Physics w/ Collins Exam 2

Problem 15. (OSC Example 6.4, p 209) (a) Calculate the Fnet providing the necessary centripetal force on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s. (b) Assuming an unbanked curve, find the minimum static coefficient of friction, between the tires and the road, static friction being the reason that keeps the car from slipping (see figure below).

(a) The centripetal acceleration needs to be ac=v2/r In the figure friction is to the left, keeping the car from slipping. The maximum static friction (of the tire surface against the road) is µsN, Here the normal force equals the car's weight, N = mg, so the net force is µsmg: Fnet = mac or µsmg = mv2/r Solving for µs, µs = v2/gr = 0.13

Problem 26. (OSC) (a) An automobile with 0.260 m radius tires travels 80,000 km before wearing them out. How many revolutions do the tires make, neglecting any backing up and any change in radius due to wear? (b) Remarkably, each of your tires leaves a onemolecule thick residue of rubber wherever it goes. If the tires in part (a) lose 1.0 cm in radius over their lifetime, what is the thickness of a single rubber molecule?

(a) The number of revolutions is x = N(2πr) ⇒ N = x 2πr = 5 × 107 rotations. (b) The thickness of a rubber molecule must be 1.0 cm/5 × 10^7 rotations = 0.2 nm

Problem 22. (FG) A car is going over the top of a hill. The hill has a radius of curvature Rc = 100 m. (a) On the figure, make a free-body diagram of the car, drawing in all forces as vectors in your picture. (As usual, do not include ma as a force!) (b) Solve F = ma in the vertical direction to find the speed v at which the car just becomes airborne. Problem 23. (OSC-CQ06-01) Race car drivers routinely cut corners as shown in the figure below. Explain how this allows the curve to be taken at the greatest speed.

(b) I'll take + downwards. Newton's law, F = ma, gives mg − N = mv2 Rc But setting N = 0 to just become airborne, mg = mv2 Rc v = p gRc = p (9.8 m/s2)(100 m) = 31.3 m/s

Problem 23. (OSC-CQ06-01) Race car drivers routinely cut corners as shown in the figure below. Explain how this allows the curve to be taken at the greatest speed.

. Friction of the tires on the road can provide only up to a certain amount of inward centripetal acceleration. Path 2 is less sharply curved than Path 1; i.e. has a larger radius of curvature r. Since centripetal acceleration goes down with r and up with vt (ac = mv2 t /r), the larger r of Path 2 allows a greater speed.

Problem 25. (OSC) Microwave ovens rotate at a rate of about 6 rev/min. What is this in revolutions per second? What is the angular velocity in radians per second?

. ω = 6 rpm = 0.1 rps = 0.63 rad/s.

Problem 24. (OSC) Semi-trailer trucks have an odometer on one hub of a trailer wheel, which counts the number of wheel revolutions, and then calculates the distance traveled. If the wheel has a 1.15 m diameter and goes through 200,000 rotations, how many kilometers should the odometer read?

. ∆θ = ∆s/r, so that ∆s = ∆θ × r = (1.257×106 rad)(0.575 m) = 7.226×105 m = 723 km.

Problem 5. (OSC-CQ06-01) There is an analogy between rotational and linear physical quantities. What rotational quantities are analogous to displacement and velocity?

5. Angle θ (in radians) is analogous to displacement. Angular velocity ω (in radians/sec) is analogous to velocity.

Problem 38. (OSC) What percentage of the acceleration at Earth's surface is the acceleration due to gravity at the position of a satellite located 300 km above Earth?

: Ratio of acceleration due to gravity at two locations: sea level r1 and 300 km up (g2/g1) = (GM/r2^2)/(GM/r1^2)=0.912=91.2% less at position 2.

Problem 8. A car coasts at a constant speed over a circular hill. Which of the free-body diagrams in the figure is correct?

C, slightly closer to top.

Problem 12. (OSC-CQ06-05) If centripetal force is directed toward the center, why do you feel that you are 'thrown' away from the center as a car goes around a curve?

Centripetal force is the inward force needed to make something move in a circle. If your car curves left, something needs to provide this inward force to make you curve with the car. You feel "thrown" outward (to the right) in your seat belt, but actually the car is accelerating inward, to the right, and the seat belt is pulling you along with the car.

Problem 43. (OSC) Calculate the centripetal force on the outermost 4 kg of mass at the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev/s

Fc = mω2r = (4 kg)(0.5×6.28 rad/s)2 (100 m) = 3.9×103 N

Problem 6. (FG) For uniform circular motion, the acceleration (Check all that are true): is parallel to the velocity. is directed toward the center of the circle. is larger for a larger orbit at the same speed. is always due to gravity. is always negative.

Is directed toward the center of the circle

Problem 20. Derive a formula for the speed at which a car can go around the banked turn in the figure below, without any friction at all. (The banked slope could be made out of slippery ice!) The FBD has been drawn in for you.

Looking at the picture, the necessary inward Fnet is provided by the horizontal component of the normal force * N, Fnet = mac i.e. N sin θ = mv2/r We need to put in N. Careful: due to the slope, N is not just mg! In the above figure, the net vertical force must cancel out (there is no up/down acceleration) so we can say N cos θ = w (= mg) ⇒ N = mg/cos θ Plug this N into the previous equation: (mg/cos θ)= sin θ = mv2/r Solve this to get a formula for the optimal velocity: v = p rg tan θ (square root)

Problem 35. (OSC) An ordinary workshop grindstone has a radius of 7.50 cm and rotates at 6500 rev/min. (a) Calculate the magnitude of the centripetal acceleration at its edge in meters per second squared and convert it to multiples of g . (b) What is the linear speed of a point on its edge?

Same as 34

Problem 11. When a car turns a corner on a level road, which force(s) provides the necessary centripetal acceleration? (Check all that are true): Normal force Static friction Kinetic friction Air resistance Gravity

Static friction

Problem 10. (OSC Example 6.3, p 207) Calculate the centripetal acceleration of a point 7.50 cm from the axis of an ultracentrifuge spinning at 7.5×104 rev/min. Determine the ratio of this acceleration to that due to gravity.

The angular frequency of the ultracentrifuge is ω = 7.5 × 104 rev min × 2p rad 1 rev × 1 min 60.0 s = 7854 rad/s. Because ω and r are given, of the two possible formulas ac = v 2 r and ac = rω2 the second one is the most convenient, and gives ac = 4.63 × 106 m/s 2 . Taking the ratio of this to g yields ac/g = 4.63×106 9.80 = 4.72×105 . The ultracentrifuge provides an astounding 472,000 g's of acceleration.

Problem 21. (OSC Example 6.5, p 211) (a) Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked, allowing them to be taken at very high speed. From the preceding problem, calculate the speed at which a 100 m radius curve banked at 65.0° (much steeper than the preceding figure) should be driven if the road is frictionless.

The answer is v = 45.8 m/s = 103 mph.

Problem 7. (OSC Example 6.2, p 206) What is the magnitude of the centripetal acceleration of a car following a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h)? Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed.

The two possible formulas are ac = v 2 r and ac = rω2 Because v and r are given, the first is the most convenient to use: ac =v2/r= 1.25 m/s2

Problem 27. (OSC) (a) What is the period of rotation of Earth in seconds? (b) What is the angular velocity of Earth? (c) Given that Earth has a radius of 6.4×106 m at its equator, what is the linear velocity at Earth's surface?

a) 24 h=86400 s. (b) ω = 1 rev 24 h × 2p rad/rev 86400 s/rev = 7.3 × 10−5 rad/s (c) v = ωr = 470 m/s (= 1050 mph)

Problem 42. (OSC) (a) A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force must she exert to stay on if she is 1.25 m from its center? (b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at 3.00 rev/min if she is 8.00 m from its center? (c) Compare each force with her weight

a) Fc = mrw^2 = ... 483 N b) Fc' = mr' w'^2 = 22x8x(3x2pi/60.. all squared) = 17.4 N c) Fc/w = Fc / mg = 2.24 Fc' = same thing but use b for Fc

Problem 34. (OSC) The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second if it spins at 1200 rev/min? (b) What is the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac? (c) What is the centripetal acceleration of the propeller tip under these conditions? Calculate it in meters per second squared and convert to multiples of g.

a) conversions rev/min * rad/rev *min/sec b) v =r*w c)ac/g

Problem 39. (OSC) Verify that the linear speed of an ultracentrifuge is about 0.50 km/s, and Earth in its orbit is about 30 km/s by calculating: (a) The linear speed of a point on an ultracentrifuge 0.100 m from its center, rotating at 50,000 rev/min.(b) The linear speed of Earth in its orbit about the Sun (use data from the text on the radius of Earth's orbit and approximate it as being circular).

a) v= rw ( final answer in km/s)(0.524) b) v= rwn (29.7)

Problem 41. (OSC) At takeoff, a commercial jet has a 60.0 m/s speed. Its tires have a diameter of 0.850 m. (a) At how many rev/min are the tires rotating? (b) What is the centripetal acceleration at the edge of the tire? (c) With what force must a determined 1.00×10-15 kg bacterium cling to the rim? (d) Take the ratio of this force to the bacterium's weight.

a) v= rw = 141.2 rad/s.. conversions... 1.35 x 10^3 rpm b) ac = rw^2 = 8.47 x 10^3 m/s c) Fc = ma = 8.47 x 10^-12 d) Fc/ m g = 865

Problem 37. (OSC) Olympic ice skaters are able to spin at about 5 rev/s. (a) What is their angular velocity in radians per second? (b) What is the centripetal acceleration of the skater's nose if it is 0.120 m from the axis of rotation? (c) An exceptional skater named Dick Button was able to spin much faster in the 1950s than anyone since—at about 9 rev/s. What was the centripetal acceleration of the tip of his nose, assuming it is at 0.120 m radius? (d) Comment on the magnitudes of the accelerations found. It is reputed that Button ruptured small blood vessels during his spins

a) w= 5 rev/s * 2pi rad / 1 rev = 31.4 rad/s b) ac = rw^2 = (0.12 m)(31.4 rad/s)=118 m/s^2 c) ac= rw^2 (d) The centripetal acceleration felt by Olympic skaters is 12 times larger than the acceleration due to gravity. That is quite a lot of acceleration in itself. The centripetal acceleration felt by Button's nose was 39.2 times larger than the acceleration due to gravity! It is no wonder that he ruptured small blood vessels in his spins

Problem 32. (OSC) A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of 30 m. If he completes the 200 m dash in 23.2 s and runs at constant speed throughout the race, what is the magnitude of his centripetal acceleration as he runs the curved portion of the track?

ac = v 2r = (200 m/23.2 s)2/30 m = 2.5 m/s2

Problem 40. (OSC) A rotating space station is said to create "artificial gravity"—a loosely-defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an "artificial gravity" of 9.80 m/s2 at the rim?

ac=rw^2 = g --> w = (square root) (g/r) = 9.80/100 m ^1/2 = 0.313 rad/s

Problem 45. (OSC) What is the ideal speed to take a 100 m radius curve banked at a 20.0° angle?

formula : tan (θ) = v^2/rg v=(squareroot) rgtanθ = 18.9 m/s

Problem 19. On a carnival ride, the rider is held up against the side of a rotating cylinder of radius R. The coefficient of static friction is µs. What must be the minimum tangential speed v of the rider to stay up? Derive a formula in terms of g, R and µs.

v = (square root) g*R/us

Problem 44. (OSC) What is the ideal banking angle for a gentle turn of 1.20 km radius on a highway with a 105 km/h speed limit (about 65 mi/h), assuming everyone travels at the limit?

v = 105 km/h * 1000 m / 1 km * 1 h / 3600 s = 29.17 m/s^2 θ= inverse tan of (v(29.17 m/s)^2 / 1200 m * 9.80 m/s ^2) = 4.14 degrees

Problem 29. (OSC) In lacrosse, a ball is thrown from a net on the end of a stick by rotating the stick and forearm about the elbow. If the angular velocity of the ball about the elbow joint is 30.0 rad/s and the ball is 1.30 m from the elbow joint, what is the velocity of the ball?

v = ωr = 39.0 m/s (= 87 mph).

Problem 14. (modif OSC Example 6.4, p 209) A car goes around a level curve of radius R = 100 m. For tires on the road, the static and kinetic friction coefficients are µs and µk. (a) Make a free-body diagram of the car, drawing in all forces as vectors in your picture. (b) Write Newton's law in two directions (vertical direction and horizontal) to get a formula for the maximum speed v the car can have before slipping. (Choose the right type of friction coefficient!)

where fs = µsn is the static friction. (b) Newton's law, F = ma, gives in the vertical (y) direction n − mg = 0, or just n = mg. In the horizontal (x) direction Newton's law gives fs = mv2 R or, since fs = µsn = µsmg, µsmg = mv2 R v= square root of µsgR

Problem 31. (OSC) A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 8.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 1.50 times that due to gravity?

ω = (square root) Ac /r ...rev --> rad x 60 s/1 min = 728 rev/s =12.9 revmin

Problem 33. (OSC) Taking the age of Earth to be about 4×109 years and assuming its orbital radius of 1.5×1011 m has not changed and is circular, calculate the approximate total distance Earth has orbited since its birth.

ω = 2p/(1 year). x = vt = rωt = (1.496×1011m)(2prad/y)(4×109 y) = 3.76×1021m. This is about half a million light years, two hundred times further away than the closest galaxy

Problem 28. (OSC) A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in the pitcher's hand is 35.0 m/s (78.4 mph) and the ball is 0.300 m from the elbow joint, what is the angular velocity of the forearm?

ω = v/r = 117 rad/s

Problem 30. (OSC) A truck with 0.420-m-radius tires travels at 32.0 m/s. What is the angular velocity of the rotating tires in radians per second? What is this in rev/min?

ω= v/r = 32.0 m/s /0.420 m =76.2 rad/s ω = 76.2 rad/s * 1 rev/ 2pi * 60sec / 1 min = 728 rev/s = 728 rpm