Chapter 3.8 & 6 Solutions

Transcription

1 Chapter 3.8 & 6 Solutions P3.37. Prepare: We are asked to find period, speed and acceleration. Period and frequency are inverses according to Equation To find speed we need to know the distance traveled by the speck in one period. Then the acceleration is given by Equation Solve: (a) The disk s frequency can be converted as follows: The period is the inverse of the frequency: (b) The speed of the speck equals the circumference of its orbit divided by the period: which rounds to (c) From Equation 3.30, the acceleration of the speck is given by : Which rounds to. In units of g, this is as follows: Assess: The speed and acceleration of the edge of a CD are remarkable. The speed, is about As you will learn in chapter 4, very large forces are necessary to create large accelerations like P3.77. Prepare: We will use Equation 3.30 to relate the acceleration to the speed. But first we need to convert the speed of the car to Solve: (a) Your acceleration is given from the equation which converts as follows: The acceleration is or (b) The formula for centripetal acceleration, can be solved for v as follows: In this form we see that if the acceleration is doubled, then the velocity is multiplied by So we multiply the speed limit by At the acceleration would be twice the acceleration at

2 Assess: As noted in the solution to Problem 40, a small change in velocity can produce a large change in centripetal acceleration. Here, with an increase in speed of less than the acceleration doubles and the friction needed for the turn also doubles. Q6.22. Reason: There must be a centripetal force acting on the car directly toward the center of the circle. There are no other forces on the car beside the normal force and the weight, which act in the vertical direction. The correct choice is E. Assess: Since the car is going around the curve with constant speed, it is not accelerating in the direction tangent to the curve. This eliminates choices A, C, and D. Choice B would represent a centrifugal force, which seems to push the car out of the circular path. As discussed in Section 6.4, such a force actually does not exist. Q6.31. Reason: The skater turns one-and-a-half revolutions in 0.5 s. One-and-a-half revolutions is Her angular velocity is radians. The correct choice is D. Assess: This result is reasonable. She makes three revolutions in one second, which is radians per second. Q6.32. Reason: We first find (in rad/s) and then the centripetal acceleration. Let s make an estimate (one significant figure) for how far the hand is from the center of the body (the axis of rotation). Either think of a meter stick, or get one. From the center of your body to your hand is about 0.8 m for an average person. Now use Equation 6.8: The correct choice is C. Assess: We kept an extra guard digit in the intermediate calculations, but rounded the final result to one significant figure since that is all the data justifies. The result of 300 m/s 2 is large about 30 g. If your whole body experienced 30 g for more than a few seconds you would black out. However, this is only the skater s hand and only for a short time. Q6.33. Reason: In Question 6.31 we found that the angular velocity of the skater is 20 rad/s. Estimating the length of her arm to be about 0.75 m from the center of rotation, we can calculate the speed of her hand with Equation 6.7. The correct choice is D. Assess: This is actually a high velocity, over 30 mph. P6.3. Prepare: To compute the angular speed takes an hour to complete one revolution. Solve: we use Equation 6.3 and convert to rad/s. The minute hand Assess: This answer applies not just to the tip, but the whole minute hand. The answer is small, but the minute hand moves quite slowly. The second hand moves 60 times faster, or 0.10 rad/s. This too seems reasonable.

3 P6.13. Prepare: The horse and rider are in uniform circular motion. We are given A preliminary calculation will determine in rad/s for part (b): Solve: (a) Solve Equation 6.4 for (b) Use Equation 6.7: Assess: A time for two revolutions of 20 s seems reasonable; a speed of 2.5 m/s also seems reasonable. Note that for part (a) the answer is independent of the radius; it takes 20 s for everything to go around twice, not just the bucking horse. P6.15. Prepare: The pebble is a particle rotating around the axle in a circular orbit. We will use Equations 6.7 and 6.8. To convert units from rev/s to rad/s, we note that 1 rev = 2π rad. Solve: The pebble s angular velocity The speed of the pebble as it moves around a circle of radius r = 30 cm = 0.30 m is The centripetal acceleration is P6.17. Prepare: Equation 6.10 tells us the tension: Because all four are moving at the same speed, we need only consider the effect of m and r on T. A small r and a large m would make for a large T, as in case 3. Solve: Assess: Case 4 is the same as case 1 because both the mass and radius are doubled. P6.20. Prepare: We are given and A preliminary calculation using Table 1.3 will give v in m/s. Solve: (a) to two significant figures. (b) With the two forces on the ball being its weight and the force exerted by the hand, apply Newton s second law at the lowest point and solve for

4 Since the hand is providing the centripetal force, the direction is up when the ball is at the bottom of the circle. Assess: We check to see that we answered all parts of the problem: We gave the centripetal acceleration and the magnitude and direction of the force exerted by the hand. The centripetal acceleration seems large (200 g), but the force exerted by the hand seems reasonable, so everything is probably correct. The units check out. P6.23. Prepare: Model the passenger in a roller coaster car as a particle in uniform circular motion. A pictorial representation of the car, its free-body diagram, and a list of values are shown below. Note that the normal force of the seat pushing on the passenger is the passenger s apparent weight. Solve: Since the passengers feel 50% heavier than their true weight, the net force at the bottom of the dip is: Thus, from Newton s second law, Assess: A speed of 12 m/s or 27 mph for the roller coaster is reasonable. P6.26. Prepare: We will use the particle model for the test tube, which is in uniform circular motion. The radius to the end of the tube from the axis of rotation is 10 cm or 0.1 m. We will use Equation 6.8 and kinematic equations and work with SI units. Solve: (a) The acceleration is (b) An object falling 1 meter has a speed calculated as follows: When this object is stopped in upon hitting the floor, This result is one-fourth of the above radial acceleration in part (a). Assess: The radial acceleration of the centrifuge is large, but it is also true that falling objects are subjected to large accelerations when they are stopped by hard surfaces. P6.31. Prepare: Model the sun (s) and the earth (e) as spherical masses. Due to the large difference between your size and mass and that of either the sun or the earth, a human body can be treated as a particle. Use Equation Solve: and Dividing these two equations gives

5 Assess: The result shows the smallness of the sun s gravitational force on you compared to that of the earth. P6.42. Prepare: We will use Equation 6.9 and Newton s second law. The electric force between the electron and the proton causes the centripetal acceleration needed for the electron s circular motion. Solve: Newton s second law is Substituting into this equation yields: Assess: This is a very high number of revolutions per second. P6.48. Prepare: Treat the coin as a particle which is undergoing circular motion. A visual overview of the coin s circular motion is shown below in the following pictorial representation, free-body diagram, and list of values. The force of static friction between the coin and the turntable, as long as the coin does not slide, causes the centripetal acceleration needed for circular motion. The force of static friction is This force is equivalent to the maximum centripetal force that can be applied without sliding. Work with SI units. Solve: That is, So, the coin will stay still on the turntable. Assess: A rotational speed of approximately 1 rev per second for the coin to stay stationary seems reasonable. P6.51. Prepare: Since the hanging block is at rest, the total force on it is zero. The two forces are the tension in the string, and the weight of the puck, Since the revolving puck is moving at constant speed in a circle, the total force on the puck is the centripetal force. We must write the equations and solve them. Solve: The total force on the block is From Newton s second law, the total force is zero so we write: The centripetal acceleration of the puck is caused by the tension in the string, so We solve this to obtain:

6 The puck must rotate at a speed of Assess: It is remarkable that a block can be supported by a puck moving horizontally. But both the puck and the block are able to pull on the string the block pulls downward on one end and the puck pulls outward on the other end. The relatively small mass of the puck is compensated by its high speed of P6.53. Prepare: Treat the car as a particle which is undergoing circular motion. The car is in circular motion with the center of the circle below the car. A visual overview of the car s circular motion is shown below in the following pictorial representation, free-body diagram, and list of values. Solve: Newton s second law at the top of the hill is This result shows that maximum speed is reached when n = 0 and the car is beginning to lose contact with the road. Then, Assess: A speed of 22 m/s is equivalent to 50 mph, which seems like a reasonable value. P6.57. Prepare: We expect the centripetal acceleration to be very large because is large. This will produce a significant force even though the mass difference of 10 mg is so small. A preliminary calculation will convert the mass difference to kg: 10 mg = kg. If the two samples are equally balanced then the shaft doesn t feel a net force in the horizontal plane. However, the mass difference of 10 mg is what causes the force. We ll do another preliminary calculation to convert Solve: The centripetal acceleration is given by Equation 6.9 and the net force by Newton s second law. Assess: As we expected, the centripetal acceleration is large. The force is not huge (because of the small mass difference) but still enough to worry about. The net force scales with this mass difference, so if the mistake were bigger it could be enough to shear off the shaft.