Solutions are posted after class.
Please put 111 in the subject line when you email me. I divert the quizzes to a file in my emailer. Thanks. Email addresses aubrecht.1@osu.edu, aubrecht@physics.osu.edu, or aubrecht@mps.ohio-state.edu work equally well.
Week 10
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. A cylinder of
identical mass and radius is subject to the same angular acceleration.
a. By how much will the angular momentum of the sphere change in
10 s?
b. By how much will the angular momentum of the cylinder change in
10 s?
c. How much torque did it take applied for 10 s to change these
angular momenta, respectively?
d. What are the final respective rotational kinetic energies,
assuming both the objects began at rest?
Solution
The sphere and cylinder have different expressions for the moments of inertia, as we see on p. 260. 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.
7 March
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 between 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.
5 March
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? NOTE: You may wish to know the moment of inertia of a 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.
Week 9
2 March
A grindstone rotates at a constant angular acceleration of 0.35
rad/s2. At time t = 0, its angular velocity is
ω0 = - 4.6 rad/s and it is at θ = 0.
a. At what time after t = 0 is the angle θ relative to the
horizon (which is an angle θ of zero) equal to 10 π?
b. What is the grindstone’s rotation angle between t = 0 and t
= 32 s?
c. At what time would the grindstone (momentarily) be stopped?
Solution
a. So 10 π rad = -4.6 rad/s x t + (1/2)(0.35 rad/s2)t2, so the solution to the quadratic equation is 32 s.
b. It starts at θ = 0, goes negative because ω0 is negative (- 4.6 rad/s), but the positive angular acceleration means it stops and reverses itself to become positive. At the end, the angle is (- 4.6 rad/s)(32 s) + (1/2)(0.35 rad/s2) (32 s)2 = 32 rad.
c. We can use α = Δω/Δt. When ω = 0, we
have Δt = t (using our usual convention that t starts when the
clock is set to zero), so
t = (0 - ω0)/α = 4.6 rad/s / 0.35 rad/s2
= 13 s.
29 February--Prelab. Activity 1, p. VIII-4; Activity 2, p. VIII-5, 57 cm, 2 revolutions; 2.72 cm, 3 revolutions; 0.25 m/s2. Activity 5, p. VIII-10 & 11
27 February
Remember that midterm 2 is due at the start of class.
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?
Solution
a. 78.0 rpm is 78.0 rev/min x 2pi rad/rev x 1 min/60 s = 156.0pi/60
rad/s = 8.16 rad/s.
b. The angular speed begins at 0 and goes to 8.16 rad/s in 0.350 s.
The change in angular velocity is thus 8.16 rad/s, and angular
acceleration is change in angular velcoity divided by the time this
change requires: (8.16 rad/s)/(0.350 s) = 23.3 rad/s2.
c. The velocity is all tangential in this case. The speed is the
angular velocity times R, the distance from the center of rotation, or
(8.16 rad/s)(8.36 cm) = 68.3 cm/s = 0.683 m/s.
d. The angular displacement is the average angular velocity times
the time required to make the change (because we define the average
angular velocity by angular displacement divided by the time required
to make the change), or (8.16 rad/s + 0)/2 x 0.350 s = 1.42 rad.
e. The tangential acceleration is R x angular acceleration, or 8.36
cm x 23.3 rad/s2 = 195 cm/s2 = 1.95
m/s2.
Week 8
24 February
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.
22 February--Lab 7. Do (I really mean it) the prelab--Activity 1 with 100 g and 1.50 m/s and Activity 3 with 120 cm/s and 80 cm/s.
February 20
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 impulse does she receive? 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.
Week 7
17 February
Jamal exerts a force of F (measured using a spring
scale similar to the ones we have used) to the right on box B. Box A, to
the left of B, has a weight of 55.0 N and is connected to box B by a
very strong massless string, box B has a weight of 37.5 N, box C
(adjacent to box B on the right) has a weight of 39.3 N, and box D
(adjacent to box C on the right) has a weight of 18.5 N. No other forces
are exerted except by the table, perpendicular to the table.
a. Determine the force F with which box B must be pushed by Jamal if
the acceleration of box D is measured by a sonic ranger to be 0.100
m/s2.
The boxes move 5.00 m to the right during some time interval.
Determine how much total work has been done during that time on
b. box A.
c. box B.
d. box C.
e. box D.
Solution
a. As the boxes are attached, all accelerations are the same. By Newton’s second law, the net force is total mass times that acceleration. The mass can be found from the weights divided by g, WTotal = 55.0 N + 37.5 N + 39.3 N + 18.5 N = 150.3 N, so mTotal = 15.03 kg (using 10 m/s2).
Hence, F = 15.03 kg x 0.100 m/s2 = 1.503 N.
b. This force is mA x 0.100 m/s2 = 5.5 kg x 0.100 m/s2 = 0.55 N. Hence the work done was 0.55 N x 5.00 m = 2.75 J.
c. This force is mB x 0.100 m/s2 = 3.75 kg x 0.100 m/s2 = 0.375 N. Hence the work done was 0.375 N x 5.00 m = 1.875 J.
d. This force is mC x 0.100 m/s2 = 3.93 kg x 0.100 m/s2 = 0.393 N. Hence the work done was 0.393 N x 5.00 m = 1.965 J.
e. This force is mD x 0.100 m/s2 = 1.85 kg x 0.100 m/s2 = 0.185 N. Hence the work done was 0.185 N x 5.00 m = 0.925 J.
15 February--Lab 6 Prelab, do activities 1 and 2 before class. In Activity 1, use 8.4 m/s (p. VI-4). In Activity 2, use 20 m (p. VI-6)
13 February
This question focuses on work and energy. Remember, energy is something that can be transferred.
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. (It is only the force times distance in
the direction moved, or, equivalently the force times the direction
moved along or against the force, that does 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. Friction involves transfer of some
kinetic and potential energy to the form thermal energy.
Week 6
February 10
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.
February 8--Do lab 5 Marion replacement, handed out on Friday. If any time remains, you can continue with the extra challenge from lab 4. Gordon is in California at a meeting of physics teachers today.
February 6--Do lab 4. You will do as prelab and record in your lab manual Activity 1; Activity 2, Exercise 1; Activity 3; and Activity 4, Exercise 1 (the part on IV-13) before class Monday. There are two lab challenges: Activity 3 (which you will check in class with the others in your group before Carol comes over to check), and the extra one handed out on Friday. Gordon is in California at a meeting of physics teachers today.
Week 5
February 3
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 there is not an acceleration.
Solution
a. Average speed is distance traveled divided by the time required to
travel that distance, so the distance traveled is the circumference, 2
pi R = 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.
February 1
A Rolling Stones CD case sits atop a Kelly Clarkson CD case that sits on a Grateful Dead CD case that sits atop a DVD case holding the movie Conspiracy Theory that in turn rests on a table. In the effort to determine the force exerted on the DVD case by the table, we do a sequence of processes that involve forces. Explain the use of Newton’s First and Third Laws in the process. List each pair of forces in this setup that are connected by the Third Law by descriptive name.
Solution
First, we need to find out what forces act. Start with non-contact forces. In this case, only the gravitational force--the weight--is acting on the objects. The Rolling Stones CD case has a force exerted by the Kelly Clarkson CD case, because that is the only thing touching the Rolling Stones CD case. Two things touch the Kelly Clarkson CD case, the Rolling Stones CD case and the Grateful Dead CD case, so there are two contact forces acting. There are two contact forces for both the Grateful Dead CD case and the DVD case.
Second, we draw free body diagrams. The forces acting on each body are shown as determined above. We see the following picture:
So, from the First Law, FKC on RS = WRS, FGD on KC = WKC + FRS on KC, FCT on GD = WGD + FKC on GD,and FT on CT = WCT + FGD on CT.
The Third Law relates forces on two bodies. Here, they are paired as magnitudes (that are equal): FKC on RS = FRS on KC, FGD on KC = FKC on GD, FCT on GD = FGD on CT, and FT on CT = FCT on T. The last item is what we wish to find. Obviously, the directions of the paired forces are opposite.
So, we use the last named equality in the Third Law to express FCT on T as WCT + FGD on CT, from the First Law. We use the second-last Third Law relation and the second-last First Law relation to rewrite this as WCT + WGD + FKC on GD, etc. It is important to remember that we are finding he numerical answer this way, and the force that is the answer is numerically equal to the sum of the weights, but it is NOT EQUAL to the actual vector sum of the weights.
January 30
Explain why people ask women not to walk on gym floors in high heel shoes. Do the analysis in terms of the shoes (if you are able).
Solution
As we have seen, when there is a force exerted on a rigid object such as a floor, there is a force exerted by that rigid object in order to prevent the object from going through the surface that is perpendicular (normal) to the surface. If the object were really totally rigid, it would be impossible, as the way a surface exerts the force is by being slightly deformed, where the deformation gives rise to the force preventing the object from penetrating the surface. If a desk rests on a surface such as linoleum, we may be able to see depressions on the floor surface where the desk rested. Maybe you have seen such evidence yourself.
Imagine a block that has a square cross-section 5 cm by 5 cm and is a meter long and that weighs 30 N. If it is at rest on the surface, the floor exerts a force of 30 N on the block (by Newton’s First Law) and the block exerts a force of 30 N on the floor (by Newton’s Third Law).That is, the normal force is 30 N. If we put the 1 m by 5 cm side on the surface, the area is 1/20 square meters. If we put instead the 5 cm by 5 cm side down, the normal force is still 30 N, but the area in contact is now 1/400 square meters.
In the first orientation, the block exerts over its area a force per unit area of 30 N/(1/20 m2) = 600 N/m2, while in the second, it exerts 30 N/(1/400 m2) = 12,000 N/m2. There is a big difference--that is how the linoleum got the dents. High heel shoes have a very small heel surface area, so a large force gets concentrated into a small surface area. This can lead to great pressure on the floor surface and the possibility of permanent dents in the surface.
Week 4
27 January
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
25 January
I will distribute the take-home first midterm today. The exam will be take home; you must attend class to pick the exam up. Nonattenders will automatically achieve zero midterm grade. The exam will be collected Monday, 30 January 2011, at the start of class.
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.(Of course, it could be compressed so much that it does become
rigid, in which case we would say it exerted a normal force.)
23 January--Prepare Lab 3 prelab (Activity 1) ahead of class; we will do some of Lab 3 and some of Newton’s Second Law and gravitation.
A compact car of mass 2000 kg is being accelerated on a flat surface
with an a of 2 m/s2.
a. What force must be being applied for that to occur? (3)
b. What acceleration would an SUV of mass 3500 kg experience
if it were to be subject to an equal amount of force? (3)
c. What force would have to be applied to that SUV in order that it
experience the same acceleration as the compact car? (4)
Solution
According to the First Law of Newton, if there is a change in motion
(an acceleration), a force must have been responsible. Newton’s
Second Law tells us that the force is given by the object’s mass
multiplied by its acceleration.
a. The force is therefore (2000 kg) x (2 m/s2) = 4000 N.
b. In this case, 4000 N = (3500 kg) x acceleration, so
acceleration = 4000 N/3500 kg = 1.143 N/kg = 1.143 m/s2.
c. Because F = ma, with m = 3500 kg and a = 2 m/s2,
F must be (3500 kg) x (2 m/s2) = 7000 N.
Week 3
20 January
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).
18 January--Lab 2 - Read the lab ahead of time and do Activity 1 as a prelab.
Week 2
13 January
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). If you used 9.8 instead of 10, you got a slightly different number, 1.15 s.
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.
We could also begin with the relation in terms of the time from the
instant it was released:
0 m = 1.70 m + (11.3 m/s) x t - (5 m/s^2) x t^2.
The solutions are t = [(11.3 m/s) +/- ((11.3 m/s)2
- 4(-5 m/s^2)(1.70 m))1/2]/10 m/s^2, or -0.14 s and 2.40 s.
11 January--Lab 1 - Read the lab and do Activity 1 as a prelab.
9 January
Gordon tosses an eraser straight upward with a speed of 7.00 m/s as the eraser leaves Gordon’s hand. Ignoring air resistance, how long is the eraser in the air before it falls back into his hand? What is the maximum height from the hand the eraser travels?
Solution
In our region of the universe (on Earth’s surface), objects experience an acceleration that is constant at about 10 m/s2, downward. The acceleration means that, in every second the clock ticks, the y-velocity changes by 10 m/s, downward. That means that we can use the definition aav,y = (vy, final - v0)/(tfinal - t0), where aav,y is constant at (taking upward as positive, say) - 10 m/s2. Then vy, final = v0 + aav,yt, taking tfinal as t and t0 = 0, so by 0.7 s, the original upward velocity, 7.00 m/s, has changed to 0 m/s. In another 0.7 s the velocity will have become - 7.00 m/s in returning to the hand at the original height. It takes a total of 2 x 0.7 s = 1.4 s to accomplish the return.
To determine the height attained, we use the definition vy,av = y-displacement / (tfinal - t0) = y-displacement / t. During the first 0.7 s, the y-displacement is 3.5 m/s, the average velocity (original velocity + final velocity)/2 = (7.00 m/s + 0 m/s)/2, times the time, 0.7 s, or 3.5 m/s x 0.7 s = 2.45 m.
Obviously, over the entire time interval the average speed will be 3.5 m/s, while the average velocity will be 0 m/s. This is due to the distance traveled, 7.00 m, divided by the total time, 2 s being 3.5 m/s. while the total displacement during that time interval is zero.
Week 1
6 January
Explain the difference between average speed and average velocity.
Solution
Average speed involves distance and the time involved in attaining that distance. Average velocity involves displacement, a vector quantity, and the time required as well. So, average velocity is a vector while speed is a scalar.
aubrecht@mps.ohio-state.edu [latest revision, 7 March 2012]