Answer :
To find the new rotation rate of the space station after reducing its length, we can use the principle of conservation of angular momentum. Angular momentum is conserved when there is no external torque acting on the system. In this case, we assume no external torque is present.
The formula for angular momentum is:
\[ L = I \omega \]
Where:
- \( L \) is the angular momentum
- \( I \) is the moment of inertia
- \( \omega \) is the angular velocity
Since the moment of inertia \( I \) depends on the distribution of mass, and the mass is not changing, we can write:
\[ I_1 \omega_1 = I_2 \omega_2 \]
Where:
- \( I_1 \) and \( \omega_1 \) are the moment of inertia and initial angular velocity, respectively
- \( I_2 \) and \( \omega_2 \) are the moment of inertia and final angular velocity, respectively
Since the space station is rod-shaped and rotating about its center, the moment of inertia can be calculated using the formula for a rod rotating about its center:
\[ I = \frac{1}{12} m L^2 \]
Where:
- \( m \) is the mass of the rod
- \( L \) is the length of the rod
We can set up the equation and solve for \( \omega_2 \):
\[ \frac{1}{12} m_1 L_1^2 \omega_1 = \frac{1}{12} m_2 L_2^2 \omega_2 \]
Given:
- \( L_1 = 907 \) m (initial length)
- \( L_2 = 525 \) m (final length)
- \( m_1 = m_2 = 5.06 \times 10^6 \) kg (mass)
- \( \omega_1 = 1.98 \) rpm (initial rotation rate)
We can solve for \( \omega_2 \):
\[ \omega_2 = \frac{m_1 L_1^2 \omega_1}{m_2 L_2^2} \]
\[ \omega_2 = \frac{(5.06 \times 10^6 \, \text{kg}) \times (907 \, \text{m})^2 \times (1.98 \, \text{rpm})}{(5.06 \times 10^6 \, \text{kg}) \times (525 \, \text{m})^2} \]
\[ \omega_2 \approx \frac{(5.06 \times 10^6 \, \text{kg}) \times (907 \, \text{m})^2 \times (1.98 \, \text{rpm})}{(5.06 \times 10^6 \, \text{kg}) \times (525 \, \text{m})^2} \]
\[ \omega_2 \approx 4.76 \, \text{rpm} \]
Therefore, the new rotation rate of the space station after reducing its length to 525 m will be approximately 4.76 rpm.
Hope this helps!
The formula for angular momentum is:
\[ L = I \omega \]
Where:
- \( L \) is the angular momentum
- \( I \) is the moment of inertia
- \( \omega \) is the angular velocity
Since the moment of inertia \( I \) depends on the distribution of mass, and the mass is not changing, we can write:
\[ I_1 \omega_1 = I_2 \omega_2 \]
Where:
- \( I_1 \) and \( \omega_1 \) are the moment of inertia and initial angular velocity, respectively
- \( I_2 \) and \( \omega_2 \) are the moment of inertia and final angular velocity, respectively
Since the space station is rod-shaped and rotating about its center, the moment of inertia can be calculated using the formula for a rod rotating about its center:
\[ I = \frac{1}{12} m L^2 \]
Where:
- \( m \) is the mass of the rod
- \( L \) is the length of the rod
We can set up the equation and solve for \( \omega_2 \):
\[ \frac{1}{12} m_1 L_1^2 \omega_1 = \frac{1}{12} m_2 L_2^2 \omega_2 \]
Given:
- \( L_1 = 907 \) m (initial length)
- \( L_2 = 525 \) m (final length)
- \( m_1 = m_2 = 5.06 \times 10^6 \) kg (mass)
- \( \omega_1 = 1.98 \) rpm (initial rotation rate)
We can solve for \( \omega_2 \):
\[ \omega_2 = \frac{m_1 L_1^2 \omega_1}{m_2 L_2^2} \]
\[ \omega_2 = \frac{(5.06 \times 10^6 \, \text{kg}) \times (907 \, \text{m})^2 \times (1.98 \, \text{rpm})}{(5.06 \times 10^6 \, \text{kg}) \times (525 \, \text{m})^2} \]
\[ \omega_2 \approx \frac{(5.06 \times 10^6 \, \text{kg}) \times (907 \, \text{m})^2 \times (1.98 \, \text{rpm})}{(5.06 \times 10^6 \, \text{kg}) \times (525 \, \text{m})^2} \]
\[ \omega_2 \approx 4.76 \, \text{rpm} \]
Therefore, the new rotation rate of the space station after reducing its length to 525 m will be approximately 4.76 rpm.
Hope this helps!
