Expert Q&A; Done Lab 6 Pre-Lab Exercise: Rotation 1 . Moment of Inertia (or Rota

Question

Expert Q&A; Done Lab 6 Pre-Lab Exercise: Rotation 1 . Moment of Inertia (or Rotational Inertia). A uniform rod ith as so·ud i pth 100 spun about a horizontal axis perpendicular to the rod and passing through the center of the rod. Two poin masses, each of mass 30 g, are attached symmetrically to the rod at 40 mm from the asis of rousion Calculate the total rotational ineria of this system. Show your work. Inclade unis Itot 0.00014 kmt +0. DOO 138kg, 2. Net Torque and Angular Acceleration. A 20 g mass is attached to a string and this string i wrapped around a pulley of radius 30 mm. The pulley is attached to the axis of the rod-mass sysiem in problem 1. What is the resulting aqularaccelczasion of the rod-mass system due to the sorque applied by the tension due to this 20 g mass? Show your work Include MKS units of angular acceleration radianss 3. The situation is like that in question 2 except that in addition to the torque from the 20 g mass, the asle of the disk? Show your work. Include MKS units. 40 Can you please check the first one and help with the rest?

Explanation / Answer

2. let mass of the pulley be m

moment of inertia of the rod mass system, I' = 0.000138 kg m^2

givben mass, M = 0.02 kg

radius of pulley, r = 0.03 m

moment of inertia of the pulley and the mass, I = I' + 0.5mr^2 = 0.00045m + 0.000138

torque on the pulley, T = M*gr = 0.02*9.81*0.03 = 0.005886 Nm

hence angular acceleration of the system, alpha = T/I = 0.005886/(0.00045m + 0.000138 )

as m is not given it can be consiedered m = 0

hence

alpha = 42.6521 rad/s

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