The pendulum is comprised of a 4.20 kg slender rod of length L=1.5m and a 9 kg t
ID: 2244178 • Letter: T
Question
The pendulum is comprised of a 4.20 kg slender rod of length L=1.5m and a 9 kg thin plate a=0.35 m x b=0.75m. Determine the location of the center of mass, moments of intertia about axes perpendicular to the page passing through O and passing through G
Xbar=_________________
I(O)=_____________________
I(G)=_______________________
*doubled the points 2 part question
The pendulum above is released from rest from the position shown. Determine the horizontal and vertical reactions at the pin O and the angular acceleration.
O(x)=______________
O(y)=_________________
(alpha)=_______________
Explanation / Answer
let m1 = 4.2 kg, m2 = 9 kg
let x1 is the center of mass of rod and x2 is the center of mass of plank
x1 = L/2 = 1.5/2 = 0.75 m
x2 = L + a/2 = 1.5 + 0.35/2 = 1.675 m
xcm = (m1*x1+m2*x2)/(m1+m2)
= (4.2*0.75 + 9*1.675)/(4.2+9)
= 1.3807 m from point O
Io = m1*L1^2/3 + m2*(a^2+b^2)/12 + m2*(L + a/2)^2
= 4.2*1.5^2/3 + 9*(0.35^2+0.75^2)/12 + 9*(1.5 + 0.35/2)^2
= 28.914 kg.m^2
IG = m1*L^2/12 + m1*(xcm - L/2)^2 + m2*(a^2+b^2)/12 + m2*(L - Xcm + a/2)^2
= 4.2*1.5^2/12 + 4.2*(1.3807-0.75)^2 + 9*(0.35^2+0.75^2) + 9*(1.5-1.3807 + 0.35/2)^2
= 3.75 kg.m^2
Torque about point O, To = m1*g*L + m2*g*(L + a/2)
To = 4.2*9.8*1.5 + 9*9.8*(1.5 + 0.35/2)
= 209.475 N.m
To = Io*alfa
alfa = To/Io
= 209.475/28.914
= 7.245 rad/s^2
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