Problem #1
The AB bar of 2 kg mass is rotated through point A, it turns out that the moment is 8 kgm2.
If it is rotated through the center point O (AO = OB), the moment of inertia becomes. . . .
A. 2 kg m2
B. 4 kg m2
C. 8 kg m2
D. 12 kg m2
E. 16 kg m2
Answer: A
Using the concept of parallel axis theorems we get,
IA = IO + Md2
With IO = moment of inertia of point O, IA = moment of inertia of point A, the length of the stem is
IA = ML2/3
8 kgm2 = 2 kg x L2/3
L = 2√3 m,
so the moment of inertia at point O is
8 kgm2 = IO + (2 kg) (√3 m) 2
IO = 2 kgm2
Or by using the formula for moment of inertia in the stem with a shaft in the middle of the stem, that is
IO = ML2/12 = 2 kg (2√3 m)2/12 = 2 kgm2
Problem #2
If the moment of inertia of a ball towards an axis through the center of the ball is I, then the moment of inertia of the other ball is similar but the radius of the second time is. . . .
A. 2I
B. 4I
C. 8I
D. 16I
E. 32I
Answer: E
The moment of inertia of the ball with the center on the axis is
IO = 2MR2/5 = I
Then a ball similar to radius 2R has an inertial moment equal to,
IO’= 2M’R’2/5
We look for M 'using the concept of density,
ρ = ρ ’
M/(4πR3/3) = M'(4πR’3/3)
M/R3 = M’/(2R)3
M' = 8M
So, the moment of ball inertia with radius 2R is
IO'= 2(8M)(2R)2/5 = 32I
Problem #3
A homogeneous circular disc with radius R starting with M with respect to an axis through the disk pisate and perpendicular to the plane of the disc, moment of inertia I0 = ½ MR2.
A hole cut in the disc as shown in the diagram as shown is ....
A. (15/32) MR2
B. (13/32) MR2
C. (3/8) MR2
D. (9/32) MR2
E. (15/16) MR2
Answer: B
Our first step is to find the mass of the hole (M ') by using,
ρ '= ρ (ρ ’= hole density, and ρ = disk density)
M’/(πR’2t) = M/(πR2t)
M’/(R/2)2 = M/(R2)
M'= M/4
Then the moment of hole inertia against the axis is
Ilubang, poros = ½ M’R’2 = ½ (M/4)(R/2)2 = MR2/32
Moment of inertia hole against disc shaft,
ILubang = IO + M’d2 = MR2/32 + (M/4)(R/2)2 = 3MR2/32
The moment of inertia of the perforated disc is given by
ICacram, hole = IO,disc - Ilubang
ICacram, hole = ½ (M)(R2) - (3MR2/32)
ICacram, hole = 13MR2/32
A. (15/32) MR2
B. (13/32) MR2
C. (3/8) MR2
D. (9/32) MR2
E. (15/16) MR2
Answer: B
Our first step is to find the mass of the hole (M ') by using,
ρ '= ρ (ρ ’= hole density, and ρ = disk density)
M’/(πR’2t) = M/(πR2t)
M’/(R/2)2 = M/(R2)
M'= M/4
Then the moment of hole inertia against the axis is
Ilubang, poros = ½ M’R’2 = ½ (M/4)(R/2)2 = MR2/32
Moment of inertia hole against disc shaft,
ILubang = IO + M’d2 = MR2/32 + (M/4)(R/2)2 = 3MR2/32
The moment of inertia of the perforated disc is given by
ICacram, hole = IO,disc - Ilubang
ICacram, hole = ½ (M)(R2) - (3MR2/32)
ICacram, hole = 13MR2/32
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