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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 angular momentum to a specific object. A mobile sculpture is suspended from the ceiling of an airport terminal building. It consists of two metal spheres, each with mass 2.0 kg, connected by a uniform metal rod with mass 3.0 kg and length s=4.0ms=4.0m. The assembly is suspended at its midpoint by a wire and rotates in a horizontal plane, making 3.0 revolutions per minute. Find the angular momentum and kinetic energy of the assembly.

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.
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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