Fall XXXX HW#22 XXXXXXX
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9 Problem 3 At t = 0 a block with mass M = 5 kg moves with a velocity v = 2 m/s at position x o = -.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? t 1 =? Solution We are given m = 5.0 kg, and k = 61.2 N/m, and so the angular frequency of oscillation is: Unit check: ω = k m = 61.2 N m 5.0 kg = 3.50 s 1 N m kg = N 1 m 1 kg = kg m s 2 1 m 1 kg = 1 s 2 = 1 s The general solution for the position as a function of time in general form is: x = A cos(ωω + φ) We are given that at t = 0, x = x 0 = 0.33 m A cos φ = x 0 (1) and differentiating with respect to time we get the velocity as a function of time in general form: v = ωω sin(ωω + φ) We are given that t = 0, v = v 0 = +2.0 m/s ωω sin φ = v 0 (2) In order to find t 1 such that x(t 1 ) = 0, we need to first solve for A and φ from equations (1), (2): Taking the sum (1) 2 + (2) 2 ω 2 gives A 2 cos 2 φ + A 2 sin 2 φ = A 2 = x v 2 0 ω A = x v 0 ω 2 = ( 0.33m) m s s 1 = 0.66 m And substituting this value for A into (1) and (2), we have (1) cos φ = x 0 A = 0.33m 0.66 m = 0.50 = 1 2 continued (2) sin φ = v 0 ωω = 2.0m/s 3 (3.50s 1 = = )(0.66 m) 2
10 cos φ = 1 2, 3 sin φ = tan φ = 3 and φ = 2π 2 3 Where φ is calculated in radians, and so we write And so x = A cos(ωω + φ) = (0.66m) cos 3.50 s 1 t 2π 3 The first time this happens x(t 1 ) = (0.66m) cos 3.50 s 1 t 1 2π 3 = 0 at Answer: t 1 = 0.15 s cos 3.50 s 1 t 1 2π 3 = s 1 t 1 2π 3 = π s 1 t 1 = π 2 + 2π 3 = 3π t 1 = π = 0.15 s s π 6 = π 6
11 Fall XXXXX HW #22 XXXXX Problem 4
12 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 θ = 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 I = mr 2 = ml 2 The torque exerted by the force of gravity is τ = rr sin φ rr = L mm sin( θ) = mmm sin θ Where we have observed that SINE is an odd function: sin( θ) = sin θ, so the equation of motion is: Which has the form d 2 θ dt 2 = τ mmm sin θ = I ml 2 = g L sin θ g L θ d 2 θ dt 2 = ω2 θ, ω 2 = g L, ω = g L = 9.81 m s2 2.57m = 1.95 s 1 And the period is related to the angular velocity/frequency by (period is the time it takes to do one cycle: i.e. ωω = 2π T = 2π ω = 2π = 3.22 s 1.95 s 1 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 tangential component F = F T = mm sin θ = 1.5 kg 9.81 m s 2 sin 9.8 = 2.50 N 3) What is the maximum speed of the pendulum? m/s In this case we started at t = 0 with And so v MMM = r ωω = L ωθ MMM θ = θ MMM = 9.8 = 9.8 π rad 180 = rad v MMM = L ωθ MMM = L = 2.57 m 1.95 s rad = m s
13 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 θ = θ MMM cos ωω θ(3.73s) = 9.8 cos(1.95 s s) = 9.8 cos(7.29 rad) = = ) What is the magnitude of the tangential acceleration as the pendulum passes through the equilibrium position? m/s 2 Equilibrium is the position at which the net force on the pendulum bob is zero. In this case this means F T = 0 Hence, by Newton s 2 nd Law: the acceleration at equilibrium is a T = F T m = 0 6) What is the magnitude of the radial acceleration as the pendulum passes through the equilibrium position? m/s 2 a r = v2 r Here r = L, and, at the equilibrium position: v = v mmm a r = v mmm 2 L = (0.859 m s)2 = m s m 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 f = ω 2π = 1 2π g L This does NOT depend on mass, or the initial displacement But it does depend on the length and the gravitational acceleration which means it will change in an accelerating elevator So correct choices are (3) and (4)
14 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 d 2 θ dt 2 α = τ I Where I is the moment of inertia of the rod about the pivot point P, given by (using Parallel Axes Theorem P.A.T.) I = I P = I CC + MD 2 Where D is the distance between the pivot and the center-of-mass, given by D = L d = 0.50m 0.17m = 0.33m 2 And the moment of inertia of the rod about its own center-of-mass is given by I CC = 1 12 ML2 The torque exerted by gravity about P is given by (r = D is the distance from the pivot to where the force of gravity acts: at the center-of-mass) So we have τ = rr sin φ rr = D MM sin( θ) MMMM d 2 θ dt 2 = τ I MMM θ = 1 ω2 θ 12 ML2 + MD 2 This then gives the angular frequency of the oscillations: Mgg gg 9.81 m s ω = = 1 = m 12 ML2 + MD L2 + D 2 (1.0m) (0.33m) 2 = 4.10s 1 And the period is related to the angular velocity/frequency by (period is the time it takes to do one cycle: i.e. ωω = 2π T = 2π ω = 2π = 1.53 s 4.10s 1
15 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 d 2 θ dt 2 α = τ I Where I is the moment of inertia of the hoop about the pivot point P, given by (using Parallel Axes Theorem P.A.T.) I = I P = I CC + MD 2 Where D is the distance between the pivot and the center-of-mass, given here by D = R = 0.40m And the moment of inertia of the hoop about its own center-of-mass is given by I CC = MR 2 The torque exerted by gravity about P is given by (r = R is the distance from the pivot to where the force of gravity acts: at the center-of-mass) So we have τ = rr sin φ rr = r MM sin( θ) MMMM d 2 θ dt 2 = τ I MMM MR 2 + MR 2 θ = ω2 θ This then gives the angular frequency of the oscillations: ω = MMR g = 2MR2 2R = 9.81 m s m = 3.50s 1 And the period is related to the angular velocity/frequency by (period is the time it takes to do one cycle: i.e. ωω = 2π T = 2π ω = 2π = 1.79 s 3.50s 1
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