144 Physics Vaishnav How Things Work
Spring 2010
Homework Set #2 Due in class Monday 2/8/06
Instructions You are responsible for your own homework solutions. Consider these problems as practice for the exam. Consult me or other students when you have questions, but try to work the problems by yourself as much as you can. Collaboration is allowed, but you must indicate your collaborators, both in the class and outside of it. For every problem, show all your work. Simply writing down a final answer without showing your reasoning and calculations is unacceptable. Draw diagrams and sketches for problems, where useful. For any essay type questions, make sure to use complete sentences. Problem 1: (a) Exercise 36 on p. 39 Look at Fig. 1.3.4, and pretend the piano is a snowflake. If the snowflake is sitting on a sloped roof, two forces act on it; its weight and the normal force from the ramp. We can sum these forces like vectors to see thatas in Fig. 1.3.4the net force acting on the snowflake is down the ramp. On a flat roof, the weight points down and the normal force points up. These two forces cancel, and there is no net force on the snowflake, which therefore simply sits there. (b) Exercise 38 on p. 39 As the truck goes up the ramp, its height is increasing. The potential energy of an object at height h is mgh. The trucks initial kinetic energy becomes potential energy as it moves up the ramp. The trucks total energy is conserved. (c) Problem 18, p. 40. The potential energy of the water is mgh where m is the mass of water, and h is the height to which we raise it.
U = mgh = (1000 kg) (9.8 m/s2 ) (200 m) = 1.96 10 6 J
Problem 2: A jogger stars a three part jog by running 0.24 km north, then 0.16 km east. She then runs back to her starting point along a straight-line path. Graphically determine the length of her path back. We can draw the vector going north, the vector going east, and determine the length of the third vector using proportionality. I have drawn the vectors to scale below using 100 pixels as 0.1 km.
Problem 3: A cockroach runs counterclockwise around the rim of a lazy Susan three and a half times in five seconds. What is the cockroachs angular speed? Suppose the lazy Susan is turning clockwise at an angular velocity of 1 rad/s. What is the cockroachs new angular velocity? The number of radians the cockroach runs is 2 rad 3.5 rad/circle = 7 rad . The cockroachs angular velocity is therefore 7 rad/5 s = 4.4 rad/s CCW . If the lazy Susan is turning CW at 1 rad/s, the cockroachs net motion is still CCW, just slower. Its new angular velocity is (4.4 - 1) rad/s CCW = 3.4 rad/s CCW.
P roblem 4: You perform a fan cart experiment like you did in Lab #2. Assume that the fan cart has mass 800 g and the fan is aligned along the 0o mark. As in that lab, well assume the fan exerts a constant force. a) You measure an acceleration of 0.08 m/s2. Determine the net force acting on this fan cart. The negative sign means that the acceleration points in the direction opposite to what we call positive; for the motion detectors, away from the detector is positive, so the acceleration is towards the detector. Since net force and acceleration point in the same direction, the net force also points towards the detector. To determine the net force, use Newtons Second Law:
r r Fnet = m a F = m a = (0.800 kg) ( 0.08 m/s2 ) = 0.064 kg m/s2
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b) You make the following two changes: you increase the mass of the fan cart to 900 g, and you adjust the angle of the fan so that it lies along the 25o mark. Predict the new acceleration of the fan cart after these changes are made. Again, well use Newtons Second Law but , this time solve for . The mass is provided as 900 g
(0.900 kg), but we need to recognize that the only part of the force that matters for the fan cart on the track is that part of the force which is parallel to the track (and to the motion). Any part perpendicular to the track wont contribute to the acceleration since the grooves that the wheels are in provide a support force that prevents the cart from moving perpendicular to the track. To find the part of the force that is parallel to the track, we use the cosine of the angle. Well assume that the force itself doesnt change in magnitude.
So
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This probably should have a negative sign on it, to
indicate the same information about direction as discussed in part a).
Problem 5: Consider a force table like you used in Lab #2, with total mass 100 g hanging at 0o and total mass 200 g hanging at 90o. Predict the total mass (in g) and the angle (measured counterclockwise from 0o) at which you should hang this mass in order to balance the central ring. You could do this graphically using rulers and protractors, or you could do this with geometry/trigonometry.
The first thing to do is to draw a vector diagram. Lets start by drawing a tail-to-tail version (Fig. A, next page), and then go to a head-to-tail version (Fig. B, next page). In the head-to-tail version, its easier to see what vector is needed to balance out the other vectors and give a vector sum that adds up to zero (Fig. C, next page). Then, (Fig. D, next page) go back to the tail-to-tail version to see what the angle needs to be as measured from 0o, though you could just do this from Fig C. Also, as discussed in that lab, we can see that the length of the vectors is proportional to the weight hanging from the string, and the weight hanging from the string is proportional to the mass hanging from the string. Once weve drawn these figures, we can use a ruler and protractor to make our measurements (we need to remember to convert length of the arrow into mass, but that is just a straight proportion), or we can use geometry/trigonometry to calculate the length of the arrow (which is the hypotenuse of a right triangle) as well as the angle. Using the Pythagorean Theorem, we can calculate the hypotenuse of the right triangle (shown in Fig. C), since we know the lengths of the two legs are 100 g and 200 g: . We can calculate the angle (shown in Fig. C) by recalling any of our trigonometric relations, since know we know all the sides of the right triangle. Ill use tangent: , so .
However, we want the angle as measured from 0o. So looking at Fig. D, we can see that the angle we want is actually 270o 26.6o = 243.4o. Putting this all together, we want to hang 224 g at an angle of 243.4o as measured from 0o. Problem 6: Two cylindrical rods are identical in mass, color, size, shape, and appearance in every way. The only difference is that in one rod, the mass is concentrated in the center, but in the other, the mass is concentrated at the ends. Describe a simple (non-destructive) method you could use to determine which rod is which. Recall the YouTube video we saw of the ice skater, where when her arms are pulled to her chest, she spins quickly, and when her arms are extended, her spin slows down. From Newtons Second Law for Rotations: , so a different rotational mass means that the same torque results in different angular accelerations. If you spin both rods, the rod with the higher rotational mass has the smaller angular acceleration and will spin more slowly. Since rotational mass depends on mass and where the mass is located, the rod with the mass concentrated at its ends will spin more slowly.