Please put 111 in the subject line when you email me. I divert the quizzes to a file in my email. Thanks. Email addresses aubrecht.1@osu.edu or aubrecht@mps.ohio-state.edu work equally well.
1 June
A solid sphere and a solid cylinder each have an identical radius of 10.0 cm and moment of inertia of 1.00 kg m2, but of course the two objects are composed of materials of differing masses. They are placed on a ramp angled at 30° to the horizontal and released from rest at the same distance from the bottom of the ramp. Which object has the greater rotational kinetic energy at the bottom of the ramp? Which has a greater total kinetic energy? (Total kinetic energy is the rotational kinetic energy about the center of mass plus the translational kinetic energy of the center of mass.) Explain.
Solution
The sphere and cylinder have different expressions for the moments of inertia, as we see on p. 262. The question is what happens. The objects both roll down the plane. They roll because there are torques acting on the objects. Drawing a free body diagram shows that the normal force and the weight contribute no torque, using the definition for torque. Only the friction force exerts a torque: fR = I(alpha), where I is the moment of inertia, alpha the angular acceleration, f the frictional force, and R is the radius.
The free body diagram, if we choose coordinates along the plane, leads (via Newton’s second law) to ma = W sin(theta) - f. Putting the two together and using a = R(alpha), we find
ma = mg sin(theta) - Ia/R2, or
a(m + I/R2) = g sin(theta), or
a = g sin(theta)/[1 + I/mR2].
For the sphere, I/mR2 = 2/5, while for the cylinder, I/mR2 = 1/2. Therefore, the sphere’s acceleration is 5/7 g sin(theta), while the cylinder’s is 2/3 g sin(theta). The sphere will be going faster at the bottom:
vsphere > vcylinder,
where vcylinder is the CM speed of the cylinder and vsphere is the CM speed of the sphere.
The respective rotational kinetic energies are Ksphere = 1/2 I(omega)2 = 1/2 x 5/7 x m [R(omega)]2 = 5/14 mvsphere2 and Kcylinder = 1/2 I(omega)2 = 1/2 x 1/2 x m [R(omega)]2 = 1/4 mvcylinder2. Clearly, the sphere has a greater rotational kinetic energy. Also, we could use that they have equal moments of inertia.
The total energy per unit mass, by conservation of energy, must be exactly the same for the two, because mgh is the same expression for both original potential energies (they started at the same height above the ground). The mass of the cylinder is larger than the mass of the sphere (10/7). Thus, the cylinder has the greatest total kinetic energy.
28 May
A sphere of mass 10.0 kg and radius 0.45 m is subject to an angular acceleration about its center of 2.00 rad/s2. What torque must have been applied to the sphere to have caused this rate of change in angular velocity?
Solution
There is a connection beyween torque and angular acceleration--it is
torque = (moment of inertia)(angular acceleration).
On p. 262, we find that a uniform sphere has a moment of inertia for rotation about an axis through its center of I = 2/5 MR2. Hence,
torque = (2/5 [10 kg][0.45 m]2)(2.00 rad/s2) = (0.81 kg m2)(2.00 rad/s2) = 1.62 kg m2/s2 = 1.62 N m.
26 May--no class. Have a good Memorial Day
23 May--also do the prelab for the torque lab, which will actually be done in class today.
A rope hangs from a cylinder over which it is wound, of radius 0.20 m and mass 40.0 kg, and is attached to a hanging mass of weight 210.0 N. What torque does the hanging mass exert on the cylinder?
Solution
This problem involves both force and torque. The hanging mass is accelerated. Therefore, we must write Newton’s Second Law as mg - T = ma, where T is the tension in the rope and m is the mass having weight mg = 210.0 N and we have taken a to be positive downward. The torque on the cylinder is thus TR, where the force T makes a right angle to the edge of the cylinder of radius R. Clearly, T < mg = 210.0 N. The applied torque is consequently less than 42.0 N m.
This above is the answer I graded for the quiz. However, we can construct a more complete answer.
Referring to Table 9.1 on p. 262, we see that a cylinder of radius R and mass M has a moment of inertia I = 1/2 MR2. You might recall that I introduced this quantity I for the rotational analog of inertia last week. In our case, I = 0.5(40.0 kg)(0.20 m)2 = 0.80 kg m2. Therefore, torque is TR = I(angular acceleration) = Iatangential/R, or atangential = TR2/I. Now the tangential acceleration of the edge of the cylinder is exactly equal to the acceleration of the hanging mass, so
mg - T = mTR2/I, or
T(1 + mR2/I) = mg, or
T = mg/[1 + mR2/I] = mg/[1 + mgR2/Ig] = 210.0 N/[1 + (210.0 N)(0.20 m)2/0.80 kg m2 x g]
= 210.0 N/[1 + (8.40 kg m3/s2)/0.80 kg m2 x g] = 210.0 N/[1 + (10.5 m/s2)/g].
Taking g = 10 m/s2, we find
T = 210.0 N/2.05 = 102 N.
The torque on the cylinder is thus 102 N x 0.20 m = 20.5 N m.
21 May--no quiz, prelab due at beginning of class.
19 May without (ugh) handling itiser?
An old phonograph record accelerates from rest to a final angular velocity of 78.0 rpm (revolutions per minute) in 0.350 s. The record has a radius of 15.0 cm.
a. Express the final angular velocity in units of rad/s.
b. What is the angular acceleration of the record during this time?
c. A dust speck lies 8.36 cm from the center of the record. What is the final speed of the speck?
d. Through what total angle does the record rotate as it accelerates?
e. What is the tangential acceleration of the speck during this time?
16 May
In the elephant exercise yard at the local zoo, there are several dung piles. A physics student estimates that a pile at a point 3.20 m west and 7.45 m north of the southeast corner has a mass of 100 kg. The student finds another pile of estimated mass 75.0 kg at 9.00 m west and 2.30 m north of the same point, and a last pile of estimated mass 45.0 kg at 5.00 m west and 3.00 m north. What would the student find as the coordinates of the center of mass? How accurate do you think the student’s estimates are? Explain your reasoning.
Hint: In each direction apply what you know about finding center of mass independently.
Solution
We have to set up some system of coordinates. The center of mass in x is found from
xCM = m1 x1 + m2 x2 + m3 x3 + ... /(sum of masses).
Likewise, the center of mass in y is found from
yCM = m1 y1 + m2 y2 + m3 y3 + ... /(sum of masses).
For the center of mass in z,
zCM = m1 z1 + m2 z2 + m3 z3 + ... /(sum of masses).
In this case, in our coordinate system the x would be the direction east and the y would be north. The origin is in the southeast corner. (You may, of course, make your own should you wish, but I have given you a perfectly good coordinate system.)
xCM = [(100.0 kg) (-3.20 m) + (75.0 kg) (-9.00 m) + (45.0 kg) (-5,00 m) ]/[100.0 kg + 75.0 kg + 45.0 kg] = (-320 kg m - 675 kg m - 224 kg m)/220.0 kg = - 1219 kg m/220.0 kg = - 5.541 m.
yCM = [(100.0 kg) (7.45 m) + (75.0 kg) (2.30 m) + (45.0 kg) (3.00 m) ]/[100.0 kg + 75.0 kg + 45.0 kg] = (745 kg m + 172.5 kg m + 135 kg m)/220.0 kg = 1052.5 kg m/220.0 kg = 4.784 m.
The CM coordinate is - 5.541 m, 4.784 m.
Who could know exactly where the center of the dung pile is without (ugh) handling it? They’re probably good to within 20 kg and (maybe) 10 cm. We certainly aren’t justified in keeping so many decimal places.
14 May--no quiz, prelab due at beginning of class.
12 May
We will have a take-home midterm handed out at the end of class today. As a result, the class will meet and tackle impulse and momentum and there is a quiz for today.
You are a passenger on a jet plane flying at constant velocity. You get up from your seat to walk forward to a restroom. Obviously, your momentum increases. Explain clearly what, if anything, happens to the momentum of the plane.
9 May
A 57.0 kg diver stands on the edge of a pool. She pushes off horizontally with an average force on the pool’s edge of 750 N for 0.20 s. What horizontal speed would you expect her to have? Explain your reasoning.
Solution
The impulse is (750 N)(0.20 s) = 150 kg m/s.
The impulse is equal to the change in momentum. Her final momentum component in the horizontal direction is (57.0 kg)vhorizontal, which is equal to 150 kg m/s. Therefore,
vhorizontal = (150 kg m/s) /(57.0 kg) = 2.63 m/s.
7 May--Do lab 3 today and finish lab 5
5 May
a. How much work is done against gravity by a 70.0-kg person climbing up a 2.00-m high staircase at constant velocity of 1.00 m/s? The person moves 4.00 m to the right as well as 2.00 m up while climbing the stairs. Explain.
b. What is the gravitational potential energy of a 32.4-kg object at rest on a plane inclined at an angle of 30° to the horizontal a distance 10.00 m up the plane? How do you know this?
c. Assuming the plane of (b) is frictionless, if the object is released, what kinetic energy will it have when it reaches the bottom of the plane? How do you know this?
Solution
a. The person is lifting herself up a distance of 2.00 m in the direction parallel to the force (upward, in the opposite direction to the weight) she is exerting. The horizontal distance is perpendicular to the force and so does no work. Only the vertical distance is connected to the work. The work done is her weight (700 N) times the distance upward she has moved (2.00 m), so work done = (700 N) x (2.00 m) = 1400 N m = 1400 J.
b. The gravitational potential energy is determined by the weight and the vertical distance above the reference level (here the ground). This is the vertical component of the 10 m plank, or 10.00 m x sin 30° = 5.00 m. The gravitational potential energy is then (32.4 kg) x g x (5.00 m) = 1620 N m = 1620 J (using 10 m/s2).
c. It must have all 1620 J of energy it began with; the energy was transformed from potential to kinetic energy. The total energy, as long as no friction acts, is fixed.
2 May
Gordon has an antique turntable that plays vinyl disks at 33 1/3 revolutions per minute.
a. What is the speed (in cm/s) at the outer edge of a record if that edge is 15 cm from the center of rotation?
b. While the turntable runs without the stylus or record (it is the same size as a record), a penny is carefully placed on the turntable and travels with it remaining in place. Explain whether or not there is any centripetal force on the penny.
c. Where on the turntable is the frictional force on the penny the greatest? Explain.
Solution
a. The speed is distance traveled divided by the time required to travel that distance, so the distance traveled is the circumference, 2 x 3.14159265 x 15.0 cm = 94.2 cm and the time is found from time for one revolution = 1 minute/(33 1/3) = 0.03 minute = 1.8 s. The speed is 94.2 cm / 1.8 s = 52.4 cm/s.
b. Because the penny does not move with respect to the turntable, it is being carried around by the turntable through static friction. This force is centripetal, that is, it is the force causing the penny to move in the circle.
c. Because v gets bigger as we move from the center of the turntable (it is proportional to R, the distance from the center) and because the centripetal acceleration is v2/R, the centripetal acceleration is greatest at the periphery.
30 April--no quiz, do Lab 5 preclass and bring to class for checkoff at start of class
28 April
A racecar goes completely around a circular track of radius 4.0 km in 600 s.
a. What is its average speed?
b. What is its average velocity?
c. Is the racecar accelerated? If so, how big is the acceleration? If not, explain why not.
Solution
a. Average speed is distance traveled divided by the time required to travel that distance, so the distance traveled is the circumference, 2 x 3.14159265 x 4.0 km = 25.1 km and the time is given. The average speed is given either in km/h or m/s. The time 600s is one-sixth of an hour, so vav = 25.1 km/(1/6 h) = 150.1 km/h. Alternatively, the distance is 25,133 m and so vav = 25,133 m / 600 s = 41.9 m/s.
b. Because the racecar returned to where it started, the displacement is zero, and thus the average velocity (displacement divided by time required to make that displacement) is zero.
c. The velocity is changing because its direction is changing, This means that there must be an acceleration. The centripetal acceleration is v2/R, so (41.9 m/s)2/ 4000 m = 0.44 m/s2.
25 April
A chandelier hangs suspended from a ceiling by two massless ropes, which connect to one rope fastened to the chandelier. One rope makes an angle of 30° below the ceiling to the right, the other an angle of 45° below the ceiling to the left. Find the forces on the ropes if the massless rope tied directly to the chandelier exerts a force of 100.0 N directly downward at the point where the three ropes come together.
Solution
This is an equilibrium problem. Let us call the ropes’ forces exerted on the knot Tchandelier, T30, and T45, respectively. There are only three forces because we are taking the ropes to be (essentially) massless, and three ropes are in contact with the rope knot. By Newton’s First Law, recognizing that the chandelier is at rest and remains at rest, the force of the rope on the chandelier plus the weight of the chandelier add to zero. Our force Tchandelier is related to the force of the rope on the chandelier by Newton’s Third Law. They are equal in magnitude and opposite in direction. That is, the force Tchandelier acts downward on the knot because the force of the rope on the chandelier must act upward to add to the chandelier’s weight and make zero; the opposite direction to this upward force must be downward.
Let us choose our arbitrary coordinate system to have perpendicular axes up and to the right. Because I haven’t said which force is right or left, you can choose either one. I will choose T30 to be up and toward the right from the knot. The components of the vector are then T30x = T30 cos 30° and T30y = T30 sin 30°. Then, recognizing that T45 points up and toward the left from the knot, T45x = -T45 cos 45° and T45y = T45 sin 45°.
The sum of the forces in the x-direction is 0 = T30 cos 30° - T45 cos 45°. This means that T30 cos 30° = T45 cos 45°, or that 0.866T30 = 0.707T45, or that T30 = 0.707T45/0.866.
The sum of the forces in the y-direction is also 0, but this time 0 = T30 sin 30° + T45 sin 45° - 100.0 N, or 0.500T30 + 0.707T45 = 100.0 N. Substituting from above, 0.500 x 0.707T45/0.866 + 0.707T45 = 100.0 N.
We may gather terms as 0.707T45 x (0.500/0.866 + 1) = 100.0 N, or (1.577) x 0.707T45 = 100.0 N, or 1.12T45 = 100.0 N, or T45 = 89.7 N.
Using the result above that T30 = 0.707T45/0.866, we find T30 = 73.2 N.
23 April--no quiz, do Lab 3 preclass and bring to class for checkoff at start of class
21 April
A student sees that a rope connected to a pulley and another block supports block B, of weight 5.0 N, which hangs unmoving in midair as a result. Block A, which weighs 80 N, rests on a rough table (that designation “rough” means that there is friction between the table and the block A). Block A is connected by an ideal rope and pulley to block B.
a. Find all the forces acting on each of the two blocks, A and B.
b. Explain what the agents are for each of these forces.
c. List the Newton’s Third Law forces corresponding to each of the forces in (b).
Solution
a. On A: weight of A, force of table (perpendicular) on A--the normal force, frictional force on A, force of rope on A.
On B: weight of B, force of rope on B.
b. These are the only forces; weight is noncontact in both cases, on B only the rope is in contact, while on A, both the table and the rope are in contact. The agent for weight is Earth; for the rope on A block B and for the rope on B block A, for the normal force, the table is the agent, and for the frictional force, the table is agent.
c. Newton’s Third Law forces: Weight of A, force of A on Earth; weight of B, force of B on Earth; rope on A, force of A on the rope; rope on B, force of B on the rope; force of table on the block A (normal force), force of block A on the table.
Note that the rope merely transmits the force between bodies. It acts to the right (say) on A and up on B. The rope puts its forces on the two blocks in different directions because of the pulley
18 April--no quiz, Midterm 1 today.
16 April--no quiz, do Lab 4 preclass and bring to class for checkoff at start of class
14 April
In mathematics, normal means perpendicular to. Explain why the force exerted by rigid bodies on other bodies can sometimes be called a “normal” force.
Under what circumstances indicated below is the force referred to a normal force?
a. A hand exerts a force on an apple.
b. A flat aluminum track exerts a force on a cart at rest on its surface.
c. A tilted aluminum track exerts a force on a cart at rest on its surface.
d. An aquarium is packed in spongy material for shipping through the mail. The spongy material exerts a force on the aquarium.
Solution
The force exerted by a rigid surface can be perpendicular to the surface (preventing the object in contact from falling through) or parallel (as in the case of the frictional force). The perpendicular part is normal in the mathematical sense, and can be called normal with clear conscience.
a. The hand is not rigid, so the force cannot be normal.
b. The track is rigid and horizontal, so the force of track on cart is normal.
c. The track is rigid and not horizontal, so the force of track on cart is partly normal (preventing the cart in contact from falling through the track) and partly parallel to the track.
d. The spongy material is not rigid, so the force is not a normal force.
11 April
A horse is pulling a cart faster and faster. A Physics 111 student says that Newton’s Third Law says that the force of the horse on the cart is the same as the force of the cart on the horse, but in the opposite direction. The horse is supposed to pull the cart, but the net force must be zero, so the horse cannot ever pull a cart. What is wrong with this logic?
Solution
The student’s mistake is to confuse Newton’s Second and Third laws. The Second Law says that the sum of the forces acting ON a body determines the acceleration (the change in motion) of the body, so the forces acting ON the cart detertmine the cart’s change in motion. The Third Law says that the force on the horse from the cart is equal and opposite to the force on the cart from the horse. One force acts on the cart, the other on the horse. These forces can be equal and opposite without affecting the fact that both horse and cart accelerate together (they must have the same acceleration because the cart is hitched to the horse)
9 April
A book rests in your hand, which is not moving. Your hand exerts a force of 30 N on the book. What is the size of any other force(s) exerted on the book?
Solution
The book is not moving, which means that it is not CHANGING ITS MOTION. According to Newton’s First Law, we know that if motion is not changing, all forces balance, i.e., they add to zero. The hand is in contact with the book; nothing else is. Therefore, the other force must be a noncontact force, the weight (caused by Earth). This must be 30 N also, and in the direction opposite that of the hand (which hand force, presumably, is upward--we know the weight is downward).
7 April
A ball is thrown straight upward with a speed of 11.3 m/s from an initial height of 1.70 m.
a. What is the speed of the ball 0.500 s later?
b. When does the ball reverse direction?
c. Where is the ball when it reverses direction?
d. When does the ball return to its original level?
e. When does the ball hit the ground?
Solution
a. The ball’s upward velocity decreases by 10 m/s for each second it’s in the air. After half a second, its upward velocity has decreased by 5 m/s. It’s moving upward at 6.3 m/s. Its speed is therefore 6.3 m/s.
b. It reverses direction when the vertical velocity is zero. This occurs at 1.13 s from the start because then the change in velocity from the acceleration due to gravity, 10 meters per second squared downward, is 11.3 m/s downward (or -11.3 m/s upward).
c. It began at 11.3 m/s upward and is (at this point) at velocity zero. The average velocity is the sum of these two divided by 2: 5.65 m/s. We know that delta-y, the change in vertical position, is the average velocity times delta-t (which here is just t = 1.13 s itself). Hence, the ball has traveled upward a distance of 5.65 m/s x 1.13 s = 6.38 m.
d. Because the acceleration is constant, it takes exactly as long to return to the initial position as it did to go from that position to 6.38 m up. This occurs at 2.26 s after we start.
e. This depends on how far above the ground my hands were when I let it go.. For example, if my hands were 1 m above the ground when I let go, the ball must travel downward that extra meter. This can be calculated. We said it was 1.70 m above the ground, so presumably you would use that. You can calculate the time it takes to go that extra distance by -1.70 m (the ball goes another 1.70 m downward) = average velocity times time. The average velocity is the velocity when it is 1.70 m above the ground plus the final velocity when it hits the ground, all divided by 2, or (-11.3 m/s + [-11.3 m/s - 10 m/s^2 x t])/2 = -11.3 m/s - (5 m/s^2) x t. This times time elapsed after reaching that point must be -1.70 m, as noted above, so
-1.70 m = (-11.3 m/s) x t - (5 m/s^2) x t^2.
This may be solved for t. We find either t is 0.14 s or it is -2.40 s. The latter number corresponds to a time before it was thrown and is nonsense. The ball falls another 0.14 s before it hits the turf, at 2.26 s + 0.14 s = 2.40 s after the start.
4 April
A pig waddles from its original position 5.00 m eastward in 5.00 s; it then waddles 4.00 m northward in 4.00 s; finally, the pig waddles 36.9° west of south 5.00 m in 5.00 s. Determine the net distance traveled; the net displacement; the average speed and average velocity for each segment of the trip; and the net average speed and average velocity for the trip.
Solution
distance:
add; 5 m + 4 m + 5 m = 14 m
displacement
We need to find the components of the motions in terms of east and north. The first is 5.00 m E. The second is 4.00 m N. The third is 5.00 m * cos 36.9° S + 5.00 m * sin 36.9° W = 4.00 m S + 3.00 m W.
Adding these, we find 0 m N + 2.00 m E.
net average speed: distance/time to travel that distance = 14.00 m/14.00 s = 1.00 m/s
net average velocity: displacement / time required to achieve that displacement = 2.00 m E / 14.00 s = 0.14 m/s E.
speed on segments:
1. 5.00 m/5.00 s = 1.00 m/s
2. 4.00 m/4.00 s = 1.00 m/s
3. 5.00 m/5.00 s = 1.00 m/s
velocity on segments:
1. 3.00 m E/5.00 s = 1.00 m/s E
2. 4.00 m N/4.00 s = 1.00 m/s N
3. 4.00 m S/5.00 s + 3.00 m W/5.00 s = 0.80 m/s S + 0.60 m/s W
2 April
In class, we found the displacement and distance for a problem just like this. Now, we look at speed.
A bicyclist races up a hill on her bike at a constant speed of 5.0 m/s. The road on the hill curves around, and the total road distance is 5.0 km. The instant she reaches the top, she turns the bike and speeds down the hill at a constant speed of 20.0 m/s. What is the bicyclist’s average speed for the entire trip as she returns to her starting point?
Solution
You must take into account the time each segment takes and apply the definition for average speed: distance traveled divided by the time required to travel that distance. The first part of the trip takes 1000 s = 5000 m/5.0 m/s; the second takes 250 s = 5000 m/20 m/s. The average speed is total distance traveled divided by the time required to travel that distance, or 10,000 m/1250 s = 8.000 m/s.
31 March--no quiz, do Lab 2 preclass and bring to class for checkoff at start of class
28 March--no quiz, Bruce Lautzenheiser, OSUM math tutor will give math pointers
26 March
What are the symbols for the International System units for distance, time, and speed?
Solution
The units are, for length (dimension [L]), the meter; for time (dimension [T]), the second; and for speed (dimension [L/T]), the meter per second (not meter/second, this is not accepted usage). Neither units nor dimensions are symbols. Symbols are not abbreviations, but letters that stand in for the unit. The BIPM (International Bureau of Weights and Measures--acronym in French) sets these for the International System (SI). They are, respectively, m for meters, s for seconds, and m/s for meters per second. The abbreviation of second may be sec, but the symbol is s; likewise, the abbreviation of hour is hr, but the symbol is h.
aubrecht@mps.ohio-state.edu [latest revision, 1 June 2008]