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Vector Nature of w. Torque as a Vector Cross Product. t = r x F | t| = rFsin q. The direction of the torque ( t ) is that of an advancing RH screw turned from the direction of r toward the direction of F. See: http://www.phy.syr.edu/courses/java-suite/crosspro.html.
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Torque as a Vector Cross Product t = r x F|t| = rFsinq The direction of the torque (t) is that of an advancing RH screw turned from the direction of r toward the direction of F See: http://www.phy.syr.edu/courses/java-suite/crosspro.html
Basics of Rotational Dynamics I = Si miri2 K = ½ I w2 t = r x F|t| = rFsinq t = I a
Rotational Inertia for Selected objects and rotation axes: HR&W Table 10-2
h 1 2 h=0 CALM Q Hoop 1 Ball 2 Each m=M Hoop rolls without slipping, ball drops. • [time: Ball first (13); hoop (1) same (5): Speed ball (7) hoop (9) same (2); KE ball (3), hoop (8) same (6)] • The ball arrives first, the hoop is moving faster just before it reaches height = 0, the hoop has greater kinetic energy just before it reaches height = 0. [This was the most common answer] • ball hits first... ball is faster... ball has more KE... all due to friction on hoop... [This got two of the three parts correct, but misses the key point about kinetic energy, so maybe the speed was fortuitous?] • The ball will arrive first on the ground, and it will also be moving faster just before it reaches h=0. However, they will have equal kinetic energy because the rotation kinetic energy of the hoop makes up for this decrease in speed. [This has it all right, but there is no explanation for the results, let’s look into that]
Parallel-Axis Theorem Ih = Icom + Mh2 h Parallel Axis a distance h away. Rotation Axis throurgh COM
The mass of the Earth is about 6x1024 kg and its radius is about 6x106 meters. Suppose you build a runway along the equator, line-up a million F-16's, bolt them down, and have them all fire their engines (eastward) simultaneously for 1/2 hour. Estimate the effect that would have on the rotational speed of the Earth. Assume the thrust of each plane's engine (these fighters have only one) is 30,000 lbs = 133,000 Newtons.[31 no answers; 11 confused] • very few people knew even how to handle the angular version of Newton II (t = I a ) • Ft = mra (??)1.33x10^11 x 6x10^24 x 6x10^6=a a= 4.788x10^42 [where did this eqn. come from? Dealing with angular motion, but mass and accel (not I and a) are the focus of this answer??] • The net torque on the earth from the jets is 6x10^6m x 133000N x 1000000 = 7.98x10^17 Nm. Using t net = inertia x angular acceleration, I found the acc. to be 9.236x10^-21 rad/s/s. After 30 minutes, the speed would only increase by 1.66x10^-17 rad/s. [4 were like this; this missed the basic point, but did handle the t = I a correctly]
The mass of the Earth is about 6x1024 kg and its radius is about 6x106 meters. Suppose you build a runway along the equator, line-up a million F-16's, bolt them down, and have them all fire their engines (eastward) simultaneously for 1/2 hour. Estimate the effect that would have on the rotational speed of the Earth. Assume the thrust of each plane's engine (these fighters have only one) is 30,000 lbs = 133,000 Newtons.[31 no answers; 11 confused] very few people knew even how to handle the angular version of Newton II (t = I a ) Ft = mra (??)1.33x10^11 x 6x10^24 x 6x10^6=a a= 4.788x10^42 [where did this eqn. come from? Dealing with angular motion, but mass and accel (not I and a) are the focus of this answer??] The net torque on the earth from the jets is 6x10^6m x 133000N x 1000000 = 7.98x10^17 Nm. Using t net = inertia x angular acceleration, I found the acc. to be 9.236x10^-21 rad/s/s. After 30 minutes, the speed would only increase by 1.66x10^-17 rad/s. [4 were like this; this missed the basic point, but did handle the t = I a correctly] HOWEVER: The key is really that the torque from the F-16’s is INTERNAL to the Earth/atmosphere system, so there is NO impact on the angular velocity
Rotational Inertia for Selected objects and rotation axes: HR&W Table 10-2
Chapter 11 Why is it “easier” to do a dive in the tuck position than the pike position? (Note: the diagram shows a dive in the pike position, not the tuck as the book suggests).
Chapter 11 problems b) The KE increases; where did this energy come from?
Rotational Inertia for Selected objects and rotation axes: HR&W Table 10-2
A B C P P P • Rank the three cases in order of the magnitude of the torque gravity exerts about the point where the post is attached to the ground (P). The cable supporting the ball is the same distance from the point P in each case, and the height at which the cable is attached to the supporting structure is also the same in all three cases. • B>A>C (6); A>B>C (5) B>C>A (3); C>B=A (3); • The rankings in order are C then B then A. This is because C forms a right angle so therefore has the greatest torque. Secondly, B has a greater distance between the ball and the hanger therefore it has a greater moment of inertia. [Be careful of what angle you use] • The order is B›A›C. In each case the torque exerts greater influence for a more horizantally directed rotation. In the case of b and a more torque is directed along b's rotational axis because there is a greater tendency to rotate outward [What direction?]
A B C P P P • Rank the three cases in order of the magnitude of the torque gravity exerts about the point where the post is attached to the ground (P). The cable supporting the ball is the same distance from the point P in each case, and the height at which the cable is attached to the supporting structure is also the same in all three cases. • C>B>A (6); B>A>C (4); B>C>A (3); A>B>C (3) • In fact: A=B=C!! (neglecting mass of cable and rods). • R is the vector from the axis to the point the force is applied; This is the same for all three, and the direction (and magnitude) of the gravitational force is the same in all.
