What would the kinetic energy of the sculpture be if the rod was three times as long but it still had the same angular momentum as in the example?
What would the kinetic energy of the sculpture be if the rod was three times as long but it still had the same angular momentum as in the example?
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Now let's apply our definition of
solution: k= 0.96J
a) What would the kinetic energy of the sculpture be if the rod was three times as long but it still had the same angular momentum as in the example?
![Tne momenl Ol inertia ol each spiere (reateu as a poini) is
Isphere = m;) = (2.0 kg)(2.0 m)? = 8.0 kg - m?
We know that the moment of inertia of a rod about an axis through its midpoint and perpendicular to its length is
Irod = Ms? =
(3.0 kg)(4.0 m)2 = 4.0 kg m?
So the total moment of inertia of the assembly is
Itotal = 2 (8.0 kg m?) + 4.0 kg m? = 20 kg m?
The angular velocity is given in revolutions per minute; we must convert it to radians per second:
3.0 rev
1 min
) (분) (주)
2n rad
w = 3.0 rev/min =
min
60 s
= 0.31 rad/s
1 rev
Now we can calculate the angular momentum of the assembly:
L = Iw = (20 kg m?) (0.31 rad/s) = 6.2 kg m2 /s
The kinetic energy is
K =
Iu = }
2 (20 kg m²) (0.31 rad/s)?
= 0.96 kg m2/s?
0.96 J
REFLECT The mass of the rod is comparable to that of the spheres, but it contributes less to the total moment of inertia
because most of its mass is closer to the axis of rotation than are the masses of the spheres.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F207288bf-51d4-4a08-b949-59b9af04eab1%2Fd4ab3baa-fb11-41df-82b0-df4dec6f9265%2Fv459nil_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Tne momenl Ol inertia ol each spiere (reateu as a poini) is
Isphere = m;) = (2.0 kg)(2.0 m)? = 8.0 kg - m?
We know that the moment of inertia of a rod about an axis through its midpoint and perpendicular to its length is
Irod = Ms? =
(3.0 kg)(4.0 m)2 = 4.0 kg m?
So the total moment of inertia of the assembly is
Itotal = 2 (8.0 kg m?) + 4.0 kg m? = 20 kg m?
The angular velocity is given in revolutions per minute; we must convert it to radians per second:
3.0 rev
1 min
) (분) (주)
2n rad
w = 3.0 rev/min =
min
60 s
= 0.31 rad/s
1 rev
Now we can calculate the angular momentum of the assembly:
L = Iw = (20 kg m?) (0.31 rad/s) = 6.2 kg m2 /s
The kinetic energy is
K =
Iu = }
2 (20 kg m²) (0.31 rad/s)?
= 0.96 kg m2/s?
0.96 J
REFLECT The mass of the rod is comparable to that of the spheres, but it contributes less to the total moment of inertia
because most of its mass is closer to the axis of rotation than are the masses of the spheres.
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