 pi = 57.3
11-2
A student sees the following question on an exam: A flywheel with mass 120 kg, and radius 0.6 m, starting at rest, has an angular acceleration of 0.1 rad/s2. How many revolutions has the wheel undergone after 10 s? Which formula should the student use to answer the question? A) w=w0 + a t 2 B) q=q0 + w0 t + (1/2) a t 2 2 C) w = w0 +2 a Dq
Answer: We have been given mass, radius, acceleration and initial angular speed (because it started "at rest") . The question is how many revs, which is equivalent to asking about "rotational displacement", i.e. delta theta. (You can measure angles in many units: degrees, rads, revolutions... they're all equivalent. It's like using inches or feet or meters for length) We have been given time too. So it looks to me like B is the best one to use. It directly gives what we're asking for, namely the final angle. Of course, that will give the angle in "rad", so we'll then have to convert by 1 rev = 2 pi rad.
11-3
BIG BEN and a little alarm clock both keep perfect time. Which minute hand has the bigger angular velocity w? A) Big Ben B) little alarm clock C) Both have the same w
Answer : Someone just told me Big Ben doesn't HAVE a minute hand. Oops! Well, if it did, it would sweep through 360 degrees in one hour. That is, omega = 360 deg/hour (or, 1 rev/hour) (or (2 pi rad/hour)... The point is, that's the same for ANY clock! The radius of the hand is irrelevant. Angular velocity "omega" describes "change in angle/change in time". The second hand, however, will have a bigger omega. (60 times faster!)
11-4
A small wheel and a large wheel are connected by a belt. The small wheel is turned at a constant angular velocity ws. How does the magnitude of the angular velocity of the large wheel wL compare to that of the small wheel?
ws wL
A: ws = wL
B: ws > wL
C: ws < wL
Answer: They are NOT the same. When the little wheel has rotated ONCE, the belt has moved a small distance 2*pi*(small radius). That small amount of belt has barely turned the big wheel very much. So the big wheel has NOT rotated a full turn, i.e. it has a much smaller omega. (Smaller by the ratio of radii, in fact) Here's another way to argue it: a point at the edge of the small wheel will move with speed v = omega_s*r. That "v" tells you the linear speed of a point at the edge - the same as the linear speed of the BELT as it passes by! (I'm assuming the belt is not slipping) Now look at the bigger circle. A point on ITS edge will move with v=omega_L*R. Since v is the same (it's the same belt with a certain speed everywhere), then omega_L will have to be smaller. This is the principle behind gears on your bike! You can turn the front (large) crank SLOWLY, and the rear (small) crank turns RAPIDLY. Otherwise, you could never get your wheels to turn faster than you can move your legs. Look at your bike - as you shift gears, you make the rear radius smaller if you want the rear wheel to spin faster...
There is a bug S on the rim of the small wheel and another bug L on the rim of the large wheel. How do their speeds compare? A: S = L B: S > L C: S < L
Answer: We've just answered this. They have the same speed, which is also the speed of the belt.
11-5
A ladybug is clinging to the rim of a spinning wheel which is spinning CCW very fast and is slowing down. At the moment shown, what is the approximate direction of the ladybug's acceleration?
A) C) E) None of these B) D)
Answer:: There are two components to her acceleration: First, an INWARD one (centripetal accel, given at any instant by v^2/R. )It's not constant, since v is changing, but at the instant shown it has some value and points straight to the center, i.e. left. There is a SECOND component, because her speed is decreasing. If she's going CCW that means she's heading UP the page at the point shown. Decreasing speed means a DOWNWARD acceleration at that point. That's the "tangential component" of her acceleration. So she has a left component and a down component, and that will add up to a "down and left" vector, i.e. C.
11-6
A student in Physics 1110 sees the following question on CAPA: An engine flywheel turns with constant angular speed of 100 rev/min. When the engine is shut off, friction slows the wheel to rest in 2 hours. What is the magnitude of the constant angular acceleration of the wheel? Give the answer in units of rev/min2. The student writes : w=w0 + a t, with w=0, so |a| = w0 /t = 2pf/t = 2 p (100 rev/min)/120 min Does the answer come out correctly with the desired units? A: Yes B: No
Answer: oh, so tricky. But, you have to watch out for this sort of thing yourself! The first formula is fine, and so is the algebraic solution for alpha. Even the next formula is o.k. alpha = 2 pi f/t. But that 2 pi really has "hidden units", what it MEANS is "There are 2 pi rad/revolution". So by writing 2 pi f, we are converting rev/min to rad/min! ("f" and "omega" are names for the same physical quantity, angular speed, simply measured in different units) The final answer above is 2 pi (rad/rev)*(100 rev/min)/120 min, which is giving us alpha in RAD/Min^2. That's fine, it's just not what the problem asked for. If you want the answer in rev/min^2, you should not (must not!) bother with the factor of 2 pi. You merely want omega/time = 100 rev/min / 120 min, that's it.
11-7
You are trying to open a door that is stuck by pulling on the doorknob in a direction perpendicular to the door. If you instead tie a rope to the doorknob and then pull with the same force, is the torque you exert increased? A: yes B: no
Answer: No, the torque is not increased. Torque is "r cross F", or the force times the perpendicular distance from point of application to the pivot. By extending a rope, you are further from the door, but not any further "in the perpendicular distance" from the hinge. Give it a try with a door, it won't help a bit. However, if you apply the force CLOSER to the hinge, you'll instantly see how much of a difference THAT makes!
11-8
You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is most effective in loosening the nut? (A=1, B=2, C=3, D=4)
Least effective in loosening the nut?
Answer: B=2 is best, because you have the maximum PERPENDICULAR distance from the force application point to the pivot. #1 and #4 are next (they're the same, 4 is the same as 1, that was the previous question!) #3 is least effective, because you have the same F but now your perpendicular distance to the nut is smallest. (If you exaggerate #3 to the point where the wrench already pointed to the right, your rightward force would do NOTHING, except pull the wrench away from the nut!
11-9
Three forces labeled A, B, C are applied to a rod which pivots on an axis thru its center.
L L/2 45o F C 2F
A F
B L/4
Which force causes the largest magnitude torque? A: A B: B C: C D: two or more forces tie for largest size torque.
Answer: A causes a torque of F*L*sin(45) = F*L*0.7 B causes a torque of F*L/2 = F*L*0.5 C cause a torque of 2F * L/4 = FL*0.5 So A is biggest, B and C tie for second place. (All are "CCW")
11-10
A mass is hanging from the end of a horizontal bar which pivots about an axis through it center, but it being held stationary. The bar is released and begins to rotate. As the bar rotates from horizontal to vertical, the magnitude of the torque on the bar.. A: increases B: decreases C: remains constant
Answer: The torque decreases! t = F^ r = F r sin q F^ decreases as the mass falls. When the bar is vertical the torque is zero.
r
q F
F
As the bar rotates from horizontal to vertical, the magnitude of the angular acceleration a of the bar.. A: increases B: decreases C: remains constant
Answer: Torque = I*alpha. I is a measure of the moment of inertia about the pivot, it is a CONSTANT given by the distribution of mass. So if torque decreases, alpha decreases. (You have to think about this... omega is INCREASING, it's still going faster and faster, but alpha is DECREASING, the RATE OF CHANGE of omega is positive but decreasing with time. It's like v and a of a ball rolling down a hill that starts off steep and ends up less steep.
11-11
A mass m hangs from string wrapped around a pulley of radius R. The pulley has a moment of inertia I and its pivot is frictionless. Because of gravity the mass falls and the pulley rotates. The magnitude of the torque on the pulley is.. A: greater than mgR
R
B: less than mgR C: equal to mgR
m
(Hint: Is the tension in the string = mg?)
Answer: The magnitude of the torque is less than mgR! The tension in the string is less than mg because the mass is accelerating down.
FT a (+) mg
Fnet = mg - FT = ma FT = mg - ma
Torque
t = FT R = Ia = I
a R
Two equations [FT = mg-ma, FT = I a/(R2)] in two unknowns (FT and a) can solve.
11-12
Consider a rod of uniform density with an axis of rotation through its center and an identical rod with the axis of rotation through one end. Which has the larger moment of inertia?
C E
axis
axis
A: IC > IE
B : IC < IE
C : IC = IE
Answer: IE is going to be larger. Remember, moment of inertia is the sum of mass*radius^2. The mass of C is more "concentrated" at small radius, the mass of E is on average much further away (and that radius gets squared!) You can also use the parallel axis theorem, because C is clearly rotating around the center of mass, so I(E) = I(C) + m(L/2)^2 is manifestly larger.
11-13
Consider two masses, each of size 2m at the ends of a light rod of length L with the axis of rotation through the center of the rod. The rod is doubled in length and the masses are halved. What happens to I?
A
2m
L/2
L/2 2m m
B
L L m
A: IA is bigger
B : IB is bigger
C: IA = IB
Answer: Just work them out, I = the sum of mass*radius^2, where radius must be measured from your pivot (center of the barbells, in this case) I(A) = 2m*(L/2)^2 + 2m*(L/2)^2 = mL^2 I(B) = m*L^2 + m*L^2 = 2 mL^2. Larger radius wins, because it gets squared!
11-14
Consider a solid disk with an axis of rotation through the center (perpendicular to the diagram). Two holes are cut out near the center and the material is placed near the rim. Which has the larger I?
A B
A: IA > IB
B: IA < IB
C : IA = IB
Answer: Moment of inertia comes from the sum of m*r^2. More mass at larger radius always means more I. Here, we've moved some m's at SMALL r and moved them out to LARGER r, so we've increased the moment of inertia...
11-15
Two wheels with fixed axles, each have the same mass M, but wheel 2 has twice the radius of wheel 1. Each is accelerated from rest with a force applied as shown. In order to impart identical angular accelerations to both wheels, how much larger is F2 than F1?
F1 R 2R Wheel 1, radius R, mass M F2
Wheel 2, radius 2R, mass M
A: F2 = F1 . D: F2 = 8F1 .
B: F2 = 2F1 . E: None of these.
C: F2 = 4F1
(Assume that all the mass of the wheels is concentrated in the rims so that the moment of inertia of each is of the form I = M R2 (hoop formula), and recall that t = I a)
Answer: Torque = R*F here. So Wheel 1: R*F1 = I*alpha = M*R^2*alpha Wheel 2: 2R*F2 = I2*alpha = M*(2R)^2 *alpha (problem asked for alpha to be same) Taking the ratio, F1/2F2 = (1/4) , so B is correct.
11-16
Consider a solid disk with an axis of rotation through the center (perpendicular to the diagram). The disk has mass M and radius R. A small mass m is placed on the rim of the disk. What is the moment of inertia of this system? A: (M+m)R2 B: less than (M+m)R2 C: greater than (M+m)R2
Answer: A solid disk has I = (1/2) M R^2, and adding a mass would give us mR^2 + (1/2)M R^2 = (m+M/2) R^2. This is clearly SMALLER than (m+M)R^2.
Two light (massless) rods, labeled A and B, each are connected to the ceiling by a frictionless pivot. Rod A has length L and has mass m at the end of the rod. Rod B has length L/2 and has a mass 2m at its end. Both rods are released from rest in a horizontal position.
11-17
B A
L 2m m L/2
Which one experiences the larger torque? A: A B: B C: Both have the same size t.
Answer: Torque is Force(external)*(perp distance to pivot point). Here, the force is just the weight of the mass, acting on the mass. So torque(A) = mg*L*sin(theta). torque(B) = 2m*L/2*sin(theta). It's the exam same torque! (By the way, you must be very careful about what theta means in the above diagram. It is the angle between the radius vector (which points down and right), and gravity (down). This is NOT the same as the angle between the rod and the roof! Draw the picture, work it out, they are "complementary" angles! )
Which one falls to the vertical position fastest? A: A B: B C: Both fall at the same rate.
t a= ) (Hint: I
Answer: We just found torque to be the same. But "I" is mass*radius^2, so for rod 1 it's mr^2, for rod #2 it's 2m(L/2)^2 = mL^2/2. Thus, rod 2 has LESS moment of inertia, and therefore MORE acceleration. The longer rod will take longer to fall!
11-18
A mass m is attached to a long, massless rod. The mass is close to one end of the rod. Is it easier to balance the rod on end with the mass near the top or near the bottom? A: easier with mass near top. B: easier with mass near bottom. C: No difference.
mg
Hint: Small a means sluggish t a= behavior, and I
m
Answer:Look at the previous question. Here, torque will go like mg*distance from pivot, while I will go like m*distance^2. So alpha = torque/I will go like distance/distance^2 = 1/distance. That means the mass near the top (bigger distance) will have smaller alpha. It's EASIER to balance. Give it a try, this is very dramatic. Find an object that's long, but heavier on one end than the other. (E.g. take a pool cue, which way is easier to balance, skinny end on top or bottom?) How about just a short stick and a long one? The CM will be in the middle. (The mass will ALSO grow with the radius for real objects, of course) For uniform sticks, torque goes like total mass*radius, and I goes like total mass*radius^2, so alpha goes like 1/radius. So you should again be able to balance a LONG stick easier than a short one. This is even easier to test out. Try balancing a pencil. It's hard! It's not hard to balance a meter stick.)
11-19
A sphere, a hoop, and a cylinder, all with the same mass M and same radius R, are rolling along, all with same the speed v.
Sphere v Hoop v Disk v
Which has the most kinetic energy? A: Sphere B: Hoop D: All have the same KE. C: Disk
Answer: K = (1/2) m v_CM^2 + (1/2) I (omega)^2. All three objects have the same m and v, so the first term is the same for each. How about the second one? I = (some number)* M*R^2. And, omega = v/R. So Rotational K term = (some number) * M*v^2. (The r's cancel out!) The number for a sphere is 2/5. For a hoop is 1. For a disk is 1/2. Thus, the HOOP has more kinetic energy! This isn't just "formulas" - just think about it a little. The hoop concentrates ALL its mass out at larger radius. And for a rotating object, your speed is GREATER when you're farther from the center. So the hoop concentrates its mass out where it goes faster - thus having more kinetic energy. (The top of the wheel goes the fastest of all, and the hoop puts MORE of its mass near the top, less in the middle.)
11-20
A disk (I = (1/2) M R ) rolls down a hill without slipping. At the bottom, when its center of mass moves with speed v, which formula below correctly gives its rotational 2 kinetic energy, KE(rot) = (1/2) I w ? A. M v B. 1/2 M v E. None of the above. Hint: v(cm) = w R
Answer: Since omega = v/R, then (1/2) I omega^2 = (1/2) (1/2) M R^2 (v/R)^2 = (1/4) M v^2. The R's cancelled out.
2
2
2
C. 1/4 M v
2
D. 2 M v
2
ii) What is the total kinetic energy of the rolling disk?
Answer: You'd have to add KE(translation)+ KE(rotation) = 1/2 M v^2 + 1/2 I omega^2. We just figured out the second term, adding the first gives (3/4) M v^2. (None of the above)
iii) Now consider a hoop (I = MR^2): What is its total kinetic energy when it moves with v?
Answer: Using the logic of parts i and ii above, we have KE(rot) = (1/2) MR^2* (v/R)^2 = (1/2) M v^2. Thus, KE(total) = (1/2) M v^2 + (1/2) M v^2 = M v^2. Notice that this is MORE kinetic energy than the disk... IF they have the same speed! (Like the previous concept test)
11-21
A hoop and a disk, each with the same mass M and same radius R, race down a hill. Who wins? (Assume they roll without slipping, and neglect rolling friction) A: Hoop wins B: Disk wins C: Tie!
Answer: Conservation of energy says MgH + 0 = 0 + KE(total). We just worked this out, and found that KE(total) = (1/2) Mv^2*(some number). The number depended on the shape, it was 3/2 for a disk, and 2 for a hoop. (MORE for the hoop) So, when you solve for v, you get v = Sqrt[2*g*H/(that number)] Since that number is bigger for the hoop, v is SMALLER. the hoop LOSES (despite the picture) It rolls slower. Why? Because it has more moment of inertia, more "sluggishness to rolling". Or, think of it this way - as it starts to roll, a larger fraction of its energy is going into the ROTATIONAL KE, and that leaves LESS behind for the LINEAR KE... It's storing lots of that KE up in rotational energy, and not as much in forward motion...
ii) A disk with mass m and radius r races another disk with mass M and radius R, as above. Who wins? A: Depends on M and R. B: Tie!
Answer: If you look at the argument above, that magic number in the formula for v(final) depended ONLY on the specific formula for I. The value of R cancelled out when we computed KE, and then the value of M cancelled out when we solved for v(final). Neither one matters! This is partly an old familiar story - the mass of a falling object has no influence over how long it takes to reach the ground! Neither does the radius (that's new here). What DOES matter is the distribution of matter. A disk of ANY mass or radius will always beat ANY ring of any size or radius (as long as they're uniform, of course) You'll have to think this one through on your own, it's not very intuitive!
CT11-x1
A bicycle repair-man is working on a bicycle's front wheel, which can spin freely because the bicycle is upside-down. The repair man gives the wheel a good, hard spin and goes off for a 5 minute coffee break. When he returns he finds the wheel has stopped. True(A) or False(B): Sometime while the repair man was on his break, there must have been a non-zero torque acting on the wheel.
Answer: You bet. The wheel accelerated angularly (that is, delta omega/delta t was definitely not zero), and net torque = I*alpha. So, if you have an alpha, you MUST have had a torque, at least at some point!
12-1
A bike rolls to the right with speed v. What is the speed of the TOPMOST point of the front tire? A) 0 B) v C) 2v D) None of these What is the speed of the hub (in the middle)? What is the speed of the BOTTOM-most point (touching the ground)?
Answer: Check out Fig 12-3 of your text for the answers to all three of these!
12-2
Two cylinders of the same size and mass roll down an incline. Cylinder A has most of its weight concentrated at the rim, while cylinder B has most of its weight concentrated at the center. Which reaches the bottom of the incline first? A: A B: B C: Tie...
Answer: It's just conservation of energy. The more moment of inertia you have (for a given M and R), the more of your energy will be "tied up" in rotational energy, leaving less behind for translation (i.e. what we see as "speed" of the object) So B will go faster (and win the race) because it has a smaller moment of inertia. A is more "sluggish". It has lots of moment of inertia, it takes longer to get it spinning fast. In more detail: You have a certain total energy (Mgh) stored at the start. They have the same mass, so they have the same TOTAL energy. As they drop, they convert that energy into KE. But the KE now has TWO terms, pure translation (1/2 M v^2) and pure rotation (1/2 I omega^2). From an earlier concept test, we noted that I = (#)*M*R^2, and omega = (v/R), so the rotational KE term becomes (#)*M*v^2. Thus, KE = MV^2*(1/2 + the number) If you concentrate mass near the center, the number will be smaller (less "I" for a given mass and radius). So Cylinder A has a BIGGER coefficient of MV^2. Since we have a fixed amount of total KE, the bigger the coefficient, the smaller V will have to be. So Cylinder B will have a bigger V, and hence wins the race. Notice that R is irrelevant. So is M (it cancels out with Mgh) All that matters is the DISTRIBUTION of mass. The more moment of inertia you have, for a given M and R, the slower you will move (because more of your energy is now "tied up" in rotational energy!)
12-3
A solid disk and a ring roll down an incline. The ring is slower than the disk if A: m ring = m disk B: r ring = r disk C: m ring = m disk and r ring = r disk. D: the ring is always slower, regardless of the relative values of m and r.
Answer: See the last one. All that matters is the number in front of MR^2 in the moment of inertia. For a ring, I = MR^2. For a disk, I= (1/2) MR^2. R doesn't matter (it cancels out, see previous explanation), and M doesn't matter (it cancels out too! See previous explanation). The ring will ALWAYS be slower than the disk, because a larger fraction of its energy is going into rotational energy, leaving a smaller fraction behind for translational. This is cool! We'll do some demos in class, you can try it on your own if you can find some disks and rings of any kind. (E.g. empty cans vs solid cans) Note that my arguments DO assume no friction anywhere. So don't use a soup can, because then there's internal friction and the story will get more complicated.
12-4
A disk is spinning about a fixed axis in the direction shown. What is the direction of the disks angular velocity vector w?
A) C) B) D)
E) The direction varies with time.
Answer: C is correct, it's the "right hand rule".
12-5
A disk is spinning as shown with angular velocity w. It begins to slow down. While it is slowing, what is the direction of its vector angular acceleration a? A: B: C: D: E: Some other direction.
Answer: If it's moving CW (as viewed from above), but slowing down, the angular acceleration is OPPOSING the omega, i.e. CCW, which is an UP arrow. (A)
12-6
A planet in elliptical orbit about the Sun is in the position shown.
planet y
S z
x
With the origin located at the Sun, the vector torque on the planet.. A: is zero. C: is in the x-y plane. B: points along +z. D: None of these.
Answer: Using the right hand rule, torque is pointing along +z. But we aren't going to use this for awhile in the course (if at all, we'll see how far we get), so don't worry about this one too much at the moment! I won't test you on this unless I explicitly tell you so near the end of the course.
PLEASE NOTE: THE REMAINING CONCEPT TESTS ARE ALL ABOUT ANGULAR MOMENTUM, WHICH WE ARE NOT COVERING (AT LEAST, NOT YET!) THEY ARE JUST HERE FOR YOUR REFERENCE
12-7
How does the magnitude of the angular momentum of the planet Lplanet (with the origin at the Sun) at positions A and B compare? A: LA=LB
B S A
B:...