Question

2. As shown below a uniform disk of radius R 10.0 cm and mass M 0.850 kg is attached to the end of a uniform rigid rod of length L = 500 mm and mass m = 0.210 kg. Take positive θ in the counterclockwise direction, and positive angular velocity ω as into the page. Assume small angle deflection (a) When the disk is suspended from a pivot as sown what is the period of its motion? (b) If this physical pendulum is lifted to an angle of o 11.5 of wo 1.15 rad/s att-0 s, then what is the resulting position (c) At the connection between the rod and disk, what is the tension as a above its equilibrium, and then given an initial angular velocity function for the SHM of this physical pendulum? 500 mm function of time? (d) If the physical pendulum was attached to the ceiling of an elevator that 10.0 cm was traveling downward at an acceleration ofa -2.15 m/s2, how many full swings would it make by the time an identical pendulum at rest completes 5.50 swings?

Answer 1

Consider the above given that formulas: moment of inertial of disk I_P = 1/2 pi r^2 + M(L + r)^2 I_R = Ml^2/12 + M(L/2)^2 = ML^2/3 I_P = I_P + 9 I_R = 1/2 Mr^2 + M(L + r)^2 + ML^2 + M(L/2)^2 = 1/2 (0.85) (001)^2 + 0.85 (0.5 + 0.1)^2 + 0.21 (0.5)^2 I_P = 0.32775 Moment of inertia of pendulum I_P = 0.32775 of pendulum h_D = M h_D + m h_r/M + m = 0.85(0.6) + 0.21 (0.25)/0.85 + 0.21 \r\n of pendulum h_P = 0.53 M time period of Motion T = 2 pi squareroot IP/(M + m) gh = 2pi squareroot 0.32775/(0.85 + 2.1) 9.8 times 0.53 = 1.533 time period of motion T = 1.533 S