Activity 1: Problem Solving Read and analyze each problem. Show your sequential solution through GUFSA Method. Attach a paper of any size for your answer and submit it together with this module. 1. What is the rotational inertia of a 3.8 kg solid disk pulley which has a diameter of 28.0 cm? 2. A solid cylinder is free to rotate about an axis through its center. It has a mass of 3.5 kg and a radius of 0.75 m. A constant force of 20.0 N is applied tangentially to the cylinder which is initially at rest. Determine the following: a. rotational inertia of the cylinder b. torque acting on the cylinder c. angular acceleration of the cylinder 3. An unbalanced torque of 7.20 N.m acts on a 30 cm diameter pulley which is initially at rest for 5.0 s. The pulley is a solid disk and has a mass of 3.1 kg and is free to rotate about an axis through its center. What is the pulley's (a) rotational inertia and (b) angular acceleration? 4. The wheel on an upside-down bicycle moves through 11.0 rad in 2.0 s. What is the wheel's angular acceleration if its initial angular speed is 2.0 rad? 5. A future astronaut lands on a planet with an unknown value of g She finds that the period of a pendulum 0.65m long is 2.8 s. What is g for the surface of this planet? Score Guide GUFSA Method Given: 1 point Unknown: 1 point Formula: 1 point Solution: 1 point Answer: 1 point Total = 5 pointsSample Problem 2: A man enters a tall tower. He needs to know the height of the tower, but darkness obscures the ceiling. He does note, however, that a long pendulum extends from the ceiling almost to the floor and that its period is 12 s. (a) How tall is the tower? (b) If the length of the pendulum is halved, what would its period of oscillation be? Following the GUFSA Method, we have: Given: T = 12 s Constant values for = 3.14160 and g = 9.8m/s2 Unknown: L Formula: L = gT242 (Derivation is applied) Solution: L = (9.8ms2)(12 s)24(3.1416)2 Answer. 35.75 m Sample Problem 3: On top of a mountain, a pendulum 1.55 m long has a period of 2.51 s. What is the acceleration due to gravity at this location? Given: T = 2.51 s L=1.55 m Constant value form= 3.14160 Unknown: Formula: g = 42T21 (Derivation is applied) Solution: g = 4(3.1416)2(2.51s)2 (1.55 m) Answer. 9.71 my's Summing up, the above concepts of rotational and simple harmonic motion is still governed by Newton's Law. And we cannot disregard the fact that in our day to day living we really and naturally observed them in our environment