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course Phy 121
I am struggling on this unit for some reason, hoping you can help me for the test!! Thanks so much, hope you had a nice thanksgiving and you are healing quickly!
Problem 1A disk is subjected to a torque of 7.5 meter Newtons. As a result, it is observed to accelerate from 6.599 radians per second to 12.09 radians per second during an observation lasting 4.7 seconds.
• What is the moment of inertia of the disk?
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???I am sort of confused on how I work this problem out if I do not have the mass???? I know the formula is : 1/2 M R^2
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You know that
net torque = moment of inertia * angular acceleration
This is Newton's Second Law for rotational dynamnics.
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Problem Number 2
A disk completes one revolution in 3 seconds.
• Through how many radians per second does it rotate?
• How fast would an object located at a point 3 meters from the center be moving?
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Problem Number 3
How many days would be required for an individual of mass 79 kg to climb a tower which begins at the surface of the Earth, at a distance of 6400 km from the center, and rises to a position 3300 kilometers further from the center? Assume that the individual has an average power output of .54 watts / kg of body mass, and can sustain this output for 8 hours daily. Approximate the work done in the climb by subdividing the height of the tower into three 1100 kilometer segments of the tower.
.79 kg*(9.8m/s^2)=774.2 Newtons
(3300km+1100km)/3300km
magnitude=(9.8 m/s^2)/[((3300km+1100km)/3300km)]^2 resulting in a force of 79 kg*(9.8 m/s^2)/[((3300km+1100km)/3300km)]^2=580.65Newtons
Average =(774.2 N+580.65N)/2
=677.425 N*1100
745168 Joules??? Not sure about this step????????
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Your calculation would be correct if the 79 kg * 9800 m/s^2 occurred at the surface of a planet of radius 3300 km.
However that isn't the case.
79 kg * 9.8 m/s^2 is the weight at the surface of the Earth, which is 6400 km from the center, not 3300 km.
3300 km is the distance you are going above the surface. Breaking this down into three 1100 km segments, you go from
6400 km to 6400 km * 1100 km
6400 km + 1100 km to 6400 km + 2 * 1100 km
6400 km + 2 * 1100 km to 6400 km + 3 * 1100 im.
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The second segment extends from (6400 + 1100) km to (6400 + 2* 1100) km:
• At altitude 2( 1100) km, gravity exerts a force of 79 kg (9.8 m/s ^ 2) / [(6400+2* 1100)/6400] ^ 2 = 2524.22 Newtons.
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Your calculation is correct up to the last step, but you don't get 2524 Newtons. More like around 400 N.
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• Exerted over a distance of 1100 km = 1100000 meters, the average of this force and the 580.65Newtons
• Newtons required at the beginning of this segment would require 1.70768E+09 Joules of energy.
???I need help with the rest of the problem! I am lost!!????
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You're pretty much on the right track. See if my notes do the job.
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Problem Number 4
A rolling ball accelerates uniformly at .3 radians/second ^ 2. At a certain instant its angular velocity is 6 radians/second.
• From this instant, how long does it require to then rotate through 14.99 radians, and what angular velocity does it attain in the process?
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Problem Number 5
A turbine accelerates uniformly at 2.5 radians/second/second.
• How long will it a to accelerate from 1.7 radians/second to 4.1 radians/second?
• Through what angular displacement will it turn in this time?
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Change in angular velocity is 2.5 radians/second
aAve=(Vf-V0)/‘dt
‘dt=(vf-v0)/a
(4.1 radians/sec)-(1.7 radians/sec)= 2.5 radians/sec
(2.5 radians/sec)/(2.5 radians/sec)= 1 second
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(2.5 radians/sec)/(2.5 radians/sec/ sec)= 1 second
or
(2.5 radians/sec)/(2.5 radians/sec^2)= 1 second.
However
(2.5 radians/sec)/(2.5 radians/sec)= 1, with no units, not 1 second.
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(1.7 radians/sec+4.1 radians/sec)/2= 2.9 radians/sec
????????I am not sure what else to do here?????
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You need to identify your quantities.
This quantity is the average angular velocity.
Given average angular velocity and time interval, how do you get angular displacement?
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Problem Number 6
The gravitational field strength at the surface of the Earth is g = 9.8 m/s^2; at the surface of the moon g is 1.5 m/s^2.
Suppose your mass is 58 kg.
• What is your weight on the surface of the Earth and on the Moon?
What is the strength of the gravitational field on the surface of a planet where you weigh 2000 Newtons?
Earth Weight=(9.8 m.s2^2)(58kg)
=568.4 Newtons
Moon Weight: (1.5 m/s^2)(58 kg)
=87 Newtons
2000Newtons/58 kg
=34.38276 m/s^2
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Problem Number 7
Imagine that you are orbiting a neutron star whose mass is 68 * 10^30 kilograms at a distance of 21 kilometers from its center.
• Find the difference between the force exerted by gravity on a one kilogram mass in this orbit and the force exerted by gravity on a one kilogram mass orbiting one kilometer further from the center of the planet.
• By how much does the orbital velocity of the first object exceed that of the second?
From your first result, determine the average force gradient (force change per unit of distance), in Newtons per meter, for one kilogram masses between 21 and 21+1 kilometers from the center of the neutron star.
• Use this gradient to estimate the difference you would therefore expect between one kilogram of your left shoulder and one kilogram of your right shoulder, assuming that one shoulder is .43 meter further from the star than the other.
From your second result, determine the average velocity gradient, in (m/s) per meter, between orbits at 21 km and at 21+1 km from the center.
• Determine how long it would take one shoulder to move ahead of the other by one complete revolution.
. 21 km=21000 meters
21 +1km=22000 meters
F = G m M / r ^ 2
G = 6.67 * 10^-11 N m ^ 2/kg ^ 2
First Distance Force:
m=68 * 10^30kg
M=????
r=21000 m
(6.67 * 10^-11 N m ^ 2/kg ^ 2)*(68 * 10^30kg)*(??) / (21000 meters)
??? I am unsure of what the “M” should be in the equation :F = G m M / r ^ 2
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The question asks about a satellite of mass one kg and the star, of mass 68 * 10^30 kg.
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Problem Number 8
Given that G is approximately equal to 6.67 * 10^-11 kg m ^ 2/s ^ 2:
• Give the strength of the gravitational attraction felt by a human being of mass 53 kg to a rock sphere with radius 2.9 km and density 3.7 times that of water (water's density is 1000 kg/m ^ 2), assuming that the person's entire mass is right at the surface of the sphere.
• Give the attraction if the sphere was compressed to a radius of 290 meters, and if it was compressed to radius 29 meters.
• To what radius would the sphere have to be compressed in order to exert a force equal to the weight of this individual on the surface of the Earth?
F = G m1 m2 / r ^ 2
9.8 m/s ^ 2 ( 53 kg) = 519.4 Newtons
r = `sqrt[G m1 m2 / F
I am lost on this problem!!! Can you help me????
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You have to find the mass of that sphere.
9.8 m/s^2 has nothing to do with this situation.
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&#Your work looks good. See my notes. Let me know if you have any questions. &#