Question

You are visiting your grandparents in Boston. There is an antique grandfather clock keeping time in one of the rooms. Your grandmother enters the room and tells you that she had it calibrated professionally in Denver and it kept perfect time for years. After they moved it here to their Massachusetts house, it has not been accurate. It runs too fast and has to be reset to the correct time every few days. You find some papers associated with the calibration of the clock when it was in Denver and find the following. The pendulum of the clock consists of a long, thin rod of length L = 1.14 m and mass m = 0.436 kg. On the rod is mounted a sliding bob, which is a uniform disk of radius R = 0.149 m and mass M = 1.82 kg. The rod is suspended from pins mounted at its upper end. The calibration data show the acceleration due to gravity in Denver to be 9.796 m/s. You measure very carefully and find that the center of the disk is located at a distance r = 1.0130 m from the pivot point at the upper end of the rod. (a) Find the period of the pendulum (in s) when it was in Denver. (b) The acceleration due to gravity in Boston is 9.803 m/s2. If the Denver-calibrated clock is set to the correct time one morning in Boston, determine by how many seconds the clock is off at the same time the next morning. Is it fast or slow? fast slow You are correct; the period in Boston is shorter, so the pendulum will make more swings in a day and run fast. (c) You decide to adjust the clock by turning a small knob located at the bottom of the bob. Turning the knob causes the bob to move up and down on the rod. Find the direction you should move the bob to restore the correct timekeeping capability of the clock: up or down. O up o down You are correct; moving the bob downward increases the effective length and lengthens the period.

Answer 1

@ Moment af inertia about hinge Bm = Mood 12 + mask R² + maskerz -0.43681.243 +1.82801492 +1.82X1. 0137 3 2 . : 3 2 2.0767 kg ma d = distance of ccenter of mass of System a mood (/2) + midiskor mood tmdisk 2 0.436 (4:44) + (1.82X40130) 0.436 +1.82 =0.92738m r IH 1 2 2tt ja msystem xgxd. 3271) 2.0767 (0.436+1-82)x9.8 %0.9273 T. = 1.999655

Bi. Time period. w bošlom = 204: _ILA msystengd 2.0787 Ezt 10u36 t/82789-840FA Mise Vis 093961&)x9-excom 731.999355 Defference in Time period att AT2 1.99935–-99965 =-3x10 4. Time gamed is ouhowne X ... = (47) ix24 = 3x104 1.99935 X 24X60X50 = 12.964 See