Exam 3 Results Approximate Grade Cutos Ø 75-1 A Ø 55 74 B Ø 35 54 C Ø 5 34 D Ø 4 F '$!" '#!" '!!" &!" %!" $!" #!"!"!"#$%&%'()*+(,-./% 13+#435%%6789:%!()" )('!" '!(')" ')(#!" #!(#)" #)(*!" *!(*)" *)($!" $!($)" $)()!" )!())" ))(%!" %!(%)" %)(+!" +!(+)" +)(&!" &!(&)" &)(,!",!(,)",)('!!" You can pick them up a/er class today or Wednesday
Chapter 14 Last Time Simple harmonic mo%on o a mass- spring system Dieren%al equa%ons and SHM Oscilla%ons in terms o amplitude, period, requency and angular requency Simple harmonic mo%on using energy Today Simple pendulums Physical pendulums Ch 14 Examples Start Ch 15 (i we get to it)
Clicker QuesEon Review rom last Monday To double the total energy o a mass- spring system oscilla%ng in simple harmonic mo%on, the amplitude must increase by a actor o A. 4. B. =.88. C.. D. E. = 1.414. 4 = 1.189. 1 ka = 1 mv + 1 kx = const. xt ( ) = Acos( ωt+ vt ( ) = Aωsin( ωt+ at A t ( ) = ω cos( ω + φ) ω = π ω = 1 T pend mgd = = I g L
SHM and Pendulums For the physical pendulum the CoM is a distance d rom the pivot, then the torque at an angle θ is:! =!d(mgsin") Using the small angle approxima%on we obtain:! =!dmg" so I!! "dmg" α Comparing to the SHM equa%on again gives: d θ dt mgd I = = θ mgd 1 mgd I ω = = T = π I π I mgd
SHM and Pendulums The simple pendulum is a special case o the physical pendulum. All the mass is located at a point atached to the end o string o length L. Using this d = L and I = ml mgd mgl g ω = = = I ml L! = g L! = 1 " g L! T = " L g
Prelecture: Physical Pendula Problem 1 Wri%ng down Newton's Second Law in the x direc%on or some system results in the equa%on on the right. What is the oscilla%on requency o this system? a d x dt =!b x a) a/b b) a*b c) b/a F x = ma x F x = m d x dt d x dt +! x =
Prelecture: Physical Pendula Problem A uniorm s%ck and a mass on a string are used to make two pendula that have the same length. Which one swings with the longer period? a) the mass on the string b) the s%ck c) can't tell without knowing how the masses compare (a)! = mgd I T =! " = = mgl ml = g L 4! L g (b)! = mgd I T =! " = = 4! L 3g mgl 1 3 ml = 3g L
Checkpoint: Physical Pendula Problem 1 A simple pendulum is used as the %ming element in a clock as shown. An adjustment screw is used to make the pendulum shorter (longer) by moving the weight up (down) along the sha_ that connects it to the pivot. I the clock is running too ast, the weight needs to be moved a)up b)down T =! L g
Checkpoint: Physical Pendula Problem A torsion pendulum is used as the %ming element in a clock as shown. The speed o the clock is adjusted by changing the distance o two small disks rom the rota%on axis o the pendulum. I we adjust the disks so that they are closer to the rota%on axis, the clock runs a) aster Angular SHM: κ = torsion constant b) slower! = " I = 1 # " I T = # I "
Checkpoint: Physical Pendula Problem 3 Consider the two pendula shown above. In Case 1 a s%ck o mass M is pivoted at one end and used as a pendulum. In Case a point par%cle o mass M is atached to the center o the same s%ck. In which case is the period o the pendulum the longest? a) Case 1 b) Case c) Same! = mgd I T =! " =! I mgd T 1 =! T =! 1 3 ml mg L =! L 3g! 1! 3 ml + m L $ $ # # & & # " " % & % =! 7L mg L 1g
Damped OscillaEons Real-world systems have some dissipative orces that decrease the amplitude. The decrease in amplitude is called damping and the motion is called damped oscillation. The igure at the right illustrates an oscillator with a small amount o damping. The mechanical energy o a damped oscillator decreases continuously. b = km When, the system is critically damped and i b is larger than this it is overdamped A critically or overdamped oscillator returns to equilibrium without oscillating.
Example Pendulum on Mars (14.48) A certain simple pendulum has a period on the earth o 1.6 s. What is its period on the surace o Mars where gravity is g mars = 3.71 m/s? xt ( ) = Acos( ωt+ vt ( ) = Aωsin( ωt+ at A t ( ) = ω cos( ω + φ) ω = π = 1 T ω pend mgd g = = I L
Clicker QuesEon A simple pendulum consists o a point mass suspended by a massless, unstretchable string. I the mass is doubled while the length o the string remains the same, the period o the pendulum A. becomes 4 %mes greater. B. becomes twice as great. C. becomes greater by a actor o. D. remains unchanged. E. decreases. 1 ka = 1 mv + 1 kx = const. xt ( ) = Acos( ωt+ vt ( ) = Aωsin( ωt+ at A t ( ) = ω cos( ω + φ) ω = π ω = 1 T pend mgd = = I g L
Example Physical pendulum (small angle approxima%on)(14.54) A 1.8 kg monkey wrench is pivoted.5 m rom its center o mass and allowed to swing as a physical pendulum. The period or small angle oscilla%ons is.94 s. a) What is the moment o iner%a o the wrench about an axis through the pivot? b) I the wrench is ini%ally displaced.4 rad rom its equilibrium posi%on, what is the angular speed o the wrench as it passes through the equilibrium posi%on? xt ( ) = Acos( ωt+ vt ( ) = Aωsin( ωt+ at A t ( ) = ω cos( ω + φ) ω = π = 1 T ω pend mgd g = = I L
Example Physical pendulum (14.57) The two pendulums shown each consist o a uniorm solid ball o mass M supported by a rigid massless rod, but the ball or pendulum A is very small compared to that o pendulum B. Find the period o each pendulum or small displacements. Which ball takes longer to complete a swing? xt ( ) = Acos( ωt+ vt ( ) = Aωsin( ωt+ at A t ( ) = ω cos( ω + φ) ω = π = 1 T ω pend mgd g = = I L
Example Pendulums and collisions(14.95) In the igure the upper ball is released rom rest, collides with the sta%onary lower ball, and s%cks to it. The strings are both 5. cm long. The upper ball has mass. kg, and ini%ally 1 cm higher than the lower ball (mass 3. kg). Find the requency and maximum angular displacement o the mo%on a_er the collision. xt ( ) = Acos( ωt+ vt ( ) = Aωsin( ωt+ at A t ( ) = ω cos( ω + φ) ω = π = 1 T ω pend mgd g = = I L
Example Energy and momentum in SHM A block o mass M is atached to a horizontal spring with spring constant k and is moving in SHM. As it passes through its equilibrium point a lump o puty o mass m is dropped rom a small height and s%ck to it. Find the new amplitude and period. 1 ka = 1 mv + 1 kx = const. ω = π ω = 1 T pend mgd = = I g L