In the packet on gravity I handed out last week:
1) Review Example Problem 1 (on page 174), also review the work we did together last Monday to calculate the length of the martian year. Do Practice Problems 2 and 4 (also on page 174).
2) On Wednesday we used Newton's Law of universal gravitation to determine the exact value of the constant in Kepler's third law. (This same calculation is performed on page 176 of your packet). You can take this calculation a few steps further to get a formula for determining the period of a planet orbiting the sun. This formula is derived and listed on page 176 of you packet. Use it to do problem problem 6 in the 7.1 Section Review on page 178.
See you Wednesday.
Monday, January 19, 2009
Tuesday, January 13, 2009
Soulutuons to problems assigned 1/10
Problems from handout:
17. a) Towards center of circle
b) Towards center of circle
c) Static friction
61.
a) Since the car is moving in a circle you now the acceleration must be ac=v2/r. You have been given the radius (r=50m) of the circle but you have to find the velocity. The car's lap time is 14.3s and since the track goes around the circumference of a circle the distance the car travels each lap is 2π(r). The velocity (distance/time) is then 2π(50m)/14.3s=22m/s.Now just plug in: ac=(22m/s)2/50m = 9.68 m/s2
b) If you know the acceleration then finding the force is just a matter of multiplying the mass: Fc=mac (or Fc=mv2/r).
67. If the pilot is moving at the right speed the only force that will be acting on him at the top of the loop will be gravity -- because that's the only force needed to maintain his motion. The pilot will feel this as a sensation of weightlessness. In this special situation centripetal acceleration is exact provided by gravitational acceleration so that g=v2/r (remember g=9.8m/s2). This problem tells you the pilot's speed and asks you to find the radius:
g=v2/r rearrange and you get: r=v2/g= (120m/s)2/(9.8m/s2)=141.2.
Thats it!
17. a) Towards center of circle
b) Towards center of circle
c) Static friction
61.
a) Since the car is moving in a circle you now the acceleration must be ac=v2/r. You have been given the radius (r=50m) of the circle but you have to find the velocity. The car's lap time is 14.3s and since the track goes around the circumference of a circle the distance the car travels each lap is 2π(r). The velocity (distance/time) is then 2π(50m)/14.3s=22m/s.Now just plug in: ac=(22m/s)2/50m = 9.68 m/s2
b) If you know the acceleration then finding the force is just a matter of multiplying the mass: Fc=mac (or Fc=mv2/r).
67. If the pilot is moving at the right speed the only force that will be acting on him at the top of the loop will be gravity -- because that's the only force needed to maintain his motion. The pilot will feel this as a sensation of weightlessness. In this special situation centripetal acceleration is exact provided by gravitational acceleration so that g=v2/r (remember g=9.8m/s2). This problem tells you the pilot's speed and asks you to find the radius:
g=v2/r rearrange and you get: r=v2/g= (120m/s)2/(9.8m/s2)=141.2.
Thats it!
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