An object is formed by attaching a uniform, thin rod with a mass of m r = 6.98 k

Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.98 kg and length L = 5.04 m to a uniform sphere with mass ms = 34.9 kg and radius R = 1.26 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 = 401 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 = 401 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

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

Explanation / Answer

Given that,

mr =6.98 kg and L = 5.04m

ms = 34.9 kg and r = 1.26m

ms = 5mr and L = 4R

(1)the moment of inertia of the object about an axis at the left end of the rod

The moment of inertia for a rod rotating around its center is Jr=1/12m(r)2

Jr = 1/12(mr)(L)2 = 1/12 (6.98)(5.04)2 = 14.77 kg-m2  

The moment of inertia of a solid sphere rotating around its center will be Js = 2/5m(r)2

Js = 2/5(ms)(rs) = 2/5(34.9)(1.26)2 = 22.16 kg-m2

As 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 Jr"=Jr+mr (d1)2 = Jr"= 59.09 kg-m2
for the sphere we get Js"=Js+ms(d2)2 =  Js"= 1407.34 kg-m2
And the total moment of inertia for the first case is
J=Jr"+Js" = 59.09 + 1407 = 1466.431

J = 1466.431 kg-m2

(2) Angular acceleration to be calculated when F = 401 N is excerted

The torque given to a system in general is = Fd Sin (theta), where (theta) is the angle bet F and d = distance from rotating axis. Here (theta) = 90 degrees

Torque = 401 x 5.04/2 = 1010.52 Nm

Acceleration can be calculated from, a = Torque / J = 1010.52 / 1466.4 = 0.689 m/sec2

Hence a = 0.689 m/sec2

(3) moment of inertia of the object about an axis at the center of mass of the object

Again we have to use Steiner theorem
In this case h1=(L+R)/2
and h2=R/2
So
Jr""=J1+mr(h1)2 = Jr""= 291.80 kg m2
Js""=J2+ms(h2)2 = Js""= 36.011 kg m2
and
J' =J1""+J2"" = 291.80 + 36.011 = 327.8 kg m2

J' = 327.8 kg m2

(4) the angular acceleration if a force F = 401 N is exerted perpendicular to the rod at the center of the rod

Again, Torque = Fx(L+R)/2 = 401 x (5.04+1.26)/ 2 = 1263.15 N m

a = Torque / J' = 1263.15/ 327.8 = 3.8

a = 3.8 m/sec2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.