Chapter 5 Homework

Fall 2007

Questions 6, 10, 12, 17 (p.129)

Problems 1, 5, 9, 16, 29, 30, 33, 34, 39, 40, 43, 49, 51, 61, 69, 77 (p.130-134)

Extra Credit: prob. 23, 41 (p.131-132)

Note: I would ordinarily draw pictures for ALL of these problems and I expect you to do so. I simply do not have the time myself to do so.

Questions:

6. You have 3 accelerators on your car: The gas pedal, the brake pedal, and the steering wheel. You should be able to explain why in each case.

10.The girl should let the ball go just as the velocity vector is pointing at her target, NOT when the ball is aligned with the target. Remember Newton’s 1^st Law!

12.If the mass of the Earth were doubled, the force of gravity on an object at or above its surface would double (see Newton’s Law of Gravity on p.119). If the force doubled, the object should have twice the acceleration, which would show up as increased velocity (by a factor of ; why?).

17.The centripetal acceleration of an object is equal to the square of the velocity divided by the radius of the orbit. Set the gravitational force equal to the centripetal force and solve for velocity:

Since R_M > R_E , V_E > V_M . Since the centripetal acceleration is proportional to the square of the velocity, the acceleration of Earth must be larger since V_E is larger.

Problems:

1. A child sitting 1.10 m from the center of a merry-go-round moves with a speed of 1.25 m/s. Calculate a) the centripetal acceleration of the child, and b) the net horizontal force exerted on the child (mass = 25.0 kg). To solve a), use the centripetal acceleration equation (p.107). To solve b), recall Newton’s 2^nd Law. (See p.A-29 for answers).

5. Suppose the space shuttle is in orbit 400 km from the Earth’s surface, and circles the Earth about once every 90 minutes. Find a_c of the space shuttle in orbit. Express you answer in terms of g as a percentage. This problem is a little trickier. You need to use the centripetal acceleration equation and substitute the fact that the velocity is related to the distance around the circle – its circumference – and the period through

Substitute into the equation for centripetal acceleration to derive the following:

Substitute the period (90 minutes = 5400 s) and the correct value for r, which would be the distance from the shuttle to the center of the Earth, or 400 km + 6380 km = 6780 km or 6.78 x 10⁶ meters. The acceleration will turn out to be 9.18 m/s², which is 0.9 g’s.

9. To solve this problem, you have to understand that the centripetal force is being supplied by the static friction between the tires and the road, or

(The vertical forces – the weight and normal forces – are equal to each other. Do the drawing and confirm this result!). The mass cancels, and solving for the velocity, you should get 24.6 m/s. This result is independent of the mass of the vehicle. (Don’t forget to take the square root!).

16. This “bucket problem” is pretty much identical to Example 5-4. A diagram reveals that the centripetal force at the bottom is equal to the tension – the weight of the bucket, or F_T – mg. Set those equal and solving for velocity should give you

At the top, the minimum speed would be when the tension goes to zero, reducing the equation to

Solving for v should give you 3.28 m/s. Note that this result is dependent on the radius only.

29. a) On either planet, the mass is still 21.0 kg. Remember – mass is a property of matter and is independent of its location.. b) Weight is simply mass times the acceleration of gravity, so on Earth, the ball would weigh (21.0 kg)(9.8 m/s²) = 206 N and on the planet, the ball would weigh (21.0 kg)(12.0 m/s²) = 256 N.

30. The weight of an object is just the gravitational attractive force applied to that object. Set Newton’s Law of Gravitation equal to the weight (mg). The mass of the object will cancel leaving

Substitute the correct values to get g = 1.62 m/s² – a result that is ~1/6^th the value of g on Earth.

33. This question is a little trickier than some of the others. The first thing you need to do is use Newton’s Law of Gravity to solve for the product of their masses; you will get 0.234 kg². You also know that the sum of the masses is 4.0 kg. This gives us two equations and two unknowns (m₁ and m₂):

Solve the top equation for m₁ and substitute it into the bottom equation. A little algebra should get you to this:

This can be solved using the quadratic formula giving two possible solutions, 3.94 kg or 0.06 kg for m₂. This is your final answer. Whichever is chosen for m₂, the other becomes m₁.

34. Use the Law of Gravity and the values for Earth’s mass (5.94 x 10²⁴ kg) and radius (6.38 x 10⁶ m) to calculate the value of g using the same formula from #30 above. Using the correct radii, a) g = 9.79 m/s² and b) g = 4.35 m/s². Note that a small change in elevation (3200 m) does not produce a large effect, hence the need for very sensitive instruments to measure variations in g.

39. We worked out the solution to this problem in class; the total force should be 3.54 x 10^-8 N at an angle of 45° above the horizontal, or toward the diagonal mass.

40. This problem is very similar to #39, except that the objects are all in a line. The first order of business is to get the distance between each planet involved, i.e. you need to know the distance between Venus and Earth, Earth and Jupiter, and Saturn and Earth. We are given the distance from the Sun to each planet and their masses, but NOT these distances. What do we do? As an example, we get the Venus-Earth distance by subtracting the Venus-Sun distance from the Earth-Sun distance to get 42 x 10⁹ m. The other distances are

Earth-Jupiter = 628 x 10⁹ m

Earth-Saturn = 1280 x 10⁹ m

Calculate the force of gravity provided by each body on the Earth:

Force of Venus on the Earth:

Similarly, the force of Jupiter on Earth and Saturn on Earth are 1.92 x 10¹⁸ N and 1.39 x 10¹⁷ N, respectively. The total force on the Earth is therefore 1.92 x 10¹⁸ N + 1.39 x 10¹⁷ N – 1.10 x 10¹⁸ N = 9.59 x 10¹⁷ N. (Why do we subtract the force provided by Venus on the Earth?). The Sun provides a much larger force, 3.53 x 10²² N, which is ~37000x greater than that of the other planets combined (!!!).

43. The velocity of the satellite can be found by equating the centripetal force to the gravitational force and solving for v:

Substituting the value of G (6.67 x 10^-11 N-m²/kg²), m_E (5.98 x 10²⁴ kg) and r (6380 km + 3600 km = 9.98 x 10⁶ m) gives a velocity of 6.32 x 10³ m/s.

49. In this case, we are looking for the period of orbit at two different distances (r_in = 7.3 x 10⁷ m and r_out = 1.7 x 10⁸ m) given the mass of Saturn (5.7 x 10²⁶ kg). Just like the previous problem, we want to set the centripetal force equal to the gravitational force. However, this time we need to substitute for the velocity and solve for the period instead:

Substitute the numbers and solve for T for the inner and outer parts of the rings to get T_in = 2 x 10⁴ s and T_out = 7.1 x 10⁴ s. The mean rotation period for Saturn is given as 10 hours 39 minutes (3.8 x 10⁴ s), so the inner parts of the ring rotate not quite twice as fast as the planet, while the outer part rotates at about half the speed.

51. This problem is similar to the “elevator problem” in previous chapters with one exception: we are now well above the surface of the Moon and not the Earth. On Earth, we can use mg as the weight, and we know that mg actually becomes less with elevation. So, we can use Newton’s gravity equation to tell us what the acceleration will be at different distances above the surface of an object, i.e. “r” is the elevation in this case. If that’s the case, then the acceleration becomes: a = GM_Moon/h², where h is the elevation. The apparent weight will then be m x the total acceleration.

a) At a constant velocity, the only acceleration is that due to the gravity field of the Moon and its directed towards the surface of the Moon. Substituting values for the Moon’s mass, G, and the elevation (4200 km), the acceleration towards the moon should be 0.28 m/s². Therefore the weight of the astronaut would be 21 N.

b) In this case, the spaceship accelerating towards the moon would make the acceleration of the ship and the astronaut in opposite directions (recall what happens if an elevator accelerates downwards – you loose weight!). The apparent weight will be less by the magnitude of the acceleration times the mass (220 N) and will in fact be ~200 N away from the Moon’s surface (21 N – 220 N).

61. Once again, you set the centripetal force equal to Newton’s gravity equation and substitute 2πr/T for the velocity. This time you solve for the mass of Jupiter to get:

For each satellite, substitute the periods and mean distances (we ignore Jupiter’s radius for this one) to get ~1.9 x 10²⁷ kg in each case (see p.A-29 for exact answers).

69. Use the radius of the Earth and rotation period (in seconds) to determine the acceleration at the equator. You should get around 0.034 m/s², which is only around 0.35% of g(!!). So this effect is very small, but when you measure g on the order of 10^-7 m/s², you have to take even this small change into account.

77. To be “weightless”, the centripetal force would have to equal the weight of the person. (Why? Think normal force Setting those equal to each other gives v = (gR_E)^1/2. Plugging in the numbers, you should get 7900 m/s². This velocity is related to the period by the fact that the Earth turns through a complete circle each “day”, or v = 2πr/T, T = 2πr/v. This should give you 5.07 x 10³ s = 84.5 minutes = 1 hr 24 min 30 sec (Wow!).

Extra Credit:

23. To work this one, you’ll need to write sum of the forces equations for each block separately:

where a_c is the centripetal acceleration, T₁ is the tension towards the center from m₁, and T₂ is the tension towards the second block, m₂. The centripetal acceleration is v²/r = (2πr/T)²/r = 4π²r/T². But recall that f = 1/T, therefore a_c = 4π²rf². Substituting into the equations for a_c gives the following:

Therefore,

and

41. Set the weight of an object on Mars (mg_mars) equal to the Law of Gravity, then solve for the mass of Mars (M_M): , with g_mars = 0.38g or 3.7 m/s².