Homework 11-14

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A vector with x component 30 cm and y component 40 cm is a length greater than 1.3 cm.

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I believe this is right. Corresponds to .25 kg m/s, which is how you will see it in the given solution.

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Very good.

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The point is said to be traveling at 30 cm/s.

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Good thinking.

The circle has a radius of 9.9 mm, which is .99 cm.

So you can figure out its circumference, which is how many cm the point travels in a rotation.

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Good, but after collision the two are moving in opposite directions so you would have relative velocity -70 cm/s.

The given solution assumes the positive direction to be in the direction of the first ball's initial velocity.

Your choice of signs would correspond to the positive direction being the direction of the second ball's initial velocity.

Either choice is valid, but the choice should be clearly declared.

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Check this against the given solution.

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Given solutions for 111107 assignment, to be compared with your solutions:

`q001: A rubber band chain is stretches from the point (10 cm, 5 cm) to the point (50 cm, 35 cm). The tension vs. length graph for this chain has horizontal intercept 42 cm and slope .15 N / cm.

What is the length of the rubber band chain?

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The 'rise' is from 5 cm to 35 cm and the 'run' is from 10 cm to 50 cm, so 'rise' is 40 cm and 'run' is 30 cm.

The distance between the points is therefore sqrt((40 cm)^2 + (30 cm)^2 ) = 50 cm.

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What therefore is its tension?

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Since y = 0 at the point (42 cm, 0), we see that when the length of the chain is 42 cm the tension is zero.

The length is 50 cm, which is 8 cm greater than the zero-tension length.

The tension is therefore 8 cm * .15 N / cm = 1.2 N. (That is, from x = 42 cm to x = 50 cm the 'run' is 8 cm, so the 'rise' is slope * run = 1.5 N / cm * 8 cm = 1.2 N).

We can get the equation of our tension vs. length graph:

The graph is a straight line with slope .15 N / cm through the point (42 cm, 0). The equation of the line is therefore found by the point-slope formula to be

(y - 0) = .15 N / cm *( x - 42 cm), or

y = .15 N / cm * x - 6.3 cm.

y represents tension and x represents length, so we could write this as

T = .15 N / cm * L - 6.3 cm.

This equation could also be reasoned out:

The horizontal intercept of the line is at 42 cm, so the point (42 cm, 0) is on the line.

If the length is L, then, the 'run' from the x intercept to x = L is (L - 42 cm) and the 'rise' is (L - 42 cm) * .15 N / cm. Since the y coordinate of the intercept is 0, this implies that when x = L, we have y = (L - 42 cm) * .15 N / cm.

y represents the tension, so

T = (L - 42 cm) * .15 N / cm.

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What are the x and y components of a vector from the (10 cm, 5 cm) point to the (50 cm, 35 cm) point?

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The x and y components are respectively 40 cm (x component) and 30 cm (y component).

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What therefore is the length of this vector?

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By the Pythagorean Theorem the length is 50 cm.

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What do you get if you divide this vector's x and y components by its length?

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We get x component 40 cm / (50 cm) = .8, and y component 30 cm / (50 cm) = .6.

Note that the unit cm divides out in both cases. Our new components are just plain .8 and .6, with no units.

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The results of the preceding are the x and y components of the unit vector in the direction of the rubber band. If you multiply these components by the tension force, you get the x and y components of the tension force vector. What are these components?

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The force is 1.2 N, so the x and y components will be

force * .8 = 1.2 N * .8 = .96 N

and

force * .6 = 1.2 N * .6 = .72 N.

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What therefore are the magnitude and angle of the tension force vector?

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The tension force has x and y components .96 N and .72 N, so the magnitude of the tension vector is

sqrt( (.96 N)^2 + (.72 N)^2 ) = 1.2 N.

The angle of the tension force is

theta = arcTan ( F_y / F_x) = arcTan(.96 N / (.72 N) ) = 53 degrees.

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How much work would have to be done to extend the rubber band one more centimeter?

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If the rubber band was 1 cm longer it would exert an additional force of 1 cm * .15 N / cm = .15 N, so its tension would be 1.2 N + .15 N = 1.35 N.

The average tension between the 50 cm length, where the tension is 1.2 N, and the 51 cm length where the tension is 1.35 N, is therefore

F_ave = (1.2 N + 1.35 N) / 2 = 1.28 N, rounded to the nearest .01 N.

The work required to extend the rubber band an additional centimeter is therefore

F_ave * `ds = 1.28 N * .01 cm = .128 N * cm = .128 N * (.01 m) = .00128 Joules.

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q002. If the rubber band chain in the preceding suspends a 40 gram mass, holding it in equilibrium against the force of gravity on the mass, then how long will the chain be?

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A 40 gram mass has a weight of about (.04 kg) * 9.8 m/s^2 = .4 Newtons

At .15 N / cm, the length of the rubber band would have to be .4 N / (.15 N / cm) = 2.7 cm longer than its 'zero-force length' of 42 cm, so it would be 44.7 cm long.

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If the 40 gram mass is pulled down until rubber band chain is 50 cm long, and the mass is released, what will be the net force on the mass the instant after release?

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At the 50 cm length the tension is 1.2 N upward and the weight of the suspended mass is still .4 N downward, so the net force is 1.2 N - .4 N = .8 N in the upward direction.

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In the preceding, with what force was the mass pulled down before being released?

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The mass was being pulled down by its .4 N weight, and by an additional force of .8 N, so that the net force on it is zero, which is the total 1.2 N downward force plus the 1.2 N upward tension.

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How much work will be done by the tension force on the mass as it returns to its equilibrium position?

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The equilibrium position for this mass occurs when the rubber band chain is 44.7 cm long, at which point it is exerting an upward force of .4 N.

So the average force exerted by the rubber band over this interval is (1.2 N + .4 N) / 2 = .8 N upward, which is exerted through an upward displacement of (50 cm - 44.7 cm) = 5.3 cm. The rubber band therefore does work

`dW_rb_on_mass = .8 N * 5.3 cm = 4.2 N cm = .042 N m = .042 Joules.

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If the rubber band tension was conservative, what would be its velocity upon return to equilibrium?

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The 40 gram mass would have kinetic energy .042 Joules.

Solving KE = 1/2 m v^2 for v we get v = +- sqrt( 2 KE / m) = +- sqrt(2 * .042 Joules / (.04 kg) ) = 1.4 m / s.

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`q003. A 40 gram mass is dropped from rest at a height of 2 meters above the floor. What is its momentum just before it strikes the floor?

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The mass has initial velocity 0 and falls 2 meters while accelerating at 9.8 m/s^2. Its final velocity during free fall is therefore about 6.3 m/s downward.

It strikes the floor at 6.3 m/s, so its momentum is

momentum = m v = .04 kg * 6.3 m/s = .25 kg m/s

in the downward direction.

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How fast is the mass going when it first loses contact with the floor on its rebound, given that it rebounds to a maximum height of 1.5 meters?

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The mass rises 1.5 m while accelerating downward at 9.8 m/s^2, reaching final velocity 0. Its initial velocity during its rise is therefore about 5.4 m/s upward.

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What therefore is its momentum when just after leaving the floor on its rebound?

By how much does its momentum change during the interval from just before contacting the floor to just after leaving the floor on its rebound? Be sure to choose a positive direction and answer relative to that direction.

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Its momentum would be (.04 kg) * (5.4 m/s) = .22 kg m/s upward.

The change in momentum from .25 kg m/s downward to .22 kg m/s upward is an upward change of .47 kg m/s. (To come to rest the mass has to lose its .25 kg m/s downward momentum, which requires a momentum change if .25 kg m/s in the opposite, i.e., upward direction; it then has to gain .22 kg m/s of additional upward momentum, for a total upward change of .47 kg m/s).

If we choose upward as the positive direction, then our momentum just before hitting the floor is -.25 kg m/s and our momentum just after leaving the floor is +.22 kg m/s, so the change in momentum is

`dp = p_final - p_initial = .22 kg m/s - (-.25 kg m/s) = .22 kg m/s + .25 kg m/s = .47 kg m/s.

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`q004. A steel ball 25 mm in diameter has mass about 70 grams. A point on that ball which lies 9.9 mm from its axis of rotation is moving in a circular path about that axis, traveling at 30 cm / second.

How many times does that point go around the circle in a second?

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That point moves in a circular path about the ball's horizontal axis of rotation. The point is 9.9 mm from that axis so it moves in a circle of radius 9.9 mm.

The circumference of the circle is therefore 2 pi r = 2 pi * 9.9 mm = 60 mm, approximately.

This is about 6 cm.

The ball moves at 30 cm/second, which matches the speed of the point around this path. So in one second the point moves through (30 cm/s) / (6 cm / revolution) = 5 revolutions.

The ball is rotating at about 5 revolutions per second.

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What is its angular velocity in radians per second?

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That gives it an angular velocity of 5 revolutions / sec = 5 * (2 pi radians) / second = 10 pi radians / second.

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What is the moment of inertia of the ball?

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The ball is a sphere, and it rotates about an axis through its center. So its moment of inertia is

I = 2/5 M R^2 = 2/5 ( 70 grams ) * (1.25 cm)^2 = 44 gram cm^2.

This can be expressed in SI units as 44 ( .001 kg ) ( .01 m)^2 = .0000044 kg m^2 = 4.4 * 10^-6 kg m^2.

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The angular velocity of the steel ball about that axis is the same as that of the point. What therefore is the angular kinetic energy of the ball?

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The point has angular velocity 10 pi radians / second, so the ball has the same angular velocity.

Its angular kinetic energy is therefore

KE = 1/2 I omega^2 = 1/2 * (.0000044 kg m^2) ( 10 pi rad/sec)^2 = .0000022 kg m^2 * 100 pi^2 rad^2 / s^2 = .0022 Joules, approximately.

Just for a ballpark idea of how much KE this would be, in terms of the behavior of the ball, it would be easy to calculate the PE change of the ball if it fell 1 cm (or rolled down an incline whose lower end is 1 cm lower than its upper end). The PE change would be .07 kg * 9.8 m/s^2 * (-.01 m) = -.007 Joules. The angular kinetic energy of the ball in the present problem is .0022 Joules, a bit less than 1/3 of the PE loss.

The translational KE of that ball is also easily calculated: KE_translational = 1/2 m v^2 = 1/2 * .07 kg * (.3 m/s)^2 = .003 Joules, approx., a bit more than its rotational KE and a bit less than half of the .007 J loss of PE.

A ball which descends 1 cm while rolling down an incline from rest loses enough PE to account for both the rotational and translational KE gained by this ball.

It remains an open question whether the PE loss is also sufficient to overcome the rolling friction.

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`q005. If you are in a car traveling at 70 mph behind another car traveling in the same direction at 60 mph, then your velocity relative to that car is 10 mph in your direction of motion. If you are traveling at 70 mph and another car is traveling at 60 mph in the opposite direction, then your velocity relative to that car is 130 mph in your direction of motion (and its velocity is -130 mph in your direction of motion).

A ball moving at 50 cm/s approaches another ball moving toward it at 20 cm/s. The balls collide and both balls reverse their directions, the first moving at 10 cm/s while the second moves at 60 m/s.

Before collision, what is the velocity of the first ball relative to the second?

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The combined speed of the two approaching balls is 70 cm/s. If we declare that the intial direction of motion of the first ball is positive, we conclude that the velocity of the second ball relative to the first is 70 cm/s toward the first, which makes the requested relative velocity -70 cm/s.

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After collision, what is the velocity of the first ball relative to the second?

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After collision the combined speed is again 70 cm/s (60 cm/s and 10 cm/s in opposite directions imply that the balls are moving apart at 70 cm/s). The first ball is moving in the negative direction and sees the second ball moving away from it. So from the perspective of the first ball, the second ball is now moving with relative velocity + 70 cm/s.

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What is the magnitude of the velocity change of the first ball, and what is the magnitude of the velocity change of the second?

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The velocity of the first ball changes from 50 cm/s to - 10 cm/s, a change of - 60 cm/s.

The velocity of the second ball changes from -20 cm/s to +10 cm/s, a change of + 30 cm/s.

The magnitude of the velocity change of the first ball is therefore double that of the second.

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Which ball do you conclude had the greater mass?

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The velocity change of the second has a lesser magnitude than that of the first, implying that it has double the mass of the first.

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While they were in contact, which ball exerted the greater force on the other? If you can't answer this question as it is posed, state why.

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While in contact the balls exerted equal and opposite forces on one another, by Newton's Third Law (the law of action-reaction, which is very well verified by experiment).

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See my notes, then check your solutions and your thinking against the discussion given above. Questions and revisions are welcome.

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