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Download Homework 22 - University of Utah Physics
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Fall XXXX HW#22 XXXXXXX Fall XXXX HW #22 XXXXX Problem 3 At t = 0 a block with mass M = 5 kg moves with a velocity v = 2 m/s at position xo = -.33 m from the equilibrium position of the spring. The block is attached to a massless spring of spring constant k = 61.2 N/m and slides on a frictionless surface. At what time will the block next pass x = 0, the place where the spring is unstretched? t1 = ? Solution We are given π = 5.0 kg, and π = 61.2 N/m, and so the angular frequency of oscillation is: π=οΏ½ Unit check: οΏ½ π 61.2 Nβm =οΏ½ = 3.50 sβ1 π 5.0 kg Nβm 1 1 kg β m 1 1 1 1 = οΏ½N β β =οΏ½ 2 β β =οΏ½ 2= kg m kg s m kg s s The general solution for the position as a function of time in general form is: π₯ = π΄ cos(ππ + π) We are given that at π‘ = 0, π₯ = π₯0 = β0.33 m β π΄ cos π = π₯0 β¦ (1) and differentiating with respect to time we get the velocity as a function of time in general form: π£ = βππ sin(ππ + π) We are given that π‘ = 0, π£ = π£0 = +2.0 m/s β βππ sin π = π£0 β¦ (2) In order to find π‘1 such that π₯(π‘1 ) = 0, we need to first solve for π΄ and π from equations (1), (2): Taking the sum (1)2 + (2)2 βπ2 gives π£0 2 π΄2 cos2 π + π΄2 sin2 π = π΄2 = π₯0 2 + οΏ½ οΏ½ π π΄ = οΏ½π₯0 2 + οΏ½ 2 π£0 2 2.0 mβs οΏ½ = οΏ½(β0.33m)2 + οΏ½ οΏ½ = 0.66 m π 3.50 sβ1 And substituting this value for A into (1) and (2), we have π₯0 β0.33m 1 (1) β cos π = = = β0.50 = β π΄ 0.66 m 2 π£0 2.0m/s β3 (2) β sin π = = = β0.866 = β β1 βππ β(3.50s )(0.66 m) 2 β¦continued 1 β3 cos π = β , sin π = β β 2 2 Where π is calculated in radians, and so we write And so tan π = β3 and π = β π₯ = π΄ cos(ππ + π) = (0.66m) cos οΏ½3.50 sβ1 β π‘ β π₯(π‘1 ) = (0.66m) cos οΏ½3.50 s β1 β π‘1 β The first time this happens at cos οΏ½3.50 sβ1 β π‘1 β 2π π =β 3 2 π 2π β3π 4π π 3.50 s β1 β π‘1 = β + = + = 2 3 6 6 6 π π‘1 = = 0.15 s 6 β 3.50 sβ1 3.50 s β1 β π‘1 β Answer: π‘1 = 0.15 s 2π οΏ½=0 3 2π οΏ½=0 3 2π οΏ½ 3 2π 3 Fall XXXXX Problem 4 HW #22 XXXXX Problem 5 A simple pendulum with mass m = 1.5 kg and length L = 2.57 m hangs from the ceiling. It is pulled back to a small angle of ΞΈ = 9.8° from the vertical and released at t = 0. 1) What is the period of oscillation? _______ s The moment of inertia about the pivot for a small particle is πΌ = ππ 2 = ππΏ2 The torque exerted by the force of gravity is π = ππ sin πππ = πΏ β ππ β sin(βπ) = βπππ sin π Where we have observed that SINE is an odd function: sin(βπ) = β sin π, so the equation of motion is: Which has the form π2 π π βπππ sin π π π = = = β sin π β β π 2 2 ππ‘ πΌ ππΏ πΏ πΏ π2π = βπ2 π, 2 ππ‘ π π = , πΏ 2 π 9.81 mβs 2 οΏ½ οΏ½ β π= = = 1.95 s β1 πΏ 2.57m And the period is related to the angular velocity/frequency by (period is the time it takes to do one cycle: i.e. ππ = 2π π= 2π 2π = = 3.22 s π 1.95 s β1 2) What is the magnitude of the force on the pendulum bob perpendicular to the string at t=0? _______ N The component of the force perpendicular to the string is also the βtangential componentβ πΉβ₯ = πΉπ = ππ sin π = 1.5 kg β 9.81 mβs2 β sin 9.8° = 2.50 N 3) What is the maximum speed of the pendulum? _______m/s π£πππ = π|ππ| = πΏ|πππππ | In this case we started at π‘ = 0 with And so π = ππππ = 9.8° = 9.8° β π rad = 0.171 rad 180° π£πππ = πΏ|πππππ | = L = 2.57 m β 1.95 s β1 β 0.171 rad = 0.859 mβs 4) What is the angular displacement at t = 3.73 s? (give the answer as a negative angle if the angle is to the left of the vertical) _______° Since we started at t=0 from rest, we have π = ππππ cos ππ π(3.73s) = 9.8° β cos(1.95 sβ1 β 3.73 s) = 9.8° β cos(7.29 rad) = 9.8° β 0.537 = 5.26° 5) What is the magnitude of the tangential acceleration as the pendulum passes through the equilibrium position? _______ m/s2 Equilibrium is the position at which the net force on the pendulum bob is zero. In this case this means πΉπ = 0 Hence, by Newtonβs 2nd Law: the acceleration at equilibrium is ππ = πΉπ =0 π 6) What is the magnitude of the radial acceleration as the pendulum passes through the equilibrium position? _______ m/s2 π£2 π Here π = πΏ, and, at the equilibrium position: π£ = π£πππ ππ = ππ = π£πππ 2 (0.859 mβs)2 = = 0.287 mβs2 πΏ 2.57m 7) Which of the following would change the frequency of oscillation of this simple pendulum? (1) increasing the mass (2) decreasing the initial angular displacement (3) increasing the length (4) hanging the pendulum in an elevator accelerating downward π 1 π οΏ½ π= = 2π 2π πΏ This does NOT depend on mass, or the initial displacement But it does depend on the length and the gravitational accelerationβwhich means it will change in an accelerating elevator So correct choices are (3) and (4) Problem 6 A rigid rod of length L= 1 m and mass M = 2.5 kg is attached to a pivot mounted d = 0.17 m from one end. The rod can rotate in the vertical plane, and is influenced by gravity. What is the period for small oscillations of the pendulum shown? T = ? seconds Solution: We use the rotational equation of motion, where the angular acceleration of the rod is given by π2π π β‘πΌ= 2 ππ‘ πΌ Where πΌ is the moment of inertia of the rod about the pivot point P, given by (using Parallel Axes Theorem β P.A.T.) πΌ = πΌπ = πΌπΆπΆ + ππ·2 Where π· is the distance between the pivot and the center-of-mass, given by π·= πΏ β π = 0.50m β 0.17m = 0.33m 2 And the moment of inertia of the rod about its own center-of-mass is given by πΌπΆπΆ = 1 ππΏ2 12 The torque exerted by gravity about P is given by (π = π· is the distance from the pivot to where the force of gravity acts: at the center-of-mass) π = ππ sin πππ = π· β ππ β sin(βπ) β βππππ So we have π2π π πππ = ββ π = βπ2 π 2 1 ππ‘ πΌ 2 2 12 ππΏ + ππ· This then gives the angular frequency of the oscillations: π=οΏ½ πππ 1 2 2 12 ππΏ + ππ· =οΏ½ ππ 1 2 2 12 πΏ + π· =οΏ½ 9.81 mβs 2 β 0.33m = 4.10s β1 (1.0m)2 β12 + (0.33m)2 And the period is related to the angular velocity/frequency by (period is the time it takes to do one cycle: i.e. ππ = 2π π= 2π 2π = = 1.53 s π 4.10s β1 Problem 7 A circular hoop of radius 40 cm is hung on a narrow horizontal hoop and allowed to swing in the plane of the hoop. What is the period of its oscillation, assuming that the amplitude is small? _______ s Solution: We use the rotational equation of motion, where the angular acceleration of the hoop is given by π π2π β‘πΌ= 2 πΌ ππ‘ Where πΌ is the moment of inertia of the hoop about the pivot point P, given by (using Parallel Axes Theorem β P.A.T.) πΌ = πΌπ = πΌπΆπΆ + ππ·2 Where π· is the distance between the pivot and the center-of-mass, given here by π· = π = 0.40m And the moment of inertia of the hoop about its own center-of-mass is given by πΌπΆπΆ = ππ 2 The torque exerted by gravity about P is given by (π = π is the distance from the pivot to where the force of gravity acts: at the center-of-mass) So we have π = ππ sin πππ = π β ππ β sin(βπ) β βππππ π2π π πππ = ββ π = βπ2 π 2 πΌ ππ 2 + ππ 2 ππ‘ This then gives the angular frequency of the oscillations: π=οΏ½ πππ π 9.81 mβs 2 οΏ½ οΏ½ = = = 3.50sβ1 2ππ 2 2π 2 β 0.40m And the period is related to the angular velocity/frequency by (period is the time it takes to do one cycle: i.e. ππ = 2π π= 2π 2π = = 1.79 s π 3.50s β1