iPad 11:20 PM flipitphysics.com Anobject is formed by attaching . urtorm, thin r
ID: 1770783 • Letter: I
Question
iPad 11:20 PM flipitphysics.com Anobject is formed by attaching . urtorm, thin rod withamass of m, «Ad5 kg and length L·536 m to,nform sphere with mass m-33.25 kg and raius R-1.34 m Note m,-5m and L-4R )What is the moment of inertia of the object about an ais at the left end of the rod 2, If the object thed at the lefted of the rod, what is the anpar acceleration if a force F·40 s exerted perpendicular to the rod at the center of the rod 31 what ts the moment of inertia of the object about an axis at the center of mass of the objectNote: 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 4) If the object is faed at the center of mass, what is the angular acceleration if a force F-400 N is eserted parallel to What is the moment of inertis of the object about an axis at the right edge of the sphere l Compare the three moments of inertia calculated aboveExplanation / Answer
Given: m1= 6.65 kg, L= 5.36 m, m2= 33.25 kg, R=1.34 m
(1) The moment of inertia for a rod rotating around its center is J1=1/12*m*(r)^2
In this case J1=1/12*m1*(L)^2
J1=15.92 kg*m2
The moment of inertia for a solid sphere rotating around its center is J2=2/5*m*(r)^2
In this case J2=2/5*m2*(R)^2
J2= 23.88 kg*m2
As the object rotates around the free end of the rod then for the sphere the axis around what it rotates is at a distance of d2=L+R
For the rod it is d1=1/2*L
From Steiner theorem
for the rod we get J1"=J1+m1* (d1)^2
J1"= 63.68 kg*m2
for the sphere we get J2"=J2+m2*(d2)^2
J2"=1516.47 kg*m2
And the total moment of inertia for the first case is
Jt1=J1"+J2"
Jt1=1580.15 kg*m2
(2) F= 400 N
The torque given to a system in general is, M=F*d*sin(a) where a is the angle between F and d and where d is the distance from the rotating axis. In this case a=90" and so
M=F*L/2
M=1072 Nm
The acceleration can be found from
e1=M/Jt1
e1=1072/1580.15
e1= 0.678 rad/s2
(3) the text to be right in the case where the center of mass is.
Again we have to use Steiner theorem
In this case h1=(L+R)/2
and h2=R/2
So
J1""=J1+m1*(h1)^2
J1""= 90.55 kg*m2
J2""=J2+m2*(h2)^2
J2""=38.81 kg*m2
and
Jt2=J1""+J2""
Jt2=129.356 kg*m2
(4) F=400 N
M=F*(L+R/2)*sin(a) In this case a=0" and so
M=0
and thus
e2= 0 rad/s2
According to Chegg guidelines I can answer these many sub parts in a question, so please post other parts in a separate question. Please rate my answer if you find it helpful, good luck...
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.