course PHY231 On the query for assignment 28, I was confused by (rEarth / r)^2 vs (r / rEarth)^2. Looking back at the notes, I've seen that the exponent was negated. I worked on question 12 in problem set 7 for some time. The trouble I had and couldn't figure out was how to calculate the KE from velocity without a given mass. I can submit the entire problem if you'd prefer, but that was my only question about it. CЩUlZոٛ
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23:24:02 `q001. Note that this assignment contains 11 questions. The planet Earth has a mass of approximately 6 * 10^24 kg. What force would therefore be experienced by a 3000 kg satellite as it orbits at a distance of 10,000 km from the center of the planet?
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RESPONSE --> The satellite experiences 11.95 kN of force: Fg = G * mE * m / r^2 With m = 3000 kg and r^2 = 1 E14 m confidence assessment: 3
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23:24:12 The force would be F = G m1 m2 / r^2, with m1 and m2 the masses of the planet and the satellite and r the distance of the satellite from the center of the planet. Thus we have F = 6.67 * 10^-11 N m^2 / kg^2 * (6 * 10^24 kg) * 3000 kg / (10,000,000 meters) ^ 2 = 12,000 Newtons.
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RESPONSE --> self critique assessment: 3
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23:28:00 `q002. What force would the same satellite experience at the surface of the Earth, about 6400 km from the center.
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RESPONSE --> Substituting RE for r in the force equation gives 2.94 kN. Using the approximation of 6400 km for RE and 9.8 m/sec^2 as the acceleration due to gravity also gives 2.94 kN confidence assessment: 3
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23:28:32 The force would be F = G m1 m2 / r^2, with m1 and m2 the masses of the planet and the satellite and r the distance of the satellite from the center of the planet. Thus we have F = 6.67 * 10^-11 N m^2 / kg^2 * (6 * 10^24 kg) * 3000 kg / (6,400,000 meters) ^ 2 = 29,000 Newtons. Note that this is within roundoff error of the F = m g = 3000 kg * 9.8 m/s^2 = 29400 N force calculated from the gravitational acceleration experienced at the surface of the Earth.
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RESPONSE --> self critique assessment: 3
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23:33:32 `q003. What would be the acceleration toward the center of the Earth of the satellite in the previous two questions at the distance 10,000 km from the center of the Earth? We may safely assume that no force except gravity acts on the satellite.
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RESPONSE --> 11.95 kn / 3000 kg = 3.98 m/sec^2 confidence assessment: 3
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23:33:38 The force at the 10,000 km distance was previously calculated to be 12,000 Newtons, the mass of the satellite being 3000 kg. Since the only force acting on the satellite is that of gravity, the 12,000 Newtons is the net force and the acceleration of the satellite is therefore a = 12,000 N / 3000 kg = 4 m/s^2.
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RESPONSE --> self critique assessment: 3
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23:48:19 `q004. The centripetal acceleration of an object moving in a circle of radius r at velocity v is aCent = v^2 / r. What would be the centripetal acceleration of an object moving at 5000 m/s in a circular orbit at the distance of 10,000 km from the center of a planet, and how this this compare to the 4 m/s^2 acceleration net would be experienced by an object at this distance from the center of the Earth?
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RESPONSE --> The centripetal acceleration of the object is 2.5 m/sec^2. This is 5/8 the net acceleration at this distance. confidence assessment: 2
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23:48:31 The centripetal acceleration of the given object would be aCent = (5000 m/s)^2 / (10,000,000 m) = 2.5 m/s^2. This is less than the acceleration of gravity at that distance.
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RESPONSE --> self critique assessment: 3
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23:53:49 `q005. What would be the centripetal acceleration of an object moving at 10,000 m/s in a circular orbit at the distance of 10,000 km from the center of a planet, and how does this compare to the 4 m/s^2 acceleration that would be experienced by an object at this distance from the center of the Earth?
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RESPONSE --> Increasing velocity to 10,000 m/sec increases the centripetal acceleration to 10 m/s. This is 2.5 times the net acceleration at this distance confidence assessment: 2
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23:53:57 The centripetal acceleration of this object would be aCent = v^2 / r = (10,000 m/s)^2 / (10,000,000 m) = 10 m/s^2, which is greater than the acceleration of gravity at that distance.
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RESPONSE --> self critique assessment: 3
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00:03:14 `q006. An object will move in a circular orbit about a planet without the expenditure of significant energy provided that the object is well outside the atmosphere of the planet, and provided its centripetal acceleration matches the acceleration of gravity at the position of the object in its orbit. For the satellite of the preceding examples, orbiting at 10,000 km from the center of the Earth, we have seen that the acceleration of gravity at that distance is approximately 4 m/s^2. What must be the velocity of the satellite so that this acceleration from gravity matches its centripetal acceleration?
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RESPONSE --> The velocity must be 6324 m/sec: acent = v^2 / r =4 m/sec^2 v^2 = 4 m/sec^2 * 10000 m sqrt (4 m/sec^2 * 10000 m) confidence assessment: 3
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00:03:22 The velocity must be such that aCent = v^2 / r matches the 4 m/s^2. Solving aCent = v^2 / r for v we obtain v = `sqrt( aCent * r ), so if aCent is 4 m/s^2, v = `sqrt( 4 m/s^2 * 10,000,000 m ) = `sqrt( 40,000,000 m) = 6.3 * 10^3 m/s.
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RESPONSE --> self critique assessment: 3
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00:11:05 `q007. The orbital velocity of a satellite in a circular orbit is that velocity for which the centripetal acceleration of the satellite is equal to its gravitational acceleration. The satellite in the previous series of examples had a mass of 3000 kg and orbited at a distance of 10,000 km from the center of the Earth. What would be the acceleration due to Earth's gravity of a 5-kg hunk of space junk at this orbital distance?
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RESPONSE --> The acceleration due to gravity is independent of the mass of the object: a = F / m F = Fg = G * ME * m/ r^2 a = (G * ME * m/ r^2) / m = G * ME / r^2 a = 4.0 m/sec^2 confidence assessment: 3
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00:11:16 The force of gravity on the junk hunk is easily found from Newton's Law of Universal Gravitation. Using F = G m1 m2 / r^2 we see that the force of gravity must be Fgrav = (6.67 * 10^-11 kg) * (6 * 10^24 kg) * (5 kg) / (10,000,000 m)^2 = 20 Newtons, approx.. Its acceleration due to gravity is thus a = Fgrav / m = 20 Newtons / 5 kg = 4 m/s^2. We note that this is the same gravitational acceleration experienced by the 3000 kg mass, and conjecture that any mass will experience the same gravitational acceleration at this distance from the center of the planet.
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RESPONSE --> self critique assessment: 3
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00:12:25 `q008. What therefore will be the orbital velocity of the 5-kg piece of junk?
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RESPONSE --> Since the acceleration due to gravity is the same, the orbital velocity must also be the same at 6.3 km/sec confidence assessment: 3
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00:12:35 Orbital velocity is calculated from distance and gravitational acceleration by solving a = v^2 / r for v, where a is the centripetal acceleration, which is the same as the gravitational acceleration. We get v = `sqrt( a * r), just as before, and v = `sqrt( 4 m/s^2 * 10,000,000 m) = 6.3 * 10^3 m/s, the same velocity as for the 3000 kg satellite.
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RESPONSE --> self critique assessment: 3
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00:16:26 `q009. Is it true that the gravitational acceleration of any object at a distance of 10,000,000 meters from the center of the Earth must be the same as for the 3000-kg satellite and the 5-kg hunk of space junk? (Hint: We have to find the acceleration for any mass, so we're probably going to have to let the mass of the object be represented by symbol. Use mObject as a symbol for the mass of the object. While dealing in symbols, you might as well leave G and r in symbols and let mEarth stand for the mass of the Earth. Find an expression for the force, then using this expression and Newton's Second Law find an expression for the acceleration of the object).
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RESPONSE --> a = (G * mEarth * mObject/ r^2) / mObject = G * mEarth / r^2 confidence assessment: 3
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00:16:46 We know that the gravitational force on the object is Fgrav = G * mEarth * mObject / r^2, where G is the universal gravitational constant, r the distance from the center of the Earth, mEarth the mass of the Earth and mObject the mass of the object. The acceleration of the object is a = Fgrav / mObject, by Newton's Second Law. Substituting the expression G * mEarth * mObject / r^2 for Fgrav we see that a = [ G * mEarth * mObject / r^2 ] / mObject = G * mEarth / r^2. We note that this expression depends only upon the following: G, which we take to be univerally constant, the effectively unchanging quantity mEarth and the distance r separating the center of the Earth from the center of mass of the object. Thus for all objects at a distance of 10,000 km from the center of the Earth the acceleration due to the gravitational force must be the same.
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RESPONSE --> self critique assessment: 3
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00:51:06 `q010. How much work would have to be done against gravity to move the 3000 kg satellite from a circular orbit at a distance of 10,000 km to a circular orbit at a distance of 10,002 km?
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RESPONSE --> 23.9 MJ of work would be required. Work can be found by taking the integral of the force function over the interval from 10,000 to 10002 km. F = m a = mObject * G * mEarth / r^2 The integral of the function for this interval: G * mEarth * mObject (1/10002000 - 1/10000000) = 23,897,906 J confidence assessment: 2
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00:51:28 As found previously the object experiences a force of approximately 12,000 N at a distance of 10,000 km. At a distance of 10,002 km, the force of gravity will be slightly less than at 10,000 km, but only by about 5 Newtons or .0004 of the force. That is, over the 2 km distance the force of gravity doesn't change by very much. Therefore to move 2 km = 2000 m further from the center of the planet would require the application of a force very close to 12,000 N in the direction away from the center. The work done by this force is therefore `dW = 12,000 Newtons * 2000 m = 24,000,000 Joules.STUDENT QUESTION: I understand this mathmatically, I'm not sure I understand practically. How do you gain KE if one object was intially stationary? It would seem that the first object would lose and the second object would gain what was lost but not more than what was lost... INSTRUCTOR RESPONSE: As stated it isn't possible for total KE to increase unless there is some other source of energy involved. For example if there is a coiled spring on one object it could uncoil on collision and add extra KE. Momentum conservation does not say anything about energy. Momentum and energy are two completely independent quantities. **
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RESPONSE --> self critique assessment: 3
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00:52:06 `q011. Does it therefore follow that the work done to move a 3000 kg satellite from the distance of 10,000 km to a distance of 10,002 km from the center of the Earth must be 24,000,000 Joules?
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RESPONSE --> Yes confidence assessment: 3
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01:46:41 It might seem so, but this is not the case. The net force does work, but when the radius of the orbit changes the velocity and hence the kinetic energy of the satellite also changes. The work done by the net force is equal to the sum of the changes in the KE and the gravitational PE of the satellite. The change in gravitational PE is the 24,000,000 J we just calculated, and if there is no KE change this will be equal to the work done by the net force. However if KE increases the net force must do more than 24,000,000 J of work, and if KE decreases the net force must do less than 24,000,000 J of work. In this case, as we move further away the KE decreases so the net force must do less than 24,000,000 J of work. (See also Conservation of Energy in Orbit under Q&A)
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RESPONSE --> self critique assessment:
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wdꘓN| assignment #028 ||^🕬yXfگ Physics I 07-21-2007
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20:49:09 Query class notes #26 Explain how we use proportionality along with the radius rE of the Earth to determine the gravitational acceleration at distance r from the center of the Earth to obtain an expression for the gravitational acceleration at this distance. Explain how we use this expression and the fact that centripetal forces is equal to v^2 / r to obtain the velocity of a satellite in circular orbit.
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RESPONSE --> The acceleration due to gravity has an inverse square proportionality to the distance from the center of Earth. Approximating the acceleration due to gravity at Earth's surface as 9.8 m/sec^2, we can find the gravitational acceleration at any distance r: g = (RE / r)^2 * 9.8 m/sec^2 To maintain a circular orbit, a satellite must experience centripetal force and gravitational force that have the same magnitude. To obtain the velocity of a satellite in circular orbit at a particular value of r, we set these values equal to each other and solve for the velocity: v^2 / r = G* mEarth/ r^2 v = sqrt ( G * mEarth /r) Every value of r correlates to a single velocity for circular orbit.
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21:04:12 ** The proportionality is accel = k r^2. When r = rE, accel = 9.8 m/s^2 so 9.8 m/s^2 = k * rE^2. Thus k = 9.8 m/s^2 / rE^2, and the proportionality can now be written accel = [ 9.8 m/s^2 / (rE)^2 ] * r^2. Rearranging this gives us accel = 9.8 m/s^2 ( r / rE ) ^2. **
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RESPONSE --> I think the given solution gives an increasing value of acceleration as the distance from earth increases.
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21:04:16 Principles of Physics and Gen Phy problem 5.30 accel of gravity on Moon where radius is 1.74 * 10^6 m and mass 7.35 * 10^22 kg.
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RESPONSE -->
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21:04:21 ** The acceleration due to gravity on the Moon is found using the equation g' = G (Mass of Moon)/ radius of moon ^2 g' = (6.67 x 10^-11 N*m^2/kg^2)(7.35 X 10^22 kg) / (1.74 X 10^6 m) = 1.619 m/s^2 **
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RESPONSE -->
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21:04:27 Query gen phy problem 5.40 force due to planets (Mv, Mj, Ms, are .815, 318, 95.1 Me; orb radii 108, 150, 778, 1430 million km). What is the total force on Earth due to the planets, assuming perfect alignment?
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RESPONSE -->
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21:04:31 ** Using F = G m1 m2 / r^2 we get Force due to Venus: F = 6.67 * 10^-11 N m^2 / kg^2 * (5.97 * 10^24 kg) * (.815 * 5.97 * 10^24 kg) / (1.5 * 10^11 m - 1.08 * 10^11 m)^2 = 1.1 * 10^18 N, approx. Force due to Jupiter: F = 6.67 * 10^-11 N m^2 / kg^2 * (5.97 * 10^24 kg) * (318 * 5.97 * 10^24 kg) / (1.5 * 10^11 m - 7.78 * 10^11 m)^2 = 1.9 * 10^18 N, approx. Force due to Saturn: F = 6.67 * 10^-11 N m^2 / kg^2 * (5.97 * 10^24 kg) * (95.7 * 5.97 * 10^24 kg) / (1.5 * 10^11 m - 1.43 * 10^11 m)^2 = 1.4 * 10^17 N, approx. Venus being 'inside' the Earth's orbit pulls in the direction of the Sun while Jupiter and Saturn pull in the opposite direction so the net force is -1.1 * 10^18 N + 1.9 * 10^18 N + 1.4 * 10^17 N = .9 * 10^18 N = 9 * 10^17 N, approx.. **
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RESPONSE -->
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21:35:30 Univ. 12.50 (12.44 10th edition). 25 kg, 100 kg initially 40 m apart, deep space. Both objects have identical radii of .20 m. When 20 m apart what is the speed of each (relative to the initial common speed, we presume), and what is the velocity relative to one another? Where do they collide? Why does position of center of mass not change?
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RESPONSE --> Linear momentum is conserved because we are considering any other forces to be negligible. Since no forces external to the system are acting, the center of mass of the system will not change. The conservation of momentum allows us to express the velocity of one mass in terms of the other: v2 = m1 * v1 / m2 The sum of the kinetic energy and potential energy will be zero: 1/2 (m1 * v1 ^2 + m2 * (m1 * v1 / m2)) = G * m1 * m2 *(1/rf - 1/r0) rf is the final distance between the centers, r0 is the initial distance between the centers. This expression is rearranged and simplified: v1 = sqrt ( (2*G * m2* (1/rf - 1/r0)/ m1 + m2) This is solved for v1 = 1.63 E^-5 m/sec, and that value of v1 is substituted into the equation found earlier : v2 = m1 * v1 / m2 = 4.08 E^-6 Relative velocity of the masses is the sum of these velocities: 2.04 E^-5 The masses have identical radii of .2m and start from 40 m apart. They will collide when x1 + x2 = 40 m -2r = 39.6 m Expression for displacement is found through relationships: m1 * v1 = m2 * v2, so m1 / m2 = v2 /v1 position is directly proportional to velocity, so x1 / x2 = m2 / m1 = 100 kg / 25 kg. x1 = 4 * x2 4* x2 + x2 = 39.6 is solved for x2 = 7.92 and x1 is therefore 31.68. The masses collide when m1 has traveled 31.68 m and m2 has traveled 7.92 m
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21:37:24 The force would be F = (6.67 * 10^-11 * 25 * 100) / 20^2 F = 4.17 * 10^-10 a1 = 4.17 * 10^-10 / 25 a1 = 1.67 * 10^-11 m/s/s a2 = 4.17 * 10^-10 / 100 a2 = 4.17 * 10^-12 m/s/s The position of center of mass doesn't change because the two spheres are the same size. ** At separation r the force is F = G m1 m2 / r^2. For any small increment `dr of change in separation the approximate work done by the gravitational force is F `dr = G m1 m2 / r^2 * `dr. We take the sum of such contributions, between the given separations, to form an approximation to the total work done by the gravitational force. We then take the limit as `dr -> 0 and obtain the integral of G m1 m2 / r^2 with respect to r from separation r1 to separation r2. An antiderivative is - G m1 m2 / r; evaluating between the two separations we get - G m1 m2 / r1 - (-G m1 m2 / r2) = G m1 m2 ( 1/r2 - 1 / r1). This expression is evaluated at r1 = 40 m and r2 = 20 m to get the change G m1 m2 ( 1/(20 m) - 1 / (40 m) ) in KE. I get around 1.49 * 10^-9 Joules but it isn't guaranteed so you should verify that carefully. Assuming a reference frame initially at rest with respect to the masses the intial momentum is zero. If the velocities at the 20 m separation are v1 and v2 we know that m1 v1 + m2 v2 = 0, so that v2 = -(m1 / m2) * v1. The total KE, which we found above, is .5 m1 v1^2 + .5 m2 v2^2. Substituting v2 = - (m1 / m2) v1 and setting equal to the KE we can find v1; from this we easily find v2. You might get something like 4.1 * 10^-6 m/s for the velocity of the 100 kg mass; this number is again not guaranteed so verify it yourself. The position of the center of mass does not change because there is no external force acting on the 2-mass system. The center of mass is at position r with respect to m1 (take m1 to be the 25 kg object) such that m1 r - m2 (40 meters -r) = 0; substituting m1 and m2 you get 25 r - 100 (40 meters - r ) = 0. I believe you get r = 4 / 5 * 40 meters = 32 m, approx., from the 25 kg mass, which would be 8 meters from the 100 kg mass.
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21:37:29 Query gen phy problem 5.50 24 m diam wheel, rot period 12.5 s, fractional change in apparent weight at top and at bottom. What is the fractional change in apparent weight at the top and that the bottom of the Ferris wheel?
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RESPONSE -->
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21:37:32 ** Centripetal acceleration is a = v^2 / r. For a point on the rim of the wheel, v = dist in 1 rev / time for 1 rev = `pi * 24 m / (12.5 sec) = 1.9 m/s, approx. Thus v^2 / r = (`pi * 1.9 m/s)^2 / 12 m = 3 m/s^2, approx. At the top the only accel is the centripetal, and it is acting toward the center, therefore downward. The forces acting on any mass at the top are the gravitational force and the force exerted by the wheel on the mass. At the top of the wheel the latter force is the apparent weight. Thus grav force + apparent weight = centripetal force - m * 9.8 m/s^2 + wtApparent = m * (-3 m/s^2 ) wtApparent = m (-3 m/s^2) + m ( 9.8 m/s^2) = m (6.8 m/s^2). A similar analysis at the bottom, where the centripetal force will be toward the center, therefore upward, gives us - m * 9.8 m/s^2 + wtApparent = m * (+3 m/s^2 ) wtApparent = m (+3 m/s^2) + m ( 9.8 m/s^2) = m (12.8 m/s^2). The ratio of weights is thus 12.8 / 6.8, approx. ** A more elegant solution obtains the centripetal force for this situation symbolically: Centripetal accel is v^2 / r. Since for a point on the rim we have v = `pi * diam / period = `pi * 2 * r / period, we obtain aCent = v^2 / r = [ 4 `pi^2 r^2 / period^2 ] / r = 4 `pi^2 r / period^2. For the present case r = 12 meters and period is 12.5 sec so aCent = 4 `pi^2 * 12 m / (12.5 sec)^2 = 3 m/s^2, approx. This gives the same results as before. **
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RESPONSE -->
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21:43:33 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> It's been very helpful to use the K2 + U2 = K1 + U1 relationship from the text. More useful is the strategy of simply reasoning things out from what I've already learned. As complex as the work seems, the basic relationships simplify it. When I get the algebra right.
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k^y}ۥ~_ assignment #029 029. Radian measure of angle; angular position, angular velocity Physics II 07-22-2007
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16:29:23 `q001. Note that this assignment contains 15 questions. If an object moves a distance along the arc of a circle equal to the radius of the circle, it is said to move through one radian of angle. If a circle has a radius of 40 meters, then how far would you have to walk along the arc of the circle to move through one radian of angle? How far would you have to walk to move through 3 radians?
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RESPONSE --> 40 m / radian * 1 radian = 40 meters along arc 40 m * 3 radians = 120 meters along arc confidence assessment: 3
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16:29:30 Since 1 radian of angle corresponds to the distance along the arc which is equal to the radius, if the radius of the circle is 40 meters then a 1 radian angle would correspond to a distance of 40 meters along the arc. An angle of 3 radians would correspond to a distance of 3 * 40 meters = 120 meters along the arc. Each radian corresponds to a distance of 40 meters along the arc.
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RESPONSE --> self critique assessment: 3
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16:33:00 `q002. On a circle of radius 40 meters, how far would you have to walk to go all the way around the circle, and through how many radians of angle would you therefore travel? Through how many radians would you travel if you walked halfway around the circle? Through how many radians would you travel if you walked a quarter of the way around the circle?
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RESPONSE --> There are 2 * 'pi radians in each circle. 40 m / radian * 2 pi radians = approximately 251 meters Halfway around the circle is 'pi radians One quarter of the way around is 'pi/ 2 radians confidence assessment: 3
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16:33:21 The circumference of a circle is the product of `pi and its diameter, or in terms of the radius r, which is half the diameter, C = 2 `pi r. The circumference of this circle is therefore 2 `pi * 40 meters = 80 `pi meters. This distance can be left in this form, which is exact, or if appropriate this distance can be approximated as 80 * 3.14 meters = 251 meters (approx). The exact distance 2 `pi * 40 meters is 2 `pi times the radius of the circle, so it corresponds to 2 `pi radians of arc. Half the arc of the circle would correspond to a distance of half the circumference, or to 1/2 ( 80 `pi meters) = 40 `pi meters. This is `pi times the radius so corresponds to `pi radians of angle. A quarter of an arc would correspond to half the preceding angle, or `pi/2 radians.
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RESPONSE --> self critique assessment: 3
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16:35:43 `q003. On a circle of radius 6 meters, what distance along the arc would correspond to 3 radians? What distance would correspond to `pi / 6 radians? What distance would correspond to 4 `pi / 3 radians?
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RESPONSE --> 6 m / radian * 3 radians = 18 meters 6 m / radian * 'pi / 6 radians = 'pi meters (approx 3.1416 m) 6 m / radian * 4 'pi / 3 radians = 8 'pi meters (approx 25 meters) confidence assessment: 3
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16:35:50 3 radians along the arc would correspond to an arc distance of 3 times the radius, or 3 * 6 meters, or 18 meters. `pi / 6 radians would correspond to `pi / 6 times the radius, or `pi / 6 * 6 meters = `pi meters. 4 `pi / 3 radians would correspond to 4 `pi / 3 * 6 meters = 8 `pi meters.
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RESPONSE --> self critique assessment: 3
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16:37:54 `q004. If you were traveling around a circle of radius 50 meters, and if you traveled through 4 radians in 8 seconds, then how fast would you have to be moving?
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RESPONSE --> 50 m / radian * 4 radians / 8 sec = 25 m / sec confidence assessment: 3
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16:38:05 If you travel 4 radians along the arc you half traveled an arc distance of 4 times the radius, or 4 * 50 meters = 200 meters. If you traveled this distance in 8 seconds your average speed would be 200 meters / (8 seconds) = 25 m/s.
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RESPONSE --> self critique assessment: 3
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16:38:51 `q005. Traveling at 3 radians / second around a circle of radius 20 meters, how fast would you have to be moving?
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RESPONSE --> 20 m / radian * 3 radians / sec = 60 m /sec confidence assessment: 3
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16:38:56 3 radians along the arc is a distance of 3 times the radius, or 3 * 20 meters = 60 meters. Moving at 3 radians/second, then, the speed along the arc must be 3 * 20 meters / sec = 60 meters /sec.
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RESPONSE --> self critique assessment: 3
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16:43:44 `q006. If you know how many radians an object travels along the arc of a circle, and if you know the radius of the circle, how do you find the distance traveled along the arc? Explain the entire reasoning process.
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RESPONSE --> Using the relationship 1 rad = distance along arc / radius of circle, the number of radians of travel is multiplied by the number of meters per radian: m / radian * radians = meters confidence assessment: 2
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16:44:31 The distance traveled along the arc of circle is 1 radius for every radian. Therefore we multiply the number of radians by the radius of the circle to get the arc distance.
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RESPONSE --> self critique assessment: 3
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16:46:41 `q007. If you know the distance an object travels along the arc of a circle, and if you know the radius of the circle, how do you find the corresponding number of radians?
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RESPONSE --> Using the relationship from the last question, we divide the distance along the arc by the radius. confidence assessment: 2
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16:47:02 An arc distance which is equal to the radius corresponds to a radian. Therefore if we divide the arc distance by the radius we obtain the number of radians.
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RESPONSE --> self critique assessment: 3
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16:50:22 `q008. If you know the time required for an object to travel a given number of radians along the arc of a circle of known radius, then how do you find the average speed of the object?
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RESPONSE --> The distance is the product of radius and radians. Dividing that quantity by the number of seconds will give the average speed of the object. confidence assessment: 3
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16:50:30 If you know the number of radians you can multiply the number of radians by the radius to get the distance traveled along the arc. Dividing this distance traveled along the arc by the time required gives the average speed of the object traveling along the arc.
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RESPONSE --> self critique assessment: 3
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16:52:17 `q009. If you know the speed of an object along the arc of a circle and you know the radius of the circle, how do you find the angular speed of the object in radians/second?
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RESPONSE --> Dividing the speed along the circle by the radius will give the angular speed in radians/sec. confidence assessment: 3
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16:52:34 The speed of the object is the distance it travels along the arc per unit of time. The angular velocity is the number of radians through which the object travels per unit of time. The distance traveled and the number radians are related by the fact that the distance is equal to the number of radians multiplied by the radius. So if the distance traveled in a unit of time is divided by the radius, we get the number of radians in a unit of time. So the angular speed is found by dividing the speed along the arc by the radius.
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RESPONSE --> self critique assessment: 3
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16:58:17 `q010. We usually let `d`theta stand for the anglular displacement in radians between two points on the arc of the circle. We usually let `omega stand for the angular velocity in radians / second. We let `ds stand for the distance traveled along the arc of a circle, and we let r stand for the radius of the circle. If we know the radius r and the arc distance `ds, what is the anglular displacement `d`theta, in radians?
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RESPONSE --> Since arc distance / radius = radians: `ds / r = `d`theta confidence assessment: 2
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16:58:24 Since an angular displacement of 1 radian corresponds to an arc distance equal to the radius, the anglular displacement `theta in radians is equal to the number of radii in the arc distance `ds. This quantity is easily found by dividing the arc distance by the radius. Thus `d`theta = `ds / r.
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RESPONSE --> self critique assessment: 3
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17:01:02 `q011. If we know the radius r of a circle and the angular velocity `omega, how do we find the velocity v of the object as it moves around the arc of the circle?
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RESPONSE --> Velocity of the object as it moves around the circle is `ds / sec: `omega * r = `ds / sec confidence assessment: 3
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17:01:12 The angular velocity is the number of radians per second. The velocity is the distance traveled per second along the arc. Since an angular displacement of 1 radian corresponds to an arc distance equal to the radius, if we multiply the number of radians per second by the radius we get the distance traveled per second. Thus v = `omega * r.
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RESPONSE --> self critique assessment: 3
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17:11:46 `q012. We can change an angle in degrees to radians, or vice versa, by recalling that a complete circle consists of 360 degrees or 2 `pi radians. A half-circle is 180 degrees or `pi radians, so 180 degrees = `pi radians. How many radians does it take to make 30 degrees, how many to make 45 degrees, and how many to make 60 degrees?
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RESPONSE --> 30 * `pi / 180 = .52 rad 45 * `pi / 180 = .78 rad 60 * `pi / 180 = 1.05 rad confidence assessment: 3
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17:12:37 30 degrees is 1/6 of 180 degrees and therefore corresponds to 1/6 * `pi radians, usually written as `pi/6 radians. 45 degrees is 1/4 of 180 degrees and therefore corresponds to 1/4 * `pi radians, or `pi/4 radians. 60 degrees is 1/3 of 180 degrees and therefore corresponds to 1/3 * `pi radians, or `pi/3 radians.
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RESPONSE --> self critique assessment: 3
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17:15:20 `q013. Since 180 deg = `pi rad, we can convert an angle from degrees to radians or vice versa if we multiply the angle by either `pi rad / (180 deg) or by 180 deg / (`pi rad). Use this idea to formally convert 30 deg, 45 deg and 60 deg to radians.
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RESPONSE --> 30 * (`pi / 180) = 'pi/6 rad = .52 rad 45 * (`pi / 180) = 'pi/4 rad = .78 rad 60 * (`pi / 180) = 'pi/3 rad = 1.05 rad confidence assessment: 2
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17:15:37 To convert 30 degrees to radians, we multiply by the rad / deg conversion factor, obtaining 30 deg * ( `pi rad / 180 deg) = (30 deg / (180 deg) ) * `pi rad = 1/6 * `pi rad = pi/6 rad. To convert 45 degrees to radians we use the same strategy: {}45 deg * (`pi rad / 180 deg) = ( 45 deg / ( 180 deg) ) * `pi rad = 1/4 * `pi rad = `pi/4 rad. To convert 60 degrees: 60 deg * (`pi rad / 180 deg) = ( 60 deg / ( 180 deg) ) * `pi rad = 1/3 * `pi rad = `pi/3 rad.
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RESPONSE --> self critique assessment: 3
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17:21:02 `q014. Convert 50 deg and 78 deg to radians.
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RESPONSE --> 50 deg *(`pi rad / 180 deg) = ( 50deg / 180 deg) * `pi rad = 5/18 * `pi rad = 15 `pi/18 rad 78 deg *(`pi rad / 180 deg) = ( 78deg / 180 deg) * `pi rad = 13/30 * `pi rad = 13 `pi/30 rad confidence assessment: 2
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17:21:09 50 deg * (`pi rad / 180 deg) = ( 50 deg / ( 180 deg) ) * `pi rad = 5/18 * `pi rad = (5 `pi/ 18) rad. 78 deg * (`pi rad / 180 deg) = ( 78 deg / ( 180 deg) ) * `pi rad = 78/180 * `pi rad = (13 `pi/ 30) rad.
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RESPONSE --> self critique assessment: 3
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17:24:48 `q015. Convert (14 `pi / 9) rad to degrees.
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RESPONSE --> (14 `pi / 9) rad * (180 / `pi) = 280 degrees confidence assessment: 2
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17:24:53 Since the angle is given radians, we need to multiply by deg / rad to get the angle in degrees. (14 `pi / 9) rad * ( 180 deg / (`pi rad)) = ( 14 `pi / 9 ) * (180 / `pi ) deg = ( 14 * 180 / 9) * (`pi / `pi) deg = 14 * 20 deg = 280 deg.
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RESPONSE --> self critique assessment: 3
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Ɲݥ Fʙ assignment #029 ||^🕬yXfگ Physics I 07-24-2007
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02:17:33 Query class notes #28. Explain how we can calculate the average angular velocity and the angular acceleration of an object which rotates from rest through a given angle in a given time interval, assuming constant angular acceleration.
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RESPONSE --> Assuming constant acceleration from rest, the final velocity is two times the average velocity. The average velocity is found by dividing the radians of rotation by the time interval. The acceleration is the final velocity divided by the time interval.
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02:17:59 **This situation is strictly analogous to the one you encountered early in the course. As before acceleration is change in velocity / change in clock time. However now it's angular acceleration. We have angular acceleration = change in angular velocity / change in clock time. The average angular velocity is change in angular position / change in clock time. This question assumes you know the angle through which the object rotates, which is its change in angular position, as well as the change in clock time. So you can calculate the average angular velocity. If angular accel is uniform and initial angular velocity is zero then the final angular velocity is double the average angular velocity. In this case the change in angular velocity is equal to the final angular velocity, which is double the average angular velocity. From this information you can calculate angular acceleration. **
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RESPONSE -->
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02:18:02 Principles of Physics and General College Physics Problem 7.46: Center of mass of system 1.00 kg at .50 m to left of 1.50 kg, which is in turn .25 m to left of 1.10 kg.
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RESPONSE -->
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02:18:05 Using the position of the 1.00 kg mass as the x = 0 position, the other two objects are respectively at x = .50 m and x = .75 m. The total moment of the three masses about the x = 0 position is 1.00 kg * (0 m) + 1.50 kg * (.50 m) + 1.10 kg * (.75 m) = 1.58 kg m. The total mass is 1.00 kg + 1.50 kg + 1.10 kg = 3.60 kg, so the center of mass is at position x_cm = 1.58 kg m / (3.60 kg) = .44 meters, placing it a bit to the left of the 1.50 kg mass.
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RESPONSE -->
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02:18:08 Query problem 7.50 3 cubes sides L0, 2L0 and 3L0; center of mass.
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RESPONSE -->
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02:18:10
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RESPONSE -->
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02:18:12 ** The mass of the second will be 2^3 = 8 times as great as the first. It takes 8 1-unit cubes to make a 2-unit cube. The mass of the third will be 3^3 = 27 times as great as the first. It takes 27 1-unit cubes to make a 3-unit cube. In the x direction the distance from left edge to center of first cube is 1/2 L0 (the center of the first cube). In the y direction the distance is from lower edge to center of the first cube is 1/2 L0 (the center of the first cube). In the x direction the distance from left edge to center of the second cube is L0 + L0 (the L0 across the first cube, another L0 to the center of the second), or 2 L0. In the y direction the distance from lower edge to center of the second cube is L0 (the center of the second cube). In the x direction the distance from left edge to center of the third cube is L0 + 2 L0 + 3/2 L0 (the L0 across the first cube, another 2 L0 across the second and half of 3L0 to the center of the third), or 9/2 L0. In the x direction the distance from lower edge to center of the first cube is 3/2 L0 (the center of the third cube). Moments about left edge and lower edge of first cube: If m1 is the mass of the first cube then in the x direction you have total moment m1 * L0/2 + 8 m1 * (2 L0) + 27 M1 * (9/2 L0) = 276 m1 L0 / 2 = 138 m1 L0. The total mass is m1 + 8 m1 + 27 m1 = 36 m1 so the center of mass is at center of mass in x direction: 138 m1 L0 / (36 m1) = 3.83 L0. In the y direction the moment is m1 * L0/2 + 8 m1 * (L0) + 27 m1 * ( 3/2 L0) = 45 m1 L0 so the center of mass is at center of mass in y direction: 45 m1 L0 / (36 m1) = 1.25 L0. **
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RESPONSE -->
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02:18:15 What is the mass of the second cube as a multiple of the mass of the first?
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RESPONSE -->
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02:18:17 ** 3 dimensions: the mass will be 2^3 = 8 times as great. It takes 8 1-unit cubes to make a 2-unit cube. **
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RESPONSE -->
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02:18:21 What is the mass of the third cube as a multiple of the mass of the first?
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RESPONSE -->
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02:18:24 ** The mass of the third cube is 3^3 = 27 times the mass of the first. **
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RESPONSE -->
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02:18:27 How far from the outside edge of the first cube is its center of mass?
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RESPONSE -->
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02:18:29 ** In the x direction the distance is 1/2 L0 (the center of the first cube). In the y direction the distance is also 1/2 L0 (the center of the first cube). **
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RESPONSE -->
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02:18:31 How far from the outside edge of the first cube is the center of mass of the second cube?
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RESPONSE -->
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02:18:34 ** In the x direction the distance is L0 + L0 (the L0 across the first cube, another L0 to the center of the second), or 2 L0. In the x direction the distance is L0 (the center of the second cube). **
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RESPONSE -->
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02:18:36 How far from the outside edge of the first cube is the center of mass of the third cube?
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RESPONSE -->
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02:18:39 ** In the x direction the distance is L0 + 2 L0 + 3/2 L0 (the L0 across the first cube, another 2 L0 across the second and half of 3L0 to the center of the third), or 9/2 L0. In the x direction the distance is 3/2 L0 (the center of the third cube). **
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RESPONSE -->
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02:18:43 How do you use these positions and the masses of the cubes to determine the position of the center of mass of the system?
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RESPONSE -->
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02:18:45 ** In the x direction you have moment m1 * L0/2 + 8 m1 * (2 L0) + 27 M1 * (9/2 L0) = 276 m1 L0 / 2 = 138 m1 L0. The total mass is m1 + 8 m1 + 27 m1 = 36 m1 so the center of mass is at 138 m1 L0 / (36 m1) = 3.83 L0. In the y direction the moment is m1 * L0/2 + 8 m1 * (L0) + 27 m1 * ( 3/2 L0) = 45 m1 L0 so the center of mass is at 45 m1 L0 / (36 m1) = 1.25 L0. **
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RESPONSE -->
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02:38:13 Univ. 8.94 (8.82 10th edition). 45 kg woman 60 kg canoe walk starting 1 m from left end to 1 m from right end, moving 3 meters closer to the right end. How far does the canoe move? Water resistance negligible.
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RESPONSE --> The center of mass of the system including the woman and the canoe is the product of her mass and her distance from the center divided by the total mass of the system: 45 kg * 1.5 m / 105 kg = .64 meters In walking twice the distance from her original location to the center of the canoe, the center of mass of the canoe moves 2 (.64 m ) = 1.28 meters.
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02:40:23 ** Since water resistance is negligible the net force acting on the system is zero. Since the system is initially stationary the center of mass of the system is at rest; since zero net force acts on the system this will continue to be the case. Assuming that the center of mass of the canoe is at the center of the canoe, then when the woman is 1 m from the left end the center of mass of the system lies at distance c.m.1 = (1 m * 45 kg + 2.5 m * 60 kg) / (45 kg + 60 kg) = 195 kg m / (105 kg) = 1.85 m from the left end of the canoe. A similar analysis shows that when the woman is 1 m from the right end of the canoe, then since she is 4 m from the left end the center of mass lies at c.m.2 = (4 m * 45 kg + 2.5 m * 60 kg) / (45 kg + 60 kg) = 310 kg m / (105 kg) = 2.97 m. The center of mass therefore changes its position with respect to the left end of the canoe by about 1.1 meters toward the right end of the canoe. Since the center of mass itself doesn't move the canoe must move 1.1 meters toward the left end, i.e., backwards. Note that since the woman moves 3 m forward with respect to the canoe and the canoe moves 1.3 m backwards the woman actually moves 1.7 m forward. The sum -1.3 m * 60 kg + 1.7 m * 45 kg is zero, to within roundoff error. This is as it should be since this sum represents the sum of the changes in the centers of mass of the canoe and the woman, which is the net change in the position of center of mass. **
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RESPONSE -->
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