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CU-Boulder PHYS 1010 - Circular Motion

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1Physics 1010:The Physics of Everyday LifeTODAY• Circular Motion• Rotational Motion• Review Exam2Exam Was…a) Way too easyb) Easyc) About Rightd) Harde) Way too hard3I believe I will get…a) Ab) Bc) Cd) D4Admin StuffSolutions will be posted on website by tomorrowGrades will be posted on website by TuesdayGrades will be cummulative (including clickersand homework up to 9/14)Clicker grades: correct: 2 pts incorrect: 1 pt no answer: 0 pts“poll” questions are not graded (but still get 1pt)5Circular Motion(at constant speed)In circular motion, acceleration isalways perpendicular to the velocity.Fc = mv2/R =mRω2θRss=Rθv=Rωω = angular velocity (rad/sec)θ in radians (there are 2π rad in a circle)ν=1/Tν = frequency, T = periodω=2πν=2π/Tac = v2/R = Rω2va6Check that equation is correct• Circumference of cirlce is 2πR• Full angle is 2π• so s=θR checks out7RotationθRss=Rθv=Rων=1/Tω=2πν=2π/Ta=Rαω = ω0 + αtθ = θ0 + ω0t + (1/2) α t2Rotational kinematic equations:θ : angular positionω : angular velocityα : angular accelerationθ,ω,α, are all VECTORS;They have both magnitudeand directionRight-had rule: Curl fingers of RIGHT HAND in direction of rotation, and thumb points in direction of vector8RotationθRss=Rθ v=Rων=1/Tω=2πν=2π/Ta=Rαω = ω0 + αtθ = θ0 + ω0t + (1/2) α t2The rotational kinematic equations,Remind me of:A pure and utter #$%*B tortureC two-dimensional motionD 1-dimensional kinematics9RotationθRss=Rθ v=Rων=1/Tω=2πν=2π/Ta=Rαω = ω0 + αtθ = θ0 + ω0 t + (1/2) αt2Rotational kinematic equations:Are mathematically the same as translational kinematic equations:v = v0 + atx = x0 + v0t + (1/2) at2x ----> θv ----> ωa ----> α10Rotational DynamicsθRsν=1/Tω=2πν=2π/Ts=Rθ v=Rωa=RαThe principal rotational dynamic equation will probably look like:A F=maB a=RαC change of velocity over timeD none of the above11Rotational DynamicsθRsν=1/Tω=2πν=2π/Ts=Rθ v=Rωa=RαT = I αF = m aForce = mass * accelerationTorque = Moment of Inertia * angular accelerationT = F dpFdRight-had rule: Curl fingers of RIGHT HAND in direction that force would tend to spin the object, and thumb pointsin direction of TorqueT, F, d are VECTORSI : property of body (distribution of mass)12Review ExamMost difficult question was:A The #@*& mule one (both of them)B the pulleyC the long questionD both A and BE anything with math13M = 1 kgm = 0.3 kgIn the above diagram, two masses are attached by a string that hangs over a pulley.The hanging mass is 0.3 kg, while the rolling mass is 1 kg. The positive direction isto the right. The rolling mass accelerates ata. 2.26 m/s2b. 9.8 m/s2c. 2.94 m/s2d. -2.94 m/s2e. 2.94 m/smg=a(M+m), a=mg/(M+m) = 0.3*9.8/(1.+0.3) m/s2 = 2.26 m/s214You throw a steel ball of 1 kg straight up. When the ball leaves your hand it has velocity = 12 m/s (take up to be the positive direction). Calculate: The time of flight to the highest point the ball will reach (hint: what is the velocity of the ball at the highest point?).a. 1.22 sb. 2.44 sc. 1.56 sd. 3.12 se. 1.88 sv=v0-gt; at the top, v=0 (cf. hint), therefore v0=gt, so t=v0/g=12/9.8 s = 1.22s15The distance to the highest point the ball will reach. a. 22.0 mb. 14.7 mc. 11.0 md. 7.35 me. 8.32 mx = x0+v0t-(1/2)gt2 = 0 + 12*1.22 m - 0.5*9.8*1.49 m = 7.35


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CU-Boulder PHYS 1010 - Circular Motion

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