Download Asim Kiani - BrainMass

1. A uniform square plate 6m on a side has a had a square piece 2m on a side cut out
of it. The center of that piece is at x=2m , y=0. The center of the square plate is at
x=y=0. Fine (a) the x coordinate and (b) the y coordinate of the center of mass of
the remaining piece
The center of mass is the average (mean) location of all the masses in a system. It’s an
important thing to know, since you can treat many problems involving solids as problems
involving point masses, located at the center of mass. There are two ways to approach a
problem like this: one is to simply go through and find the center of mass algebraically,
that is, to actually find the mean position of everything. The other way is to use symmetry
arguments to save some work.
So, let’s use the second method. If the square were whole (with the piece in place), the
center of mass is obviously at the origin (0,0). Now, we remove a piece. That piece
obviously has its center of mass on the x-axis, since it is symmetrical along y. That means
that the overall center of mass will only move along the x-axis. The only trick left is to
find out how far.
Assume that the density of the plate is a constant. That is, the mass is proportional to the
area. Then we have 36 mass units for the whole plate (6 m x 6 m). The chunk we cut out
has 4 mass units (an area of 4 m2).
The center of mass of the chunk is located at y=0, and at 1 m from either edge, again
from symmetry considerations, which makes the center of mass at x = 2 m.
Now comes the real trick: A square with a block cut out has the same center of mass as a
whole square plus a square equal to the cut-out with negative mass. Now, that’s not to say
that there actually is such a thing as negative mass – it’s just another handy symmetry
trick to make the math easier.
So, our center of mass is then the weighted average of the centers of masses of our two
squares that make up the overall whole:
CoM = [0(36) + 2(-4)]/[36-4] = -8/32 = -1/4
Therefore, the center of mass of the square with the block cut out will be at -.25 m along
x, and at y=0.
And if you don’t like that, then you can just do it the straightforward way:
CoM = [0(6x2) + 0(6x2) + (-1)(2x4)]/32 = -8/32 = -1/4.
2. A right cylindrical can with mass M , height H, and uniform density is initially
filled with soda of mass m. We punch small holes in the top and bottom to drain
the soda; we then consider the height h of the center of mass of the can and any
soda within it. What is h (a) initially and (b) when all the soda has drained (c)
What happens to h during the draining of the soda ? (d) If x is the height of the
remaining soda at any given instant, find x in terms of M , H, and m when the
center of mass reaches its lowest point
Ok, the symmetry is obvious here, so they only ask you to consider the one-dimensional
case (the height of the center of mass).
So, first off, let’s find the center of mass of the can alone (actually part (b)): It’s easy,
since the can is empty, it’s exactly halfway up. Same for the case when the can is full –
there’s a nice line of symmetry through the middle. Which is another way of looking at
the center of mass – it’s the point at which you have exactly half the mass on either side
of a symmetric object (for asymmetric objects, you have to do the weighting business).
Now, for (c): the center of mass while the soda is draining. Well, we can look at it this
way: we will have the weighted average of the center of mass of the soda can, which will
always be at H/2, weighted by the mass of the can M, with the center of mass of the soda.
It should be obvious that that will be at x/2 – halfway up the height of the soda, weighted
by the weight of the soda. As the soda drains out, the center of mass will drop. But, here’s
the funny part: the center of mass will dip below the center of the can for a while, but
when the can is nearly empty, it will rise up again so that it will be at the center (H/2), as
it should be for an empty can.
Following me?
That means that there is a point where the center of mass is lowest, hence the next part of
the question.
So, first off we need an expression for the mass of cola left in the can. If a full can of
soda has a mass of m, at a height H, then if the level drops to x the mass left is:
xm/H.
So, the CoM = [M(H/2) + (x/2)(mx/H)]/[M + xm/H] = [MH/2 + mx2/2H]/[M + xm/H]
Now you need to find the point at which the CoM is lowest, and rearrange that to get an
expression for x. Some basic calculus skills and a little perseverance will get you through
it, but I’m sad to say it won’t be pretty. Email [email protected] if you still have
trouble with this one, and they’ll put you in touch with me.
3. A shell is shot with an initial velocity vector vo of 20 m/s, at an angle of 60 degrees
with the horizontal. At the top of the trajectory, the shell explodes into two fragments of
equal mass. One fragment, whose speed immediately after the explosion is zero, falls
vertically. How far from the gun does the other fragment land, assuming that the terrain is
level and that air drag is negligible ?
Here’s what we can say: at the top of the trajectory, the vertical component of the
velocity is zero. Throughout (prior to the explosion), the horizontal component of the
velocity is 20cos60° = 10 m/s.
Due to the conservation of momentum, if one fragment falls to a dead stop, the other
must go at twice the speed. That is, at the top of the trajectory, the moving piece will
suddenly start to go at 20 m/s.
Now, all you have to do is go back to your basic equations of motion to find how far the
fragment will travel horizontally in the time it takes to go up and then down. Hint: it will
go twice as far on the way down as it did on the way up, so just find how far it goes while
heading up with an initial velocity of 20sin60° = 17.3 m/s while being slowed by gravity
at 9.8 m/s2, then multiply by 3 for the total distance (or by 2 for the distance after the
explosion).
4. Ricardo, of mass 80kg and Carmelita, who is lighter are enjoying Lake Mercen at
dusk in a 30 kg canoe. When the canoe is at rest in the placid water, they exchange seats,
which are 3.0 m apart and symmetrically located with respect to the canoe’s center.
Ricardo notices that the canoe moves 40 cm relative to a submerged log during the
exchange and calculates Carmelita’s mass, which she has not told him. What is it ?
This is another conservation of momentum problem, but it takes a little unraveling.
You see, during the exchange, Ricardo (80 kg) moves one way, while Carmelita (m)
moves the other way. You can assume that they move the same distance (3 m) in the
same amount of time. Since there is then a net difference in their momentum, it is made
up for by the motion of the canoe to compensate.
Now, we take another leap of logic: the center of mass of the whole system will stay in
the same place.
That’s really what we mean when we talk about conservation of momentum – the center
of mass of a system moves with constant velocity unless a net force is applied. Here,
we’re assuming that the lake is frictionless, so there is no net force to move the canoe.
Thus, the overall center of mass must stay in the same place. Initially, let’s say that we
had:
So, when Carmelita and Ricardo switch positions, the canoe must move .4 m (40 cm) to
keep the center of mass in the same place. Knowing this, we can find Carmelita’s mass!
CoM = (-1.5)(m) + (-.4)(30) + 1.5(80) = 0 (assume the center of mass is at the origin).
After the switch:
CoM = (+1.5)(m) + .4(30) – 1.5(80) = 0
So, solving for m = 72 kg is easy!
5. The last stage of a rocket, which is traveling at a speed of 7600 m/s, consists of two
parts that are clamped together: a rocket case with a mass of 290.0 kg and a payload
capsule with a mass of 150.0 kg. When the clamp is released, a compressed spring causes
the two parts to separate with a relative speed of 910.0 m/s. What are the speeds of (a) the
rocket case and (b) the payload after they have separated? Assume that all velocities are
along the same line. Fine the total kinetic energy of the two parts (c) before and (d) after
they separate; account any difference.
Ok, imagine that the rocket is sitting in space not moving. The center of mass has 0
velocity, so we must have conservation of momentum:
m1v1 + m2v2 = 0
And we also know that the relative velocity is 910 m/s, so:
v1 - v2 = 910 m/s
Now that’s a system of two equations with two unknowns (the masses are given in the
problem). You should be able to solve for the velocities. Then, keep in mind that the
center of mass is actually traveling with a constant velocity of 7600 m/s, so add that value
to each of your other velocities when finding the kinetic energy.
Also, assume that the payload capsule is the one with the positive extra velocity (the
point is to shoot it further, at the cost of slowing down the case). You should find that the
kinetic energy after the release is higher than the kinetic energy of the center of mass –
that will be due to the release of the stored energy in the spring.
6. A vessel at rest explodes, breaking into three pieces. Two pieces having equal mass fly
off perpendicular to one another with the same speed of 30 m/s. The third piece has three
times the mass of each other piece. What are the magnitude and direction of its velocity
immediately after the explosion ?
Start by finding a way to treat the two pieces as one – find a line of symmetry where their
velocity in one direction cancels out. Since they have the same mass and speed, it’s
obviously the midline between them.
So, they each move at 45° relative to this midline. Each contributes 21.2 m/s along that
axis. So the overall momentum of the two identical bits is 2m(21.2).
Obviously then, the third piece will travel along this axis in the opposite direction. For
conservation of momentum, we have:
3m(v) = 2m(21.2)
v = 2(21.2)/3 = 14.1 m/s as the velocity of that piece.
7. A 6100 kg rocket is set for vertical firing from the ground. If the exhaust speed is 1200
m/s, how much gas must be ejected each second if the thrust (a) is to equal the magnitude
of the gravitational force on the rocket and (b) is to give the rocket an initial upward
acceleration of 21 m/s ?
Ignoring the mass loss due to the gas shooting out the back, the increase in speed of the
rocket is due to the conservation of momentum:
F = dp/dt which simplifies to F = ma
That is, force is the rate of change of momentum.
So, if we have some mass of gas M coming out every second, then the momentum of the
rocket will change by 1200M every second. Thus:
F = 1200M.
In order for the force to compensate for that of gravity, mg, we have:
mg = 1200M
m is the mass of the rocket, 6100 kg, g = 9.8 m/s2. So:
M = (6100 kg)(9.8 m/s2)/1200 m/s = 49.82 kg/s.
Since this is much smaller than the mass of the rocket, we were justified in assuming that
the overall mass remains roughly constant (at least for the first second or two).
And, in order to give the rocket an initial acceleration of 21 m/s/s, you must add an
additional force of:
F = 6100(21) = 1200M
M = 106.75 kg/s.
Since you need to also overcome gravity at the same time, the overall mass of gas that
must be ejected per second to give the rocket an acceleration of 21 m/s2 is 156.57 kg/s.
8. Resistance to the motion of an automobile consists of road friction, which is almost
independent of speed and air drag which is proportional to speed-squared. For a certain
car with a weight of 12,000 N, the total resistant force F is given by F = 300 + 1.8 v^2 ,
where F is in Newtons and v is in meters per second. Calculate the power (horsepower )
required to accelerate the car at 0.92 m/s^2 when the speed is 80 km/h
This is relatively straightforward: To accelerate you need to overcome both the friction
and air resistance, as well as provide force to change your velocity. So:
Ftotal = ma + 300 + 1.8v2.
First, convert v into m/s:
80 km/h = 80000 m/h = 22.22 m/s
And weight into mass:
12 000 N = mg
m = 1224 kg
Thus, F = 1224(.92) + 300 +1.8(22.22)2 = 2315 N.
Finally, convert to horsepower:
1 Horsepower = 735.499 W
And 1 W = 1 Nm/s
So, we see that we’re missing one step here: the conversion from force to work and
power. So, in one second the car will go:
D = v0 + .5at2 = 22.22 + .5(.92)(1) = 22.68 m.
The power is then: Fd/t = (2315 N)(22.68m)/1s = 52504 W
Which is 71.4 horsepower.
9. An electron ( mass m1 = 9.11 X 10^ -31 kg ) and a proton (mass m2 = 1.67 X 10^ -27
kg ) attract each other via an electrical force. Suppose that and electron and proton are
released from rest with and initial separation d = 3.0 X 10 ^-6 m When their separation
has decreased to 1.0 X 10^-6 m , what is the ratio of (a) the electrons linear momentum
to the protons linear momentum (b) the electrons speed to the protons speed and (c) the
electrons kinetic energy compared to the protons kinetic energy (d) as the separation
continues to decrease do you values from (a) thru (c) decrease or remain the same ?
Now, the force on both should be the same, so their relative momentums should be the
same (remember, F = dp/dt), at least as far as the magnitude is concerned (they will have
opposite directions). But, since the electron is so much lighter than the proton, it should
have a much greater velocity, and that ratio should be the same as the (inverse) ratio of
their masses.
You should expect that these two values, at least, will remain the same no matter what the
separation is.
The kinetic energy will be a little trickier.
K = ½ m v2
Now, we have F = ma as well, but the acceleration for each one will be different. The
forces are the same, though, so:
meae = mpap
(electric forces depend only on charge and not on mass)
Thus me/mp = ap/ae – and the velocities are directly related to the acceleration:
v = at
So the velocities are inversely proportional to the ratio of the masses, which makes sense.
Ke  me(ve)2  me(ae)2
Kp  mp(ap)2
So the ratio of the kinetic energies is:
 
me a e2
m p a 2p
ap = meae/mp so:
mp
me ae2
- that is, inversely proportional to the ratio of the masses. It

(1 / m p )me2 ae2 me
is also independent of time.
Ke/Kp =