Answers, Physics 113 Exam #2

2D Kinematics, Newton’s Laws, Circular Motion

This sheet contains the answers to the first exam and hints about how to solve the problem. Answers and notes are given in italics. The complete questions are not given--see the original exam for the complete question.

1. Roller coasters work because of physics.
1. Draw a free body diagram for a person riding a roller coaster.

Normal force, n: normal force of rail holding up cart/person

See video on free body diagrams (Guthold) for how to draw diagram, video on Newton’s First Law (Macosko) for the types and definitions of forces and in-class exercises on Newton’s laws, where we drew free body diagrams.

For the cart/person as the “body”:

Note: If you defined your “body” as just the person, then the forces are gravitational and normal forces. If you made the diagram with the coaster still being pulled by a mechanical force, then that could be included. The exact answer depends on how you define the “body” for which you are drawing the diagram.

2. If all the forces that you identified in (a) are the “action” part of the pair, identify the “reaction force” that forms the action-reaction force pair.

See videos on Newton’s Third Law (Macosko) and Newton’s Law (Charnock) for the discussion of third law. Also see the in-class exercise on Newton’s third law.

“action” force	“reaction” force
Gravitational force, mg: gravitational force of earth pulling on cart and person; m is mass of cart and person	Gravitational force, -mg, gravitational force of cart and person pulling on earth; m is mass of earth
Normal force, n: normal force of rail holding up cart/person	Normal force, -n: normal force of cart/person on rail
Friction force, f: friction force of contact of rail on cart/person	Friction force, -f; friction force of cart wheels on rail

2. A skier skies down a slope towards a ski jump, like the one shown in the above picture. Just before he leaves the jump, the ramp flattens to an angle of 0.00° to the horizontal. He leaves the end of the ski jump with a velocity of 10.0 m/s. The slope below him (that he eventually lands on) is inclined at 35.0° to the horizontal.

This is a standard projectile motion problem. See in-class exercise on Galileo’s projectile, videos on 2-dimensional kinematics (Macosko) and projectile motion (Fetrow), and homework problems on projectile motion. See also, example problem 4.7, p. 90, Serway and Jewett, sixth edition.)

1. How far down the slope does the ski jumper land? 17.3 m
2. Calculate the magnitude of his final velocity just as he lands. 17.1 m/s
3. In terms of the physics you have learned about projectile motion, friction, and Newton’s Laws, describe why this [the skier in the shape of an airfoil] works. Air resistance is not negligible. In a free body diagram, it would be a resistive force in the direction opposing the skiers forward motion (opposing the x-velocity and thus causing the x-velocity to decrease, rather than remain constant). In an airfoil position, the skier is decreasing the air resistance, thus decreasing this force that opposes his/her forward motion.

3. Imagine a tennis ball whirling in a vertical circle on a string that is 0.60 m in length. The tennis ball has a mass 57 g and is whirled clockwise in a vertical circle in the x-y plane.

This problem contains concepts from both Newton’s laws, circular motion, and projectile motion. Most questions are taken directly from the in-class exercise on circular motion. See videos on free body diagrams (Guthold) for how to draw diagram;circular motion (Kerr); video on Newton’s First Law (Macosko) for the types and definitions of forces.

1. -0.6

   0.6

   0.6

   0.6

   Draw the x- and y-coordinates for the tennis ball, if it starts at the 6 o’clock position and travels 2 revolutions.
   
   
2. Draw the direction of the velocity vector and the direction of the centripetal acceleration vectors.

V

3. Draw a free body diagram for the ball in this position. Label and define your forces.

4. Write the force equations for the x- and y-components of the forces on the ball at the 9 o’clock position.

SFx = max = T	SFy = may
In x-direction, a = ac	In y-direction, a = -g
m ac = T	may = -mg = (0.057 kg)(-9.8 m/s^2)
V (v^2/R) = T
V = sqrt(T * 10.5 m/kg)

5. While whirling the ball, the string breaks when the ball is at the 9 o’clock position. Use an arrow to indicate the direction that the ball will travel if the string breaks just as the ball reaches the 9 o’clock position. Using a dotted line, indicate the path of the ball as you watch it travel until it hits the ground.

6. Justify your response to part (e) in terms of your answers to parts a-d and Newton’s laws.
   Direction of travel is same as velocity vector in part a—tangent to circle, or straight up when ball is at 9 o’clock position. After it is released, only gravitational force continues to act on it, so it goes straight up, and comes back down (1D motion with acceleration due to gravity).

4. Ryan Newman is driving his #12 Dodge race car on one of the NASCAR tracks. This track is flat (horizontal). The curve on the race track has a constant radius of 100 m. The coefficient of static friction between a fresh set of tires and the race track pavement is 1.0.
1. If Ryan’s car has a mass of 1542 kg, what is the maximum speed at which Ryan can make the curve without crashing into the wall? 31.3 m/s (70 mi/hr)
2. The coefficient of static friction with the pavement has decreased to 0.74. How fast can Ryan now negotiate this turn? 26.9 m/s (60.2 mi/hr)
3. Why are we calling the friction coefficient the coefficient of static friction, rather than the coefficient of kinetic friction? Under what condition would it be correct to call it the coefficient of kinetic friction? It is static friction because the car is not sliding. It becomes kinetic friction when his car starts sliding.
4. Draw a free body diagram illustrating Ryan taking the turn on a banked curve, banked at some angle, q, to the horizontal. Label all forces and define them.

Force	Equation	Components	Description
Gravitation	Fg = -mg	No x-component; Fg,y = -mg	Gravitational pull by earth on car
Normal	Fn = n	Fn,x = -n sinq; Fn,y = n cosq	Normal force between road and car
Friction	Ff = us n	Ff,x = -ms n cosq; Ff,y = -ms n sinq	Friction force between road and car

5. Describe how banking affects the maximum speed at which Ryan can take the turn. On the flat track, the normal force only has a y-component. There is no x-component to help prevent sliding. Only friction force prevents sliding. On a banked turn, the x-component of the normal force provides an additional force that is opposite the sliding. (See Example 6.5, page 155, Serway and Jewett, sixth edition.)