Question

Problems 1) and 2) refer to the following: A yo-yo spins at an almost constant angular speed of 151 rad/s for many seconds after it reaches the bottom end of its string. The yo-yo will rise when you give the string a very brief tug. Assume no slipping as the yo-yo climbs the string. The moment of inertia of the vo-vo about its own central spin axis is 1-1.10x10- kg·m 1) (5 points) Which value best represents the height to which the yo-yo will rise above its low point? A) 44 cmm B) 27 cm C) 85 cm D) 19 cm E) None of these. (5 points) The yo-yo can be modeled as two uniform disks (cylinders), each of unknown radius R, length L = 1.2 cm, and mass M = 0.14 kg, joined by short cylindrical rod of length l - 1.0 cm, radius r-0.25 cm, and mass m 0.010 kg. Which choice best represents the radius of each disk, R? 2) A) 3.5 cm B) 3.1 cnm C) 4.0 cnm D) 2.5 cnm E) 2.8 cm

Answer 1

The answer for above problem is explained below

The answer for problem 1) is option "A"

The answer for problem 2) is option " E ".

By using the law of conservation of energy, we know that, Initial kinetic energy = Final potential energy (1/2) I w^2 = [2 (M) + m] gh implies h = (1/2) I w^2/[2 (m) + m] g implies h = 1/2 times 1.1 times 10^-4 kg m^2 times (151 rad/sec)^2/[2 (0.14 kg) + 0.010 kg] 9.8 m/s^2 implies h = 1.254 kg m^2 rad^2/sec^2/2.842 kg m/s^2 implies h = 0.44 m (or) h = 44 cm \r\n We know that, I = 2 (1/2) MR^2 + m r^2 implies I = M R^2 + m r^2 implies 1.1 times 10^-4 kg m^2 = 0.14 lg times R^2 + 0.01 kg times (0.25 times 10^-2 m)^2 implies 101 times 10^-4 kg m^2 = 0.14 kg R^2 + 6.25 times 10^-8 kg m^2 implies 0.14 kg R^2 = 1.1 times 10^-4 kg m^2 - 6.25 times 10^-8 lg m^2 implies 0.14 kg R^2 = 1.09937 times 10^-4 kg m^2 implies R^2 = 1.09937