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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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?
![**Text Transcription for Educational Website:**
The moment of inertia of each sphere (treated as a point) is
\[ I_{\text{sphere}} = m\left(\frac{s}{2}\right)^2 = (2.0 \, \text{kg})(2.0 \, \text{m})^2 = 8.0 \, \text{kg} \cdot \text{m}^2 \]
We know that the moment of inertia of a rod about an axis through its midpoint and perpendicular to its length is
\[ I_{\text{rod}} = \frac{1}{12}Ms^2 = \frac{1}{12}(3.0 \, \text{kg})(4.0 \, \text{m})^2 = 4.0 \, \text{kg} \cdot \text{m}^2 \]
So the total moment of inertia of the assembly is
\[ I_{\text{total}} = 2 \times (8.0 \, \text{kg} \cdot \text{m}^2) + 4.0 \, \text{kg} \cdot \text{m}^2 = 20 \, \text{kg} \cdot \text{m}^2 \]
The angular velocity is given in revolutions per minute; we must convert it to radians per second:
\[ \omega = 3.0 \, \text{rev/min} = \left(\frac{3.0 \, \text{rev}}{1 \, \text{min}}\right)\left(\frac{1 \, \text{min}}{60 \, \text{s}}\right)\left(\frac{2\pi \, \text{rad}}{1 \, \text{rev}}\right) = 0.31 \, \text{rad/s} \]
Now we can calculate the angular momentum of the assembly:
\[ L = I\omega = (20 \, \text{kg} \cdot \text{m}^2)(0.31 \, \text{rad/s}) = 6.2 \, \text{kg} \cdot \text{m}^2/\text{s} \]
The kinetic energy is
\[ K = \frac{1}{2} I \omega^2 = \](/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:**Text Transcription for Educational Website:**
The moment of inertia of each sphere (treated as a point) is
\[ I_{\text{sphere}} = m\left(\frac{s}{2}\right)^2 = (2.0 \, \text{kg})(2.0 \, \text{m})^2 = 8.0 \, \text{kg} \cdot \text{m}^2 \]
We know that the moment of inertia of a rod about an axis through its midpoint and perpendicular to its length is
\[ I_{\text{rod}} = \frac{1}{12}Ms^2 = \frac{1}{12}(3.0 \, \text{kg})(4.0 \, \text{m})^2 = 4.0 \, \text{kg} \cdot \text{m}^2 \]
So the total moment of inertia of the assembly is
\[ I_{\text{total}} = 2 \times (8.0 \, \text{kg} \cdot \text{m}^2) + 4.0 \, \text{kg} \cdot \text{m}^2 = 20 \, \text{kg} \cdot \text{m}^2 \]
The angular velocity is given in revolutions per minute; we must convert it to radians per second:
\[ \omega = 3.0 \, \text{rev/min} = \left(\frac{3.0 \, \text{rev}}{1 \, \text{min}}\right)\left(\frac{1 \, \text{min}}{60 \, \text{s}}\right)\left(\frac{2\pi \, \text{rad}}{1 \, \text{rev}}\right) = 0.31 \, \text{rad/s} \]
Now we can calculate the angular momentum of the assembly:
\[ L = I\omega = (20 \, \text{kg} \cdot \text{m}^2)(0.31 \, \text{rad/s}) = 6.2 \, \text{kg} \cdot \text{m}^2/\text{s} \]
The kinetic energy is
\[ K = \frac{1}{2} I \omega^2 = \
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