The three problems attached having to do with moment of inertia of things like a yo-yo.
I need help setting them up. Please send explanations with the answer so I can see how your answer was achieved.© BrainMass Inc. brainmass.com October 24, 2018, 6:20 pm ad1c9bdddf
Let us assume the tension in the string be T, acceleration be a, angular acceleration be alpha, and time taken be t.
r = 5 in = 5/12 ft
R = 10 in = 10/12 ft
moment of inertia I = 0.24 slug.ft^2 = 0.24*32.2 lb.ft^2
mass m = 50 lb
gravitational acceleration g = 32.2 ft/s^2
m*g - T = m*a .........(1)
Moment about the center,
T*r = I*alpha = I*a/r
=> T = I*a/r^2 .........(2)
From Equations. (1) and (2),
m*g = (m + I/r^2)*a
=> a = m*g/(m + I/r^2)
Kinematics: s = 4 ft; u = 0 (rest)
s = u*t + (1/2)*a*t^2
=> s = (1/2)*t^2*m*g/(m + I/r^2)
=> t^2 = 2*s*(m + I/r^2)/(m*g)
=> t^2 = 2*4*(50 + ...
The solution provides comprehensive workings-out of the three moment of inertia problems given, including step-by-step break-downs of the necessary calculations and snippets of written explanation where needed.
Moments of Inertia: Example Problem
Would you help me with the following problem?
Two metal disks, one with radius R1 = 2.44 and mass M1= 0.840 and the other with radius R2= 4.92 and mass M2= 1.70 , are welded together and mounted on a frictionless axis through their common center. . picture attached
A) What is the total moment of inertia of the two disks?
B) A light string is wrapped around the edge of the smaller disk, and a 1.50-kg block, suspended from the free end of the string. If the block is released from rest at a distance of 2.06 m above the floor, what is its speed just before it strikes the floor?
C) Repeat the calculation of part B, this time with the string wrapped around the edge of the larger disk.View Full Posting Details