Part A Calculate the moment of inertia of the array of point objects shown in th
ID: 1287729 • Letter: P
Question
Part A Calculate the moment of inertia of the array of point objects shown in the figure(Figure 1) about the vertical axis. Assume m = 1.2kg , M = 3.0kg , and the objects are wired together by very light, rigid pieces of wire. The array is rectangular and is split through the middle by the horizontal axis. Express your answer using two significant figures.
Part B
Calculate the moment of inertia of the array of point objects about the horizontal axis.
Express your answer using two significant figures.
About which axis would it be harder to accelerate this array?
About which axis would it be harder to accelerate this array?
about the vertical axis
or
about the vertical axis
or
about the horizontal axis Part A Calculate the moment of inertia of the array of point objects shown in the figure(Figure 1) about the vertical axis. Assume m = 1.2kg , M = 3.0kg , and the objects are wired together by very light, rigid pieces of wire. The array is rectangular and is split through the middle by the horizontal axis. Express your answer using two significant figures. Part B Calculate the moment of inertia of the array of point objects about the horizontal axis. Express your answer using two significant figures. About which axis would it be harder to accelerate this array? About which axis would it be harder to accelerate this array?Explanation / Answer
The moment of inertia for a point-mass is defined as follows:
J=m*r^2
The total moment of inertia is the sum of all point-mass moments of inertia.
where m is the mass of the body, and r is the distance from the axis of rotation(calculation).
So, in your case, we have 2 axes. The axis X and the axis Y.
Let us begin with axis X.
The distance of the masses are as follows:
r_m1=r_m2=0.50/2 = 0.25m
r_M1=r_M2=0.50/2 = 0.25m
So, the moment of inertia around x axis is:
(b) Jx=m1*r_m1^2+m1*r_m1^2+M1*r_M1^2 + M2*r_M2^2=
=2*1.2*0.25^2+2.3*0.25^2=0.525kgm^2 (answer b)
The similiar goes for the inertia around the y axis:
The distance of the masses are as follows:
r_m1=r_M1=0.50 = 0.5m
r_m2=r_M2=1 = 1m
So, the moment of inertia around y axis is:
Jy=m1*r_m1^2+m1*r_m1^2+M1*r_M1^2 + M2*r_M2^2=
=0.5^2*(1.2+3) + 1^2*(1.2+3)=1.05 + 4.2=5.25 kgm^2 (answer A)
So, since Jy>Jx, it would be harder to rotate the system around the y axis.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.