An object is formed by attaching a uniform, thin rod with a mass of m r = 7.09 k
ID: 1479102 • Letter: A
Question
An object is formed by attaching a uniform, thin rod with a mass of mr = 7.09 kg and length L = 5.24 m to a uniform sphere with mass ms = 35.45 kg and radius R = 1.31 m. Note ms = 5mr and L = 4R.
1)
What is the moment of inertia of the object about an axis at the left end of the rod?
kg-m2
2)
If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 431 N is exerted perpendicular to the rod at the center of the rod?
rad/s2
3)
What is the moment of inertia of the object about an axis at the center of mass of the object? (Note: the center of mass can be calculated to be located at a point halfway between the center of the sphere and the left edge of the sphere.)
kg-m2
4)
If the object is fixed at the center of mass, what is the angular acceleration if a force F = 431 N is exerted parallel to the rod at the end of rod?
rad/s2
5)
What is the moment of inertia of the object about an axis at the right edge of the sphere?
kg-m2
6)
Compare the three moments of inertia calculated above:
ICM < Ileft < Iright
ICM < Iright < Ileft
Iright < ICM < Ileft
ICM < Ileft = Iright
Iright = ICM < Ileft
Explanation / Answer
mass of hte rod
mr = ms/5
(1)
moment of inertia of the sphere i
I = Is + IL = ms ( L+R)^2
I = 2/5 ms R^2 + 1/3 mr L^2 + ms ( L+ R)^2
= 2/5 ms R^2 + 1/3 ( ms/5) ( 4R)^2 + ms ( 4R + R) ^2
= ms R^2 ( 2/5 + 16/15 + 25)
= 35.45 ( 1.31)^2 ( ( 2/5 + 16/15 + 25)
=1610.1 kg m^2
-----------------------------------------------------------------------------------------------
Im = I s + Ir + ms ( L/2 + R)^2
= 2/5 * ms R^2 + 1/12 mrL^2 + ms ( L/2 + R)^2
= 2/5 ms R^2 + 1/12 ( ms/5) 4R)^2 + ms ( 4R/2) + R)^2
= ms R^2 ( 2/5 + 16/60 + 4)
=35.45 ( 1.31)^2 ( 2/5 + 16/60 + 4)
=283.9 kg m^2
T = Im alpha
F ( R+ L/2) = Im alpha
alpha = F ( 3R) Im
= 431 ( 3 ( 1.31)/283.9
=5.96 rad/s^2
(3)
x cm = ms xs + mr + xr/ ms + mr
= ms ( 0) + 7.09 ) ( 5.24/2)/7.09 + 35.45
=0.436 m
I cm = Ir + mr ( R+ L/2) - x cm)^2
= 1/12 ( 7.09 ( 5.24 ) ^2 + ( 7.09 ( 1.31 + 5.24/2) -0.436 m)^2
= 102.77 kg m^2
I s cm = I s + ms ( xcm)^2
= 2/5 ms R^2 + ms ( x cm)^2
= 2/5 ( 35.45 ) ( 1.31)^2 + 35.45 ) ( 0.436 )^2
= 31.07 kg m^2
Icm = I r cm+ I s cm
= 102.77 + 31.07 = 133.84 kg m^2
(4)
if the force is parallel to the rod the angualr accleration would be zero
alpha = 0
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.