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course Phy 121
12/10 10 pmThis is my practice exam for the final.
Thanks,
Jeremy" "
Problem Number 1
A circular disk initially rotating at 2 radians/second accelerates uniformly to 8 radians/second while rotating through an angular displacement of 17.99 radians.
How long does the acceleration require and what is the angular acceleration?
`omegaf^2 =1omega0^2 + 2`a * `theta
`a = (`omegaf^2 - `omega0^2)/ (2* `theta)
`a = [(8 radians/s)^2 - (2radians/s)^2)/ (2*17.99 radians)
`a = 60 radians^2/sec^2 / 36 radians
`a = 1.67 radians/sec^2 (approx)
`dt= `domega/`a
`dt = (6 radians/s )/ 1.67radians/sec^2
`dt = 3.59 seconds (approx)
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Good.
You could also have reasoned this out:
ave ang vel is 5 rad / sec, so it takes 17.99 rad / (5 rad/sec) = 3.6 sec to accelerated; acceleration is therefore `dOmega / `dt = 6 rad / s / (3.6 s0 = 1.67 rad / sec.
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Problem Number 2
Using proportionality, the acceleration of gravity at the surface of the Earth and the fact that the radius of the Earth is approximately 6400 km, use proportionality to find:
The field strength at twice the radius of the Earth from its center.
The strength at 4 times the radius of the Earth from its center.
The strength at a distance of 27500 kilometers from its center.
The distance from Earth's center where the gravitational field of the Earth is half its value at the surface.
At twice the radius, acceleration of gravity = 9.8m/s^2 * 2^-2 = 2.45 m/s^2
At 4 times, accel of gravity = 9.8m/s^2 * 4^-2 = .6125 m/s^2
At 27500km accel of gravity = 9.8m/s^2 * (6400km/27500km)^2 = .531 m/s^2 (approx)
4.9m/s^2 = 9.8m/s^2 * (6400km/R)^2
.5 = (6400km/R)^2
`sqrt(.5) = 6400km/R
.707 = 6400km/R
R= 6400km/.707
R= 9052 km
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Problem Number 3
What are x and y the components of the displacement vector obtained when we add the two following displacement vectors:
vector A, with x and y components -7.6 meters and -2 meters, and
vector B whose x and y components are 4.3 meters and 2.2 meters?
What are the magnitude and angle of the resultant vector?
X component is -7.6 m + 4.3 m = -3.3m
Y component is -2 m + 2.2 m = .2m
Mangitude = `sqrt(x^2 + y^2) = `sqrt(10.93m^2) = 3.31 m (approx)
Angle = Tan^-1 (.2m/-3.3m) -3.47 degrees or -3.47 deg + 180 deg.
Since x is negative and y is positive angle = 177 degrees (approx)
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Problem Number 4
If velocity increases by 8 meters per second per second, by how many meters per second does it increase in 4 seconds?
If velocity is initially 8 meters per second, then what is the velocity after 4 seconds?
What will be the average velocity and the distance traveled during the 4 seconds?
8m/s^2 * 4 sec = 32m/s
The velocity increases 32m/s
After 4 seconds the velocity is 8m/s + 32m/s. Therefor it is 40m/s
Average velocity is (v0+ vf)/2 = (8m/s + 40m/s )/2 = 24m/s
`ds= vAve* `dt
`ds = 24m/s * 4 sec = 96 meters
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Problem Number 5
What will be its velocity 6.2 seconds after being released from rest if an object accelerates freely under the influence of the gravitational field at the surface of Earth?
6.2 seconds * 9.8 m/s^2 = 60.8 m/s (approx)
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Problem Number 6
A rubber band exerts forces of 46.27201 Newtons, 76.21842 Newtons and 102.0577 Newtons when stretched by 5 cm, 10 cm and 15 cm.
If the rubber band is stretched a distance of 8.75 cm and used to accelerate a mass of .133 kg, what velocity will the mass attain if there is no friction acting on the mass?
If the coefficient of friction between the mass and the horizontal surface over which it travels is .04, then how far will the object travel before coming to rest?
If the mass travels without friction up a ramp inclined at angle 5.2 degrees with horizontal, how far up the incline (i.e., in the vertical direction) will it travel? How far along the incline will it travel?
Using a graph we estimate the Force to be .5*b*altitude = .5 * .0875m * 65Newtons = 2.84 Newton meters or Joules
Without friction all of the PE goes into the KE of the mass
2.84 J = KE = m*v^2
v= `sqrt (2.84J/.133kg)
v = 21.33 m/s
With Friction: `ds= `dW/fFriction
fFriction = .04 * .133kg * 9.8m/s^2 = .0521 N
`ds = 2.84 N m / .0521 N = 54.5 meters
Without friction, up the incline:
`dPE = weight * `ds = 1.30N *`ds
2.84 N meters = 1.30N * `ds
`ds = 2.84N m /1.30 N = 2.18 meters
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If `ds_parallel is the distance traveled along the incline, then our displacement along the incline is a vector of magnitude `ds_parallel at angle 5.2 degrees. Its vertical component is therefore `ds_parallel * sin(5.2 degrees).
The vertical distance is the 2.18 meters you've calculated.
So
`ds_parallel * sin(5.2 deg) = 2.18 meters
and
`ds_parallel = 2.18 m / sin(5.2 deg) = 2.18 m / (0.09) = 24 meters, very approximately.
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Problem Number 7
What vector of magnitude 9.19 must be added to the force vector A = < -6.77 Newtons, 5.84 Newtons> in order to obtain a vertical vector R? Answer by giving the magnitude and angle of the vector to be added.
(Note on notation: stands for a vector whose x component is u and whose y component is v.)
For a vertical vector, x must equal 0 so x component of added vector must be 6.77N
(6.77N)^2 + y^2 = (9.19N)^2
y^2 = 38.6N^2
y = +-6.21N
if y = +6.21 angle = tan^-1 (6.21/6.77) = 42.5 deg. This checks out because x and y are positive (no need to add 180 deg)
if y = -6.21 angle = angle = tan^-1 (-6.21/6.77) = -42.5 deg. This checks out because x is positive and y is negative (no need to add 180 deg)
Magnitude is given as 9.19 (assuming Newtons)
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Problem Number 8
An object of unknown mass moving at 5 m/s collides with another object.
By observing the behavior of the second object we conclude that the first object must have had momentum 15 kg m/s.
What is the mass of the first object?
p=m*v
m=p/v
m=(15kg m/s)/5m/s
m = 5kg
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Problem Number 9
An object whose mass is 4 kilograms is accelerated from rest to 25 meters/second over an unspecified time interval. It continues accelerating at the same average rate for an equal time interval. The result is that its previous velocity is doubled.
What will be the kinetic energy after the first time interval?
What will be the kinetic energy after the second time interval?
How many times as much kinetic energy does the object have after the second time interval, compared with the kinetic energy at the end of the first time interval?
KE =.5m * `dv^2
KE = .5 *4kg *( 25m/s)^2
KE = 1250 kg m^2/s^2 = 1250 Newton meters or Joules
After second interval
KE = .5 *4kg *( 50m/s)^2
KE = 5000 kg m^2/s^2 = 5000 Newton meters or Joules
5000J/1250J = 4
The KE quadruples.
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Problem Number 10
What is the acceleration of an object of mass 300 kilograms in circular orbit about a planet of mass 9 *10^ 24 kilograms at a radius of 20000 kilometers?
How fast must be object be traveling if it is to remain in a circular orbit about the planet?
How long will it take the object to complete one orbit? Neglect any difference between the radius of the orbit and the distance between the object and the center of the planet.
mv^2/r = GMm/r^2
v^2/r = GM/r^2
v^2 = GM/r
v = `sqrt(GM/r)
v= 5480m/s (approx
aCent = v^2/r = (5480m/s)^2/20,000,000 meters
aCent = 1.50m/s^2 (approx)
orbit = 20,000,0000meters *2 * `pi = 125663706meters
t= `ds/v = 125663706meters/ 5480m/s = 22900 seconds (approx)
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Good.
Notice that this agrees with what you would have obtained by calculating G M / r^2 directly.
Nothing wrong with your solution but the connection is worth thinking about.
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Problem Number 11
A rubber band does 58 J of work as it launches a mass of 3.3 kg.
By how much does the elastic PE of the system change, and if no energy is dissipated what will be the KE change of the mass?
Assuming that no other forces act on the mass, if the mass started from rest, what velocity will attain?
The PE of the system decreases by 58J .
The KE increases 58J
KE = m`dv^2
58J = 3.3kg *`dv^2
`dv^2= (59kg m/s^2 m)/3.3kg
`dv^2 = 17.87878788
`dv= +-4.23 m/s
Use the postive value since positive KE is in positive direction of motion.
Since the mass starts from rest the mass acheives a velocity ofapprox. 4.23m/s
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Problem Number 12
What is the centripetal acceleration of an object on a circle of radius 5 meters, if the object's speed is 9 meters per second?
If the object has mass 63.99 kg, how much force is required to provide the acceleration?
aCent = v^2/r
aCent = (9m/s)^2/5 meters = 16.2m/s^2
F=m*a
F= 63.99 kg* 16.2m/s^2
F = 1040kg m/s^2 or 1040 Newtons (approx)
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Problem Number 13
The velocity of an object increases by 18 meters per second during its lifetime.
If the rate of increase is constant, and if during a 1 minute observation the velocity of the object is seen to increase by 6 meters/sec, then how long does it live?
For the one minute observation
a = `dv/`dt
a =( 6m/s)/60sec = .1m/s
over the lifetime
`dt= `dv/a
`dt = (18m/s)/ .1m/s^2
`dt = 180 seconds
The object lives for 180 seconds
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Problem Number 14
What is the centripetal acceleration, in m/s ^ 2, of the reference point on the circle which models the vertical simple harmonic motion of a mass of 7.7 kilograms, released from rest at a distance of .55 meters from its equilibrium position, when a restoring force of 100 Newtons/meter acts to pull the object back to its equilibrium point?
What acceleration would you expect for the object at the instant it reaches an extreme point?
Explain in your summary why you would expect the object itself to undergo this acceleration at the extreme positions in its cycle.
`omega = `sqrt(k/m) = `sqrt [(100kgm/s^2/m)/7.7kg) = 3.60radians/s
speed of reference point = 3.60radians/s * .55 meters = 1.98 meters/sec
aCent = v^2/r = (1.98m/)s^2/.55 meters = 7.14 m/s^2
At the extreme position max force = .55 meters * 100 Newtons/meter = 55 N
F= m*a
a = F/m = 55 kgm/s^2 / 7.7k g= 7.14 m/s^2
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Problem Number 15
An ideal spring has restoring force constant 420 Newtons/meter. An unknown mass on a spring is observed to complete a cycle of oscillation in .42 seconds. What is the mass, in kilograms?
Angular frequency = 2 `pi radians/.42 sec = 15.0 radians/sec
Angular frequency = `sqrt(k/m)
15.0 radians/sec = `sqrt(k/m)
225 radians^2/sec^2 = 420 Newtons/meter /mass
(420 Newtons/meter)/(225 radians^2/sec^2) = mass
mass = 1.87 kg (approx)
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Very good.
I've inserted a few notes, but you appear to be in excellent shape for the final.
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