Unformatted text preview: Main Results of This Chapter
!=
d" dt f = ! # ! = 2" f 2" T= 1 f ! = 2" f = 2" T Physics for Scientists & Engineers 1
Spring Semester 2006 Lecture 32 != v = r! d" d 2# = 2 dt dt r ^ a = r! t^ " r# 2 r at = r!
ac = r! 2 = v2 r March 19, 2006 Physics for Scientists&Engineers 1 1 March 19, 2006 Physics for Scientists&Engineers 1 2 Constant Angular Acceleration
! Kinematical equations for constant angular acceleration are obtained in complete analogy to those for linear motion with constant acceleration Example: Hammer Throw
! Throw the "hammer", a 12 cm diameter ball attached to a grip by a steel cable, a maximum distance. The hammer's total length is 121.5 cm, and its total weight is 7.26 kg (4 kg for women). The athlete has to accomplish the throw while not leaving a circle of radius 7 feet (= 2.135 m) ! = ! 0 + " 0t + 1 # t 2 2 ! = ! 0 + "t " = " 0 + #t " = 1 (" + " 0 ) 2
2 " 2 = " 0 + 2# (! $ ! 0 ) Sergei Litvinov (USSR) 84.80m, Seoul `88 Yipsi Moreno (Cuba) 73.33m, Paris `03 http://www.hammerthrowing.co.uk/
March 19, 2006 Physics for Scientists&Engineers 1 3 March 19, 2006 Physics for Scientists&Engineers 1 4 Hammer Throw (2)
! Litvinov took seven turns before releasing hammer. These took 1.52 s, 1.08 s, 0.72 s, 0.56 s, 0.44 s, 0.40 s, and 0.36 s ! Assume frequency increases linearly with time => constant angular acc. ! Question: what is value of !? ! Answer: total time = 1.52 s+...+ 0.36 s = 5.08 s; total angle: Constant angular acceleration:
! = 1 " t2 # " = 2
2! 2 $ 44.0 = = 3.41 s2 2 2 t (5.08 s)
Physics for Scientists&Engineers 1 5 Example: Hammer Throw (3)
! Question: Assuming that the radius of the circle on which the hammer moves is 1.67 m (= length of hammer + arms of the athlete), what is the linear speed with which the hammer gets released? ! Answer: Under constant angular acceleration from rest for a period of 5.08 s, the final angular velocity reached is ! all = 7 " 2# = 14# $ 44.0 ! = " t = (3.41 s2 ) # (5.08 s)=17.3 s1 ! Velocity:
v = r! = (1.67 m) " (17.3 s1 ) = 28.9 m/s
March 19, 2006 Physics for Scientists&Engineers 1 6 March 19, 2006 Example: Hammer Throw (4)
! Question: What is the centripetal acceleration and centripetal force that the hammer thrower has to exert on the hammer right before the hammer gets released? ! Answer: The centripetal acceleration right before release is given by:
ac = v! = (28.9 m/s) " (17.3 s1 ) = 500. m/s 2 Quiz 32.1
! A Ferris wheel is rotating with a constant angular speed. Which person feels the largest net centripetal force (magnitude)? ! With a mass of 7.26 kg for the hammer, the centripetal force required is then
Fc = mac = (7.26 kg) ! (500 m/s 2 ) = 3630 N A) At the top B) At midheight C) At the bottom D) They all feel the same net centripetal force. ! Same force as weight of 370 kg object!
March 19, 2006 Physics for Scientists&Engineers 1 7 March 19, 2006 Physics for Scientists&Engineers 1 8 Quiz 32.2
! A Ferris wheel is rotating with a constant angular speed. Which person feels the largest normal force from the seat (magnitude)? Roller Coaster Loop
! What is the physics behind this? A) At the top B) At midheight C) At the bottom D) They all feel the same normal force. March 19, 2006 Physics for Scientists&Engineers 1 9 March 19, 2006 Physics for Scientists&Engineers 1 10 Roller Coaster Loop
! Why does the cart not fall off the track at the top? ! For a given speed v at the top, the centripetal force is v2 Fc = m r Quiz 32.3
Dale Earnhardt Jr drives 150 mph around a circular track at constant speed. Which vector might describe the force the tires experience from contact with the pavement? a) Vector A b) Vector B c) Vector C B A r v ! This force is provided by the normal force + weight v2 m = N + mg r ! In the limit of zero normal force, we obtain the minimum speed at the top needed to stay on the track: r r N Fg r Fc C v = rg
March 19, 2006 Note: larger radius requires larger speed
Physics for Scientists&Engineers 1 11 Quiz 32.4
Dale Earnhardt Jr drives 150 mph around a circular track at constant speed. Which vector might describe the force Dale experiences from the seat? a) Vector A b) Vector B c) Vector C B A Cutting Curves
C ! Cars move through Uturn shown at constant speed. Coefficient of friction between the tires and the road is s= 0.9. Inner radius of curve r = 8.3 m, outer radius R = 22.2 m. How much time will it take to move from point A to A', and how much from point B to B'? ! Answer: Friction force needs to Provide centripetal force ! ! ! ! ! v2 " v = sgR R Red curve: vmax=14 m/s (30 mph) Blue: vmax=8.56 m/s (20 mph) (Tight curve!) Now we need to figure total time for both paths! s g = March 19, 2006 Physics for Scientists&Engineers 1 14 Cutting Curves (2)
! Length of red and blue trajectories:
l blue = ! RB + 2(RA " RB ) = 53.9 m l red = ! RA = 69.7 m Cutting Curves (3)
! More realistic: blue trajectory can accelerate on straight segments Driver will come to B with speed that allows to decelerate to 8.56 m/s at entrance of circular piece v B = v 2 + 2sg(RA " RB )
= 17.8 m/s ! More realistic blue line time: ! Time required:
t red = l red 69.7 m = = 4.98 s vred 14.0 m/s !
! tblue = tblue l 53.9 m = blue = = 6.30 s vblue 8.56 m/s ! RB 2(RA " RB ) + = 5.15 s vblue 1 (vB + vblue ) 2 ! Best approach: elliptical cut through curve March 19, 2006 Physics for Scientists&Engineers 1 15 March 19, 2006 Physics for Scientists&Engineers 1 16 Banked Curves
! Question: ! If the coefficient of friction between the racetrack surface and the tires of the racecar is 0.62 and the radius of the turn is 110 m, what is the maximum speed with which the driver can take this curve banked at 21.1? ! Answer: all forces (friction, gravity, normal force) add up to centripetal force, which determines the maximum speed.
March 19, 2006 Physics for Scientists&Engineers 1 17 Banked Curves (2)
! Equations for components:
y : ! mg cos" + N = m v2 v2 sin " # N = m sin " + mg cos" R R v2 x: mg sin ! + s N = m cos ! R v2 v2 mg sin ! + s m sin ! + s mg cos ! = m cos ! " R R v2 g (sin ! + s cos ! ) = (cos ! # s sin ! ) " R
v= Rg (sin ! + s cos ! ) cos ! " s sin ! Mass m cancels out again!
March 19, 2006 Physics for Scientists&Engineers 1 18 Banked Curves (3)
! Insert numbers:
v= (110. m) ! (9.81 m/s 2 ) ! (sin 21.1 + 0.62 ! cos 21.1) = 41.9 m/s cos 21.1 " 0.62 ! sin 21.1 ! Compare without banking:
v = (110. m) ! (9.81 m/s 2 ) ! 0.62 = 25.9 m/s March 19, 2006 Physics for Scientists&Engineers 1 19 ...
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This note was uploaded on 03/19/2008 for the course PHY 183 taught by Professor Wolf during the Spring '08 term at Michigan State University.
 Spring '08
 Wolf
 Physics

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