- #1
lizzyb
- 168
- 0
This is regarding that planet question. I set up the equation as:
[tex]L_{pi} = L_a + L_{pf} = L_a + I_p \omega_{pf} \Longleftrightarrow \omega_{pf} = \frac{L_{pi} - L_a}{I_p}[/tex]
where [tex]I_p = \frac{2 M R^2}{5}[/tex]
so [tex]\omega_{pf} = \frac{L_{pi} - L_a}{\frac{2 M R^2}{5}}[/tex]
In the original problem, we're given T = 13 hours, so
[tex]T_i = \frac{13 "hours"}{"rev"} \cdot \frac{60 "min"}{1 "hour"} \cdot \frac{60 "sec"}{1 "min"} = \frac{46800 "sec"}{"rev"}[/tex]
Using the final [tex]\omega[/tex] I came up with [tex]T_f = \frac{46801.3 "sec"}{"rev"}[/tex] - isn't that a longer day? Yet the question states "But, thanks to the asteroid's angular momentum, the planet rotates faster after the imapact than it did before."
[tex]L_{pi} = L_a + L_{pf} = L_a + I_p \omega_{pf} \Longleftrightarrow \omega_{pf} = \frac{L_{pi} - L_a}{I_p}[/tex]
where [tex]I_p = \frac{2 M R^2}{5}[/tex]
so [tex]\omega_{pf} = \frac{L_{pi} - L_a}{\frac{2 M R^2}{5}}[/tex]
In the original problem, we're given T = 13 hours, so
[tex]T_i = \frac{13 "hours"}{"rev"} \cdot \frac{60 "min"}{1 "hour"} \cdot \frac{60 "sec"}{1 "min"} = \frac{46800 "sec"}{"rev"}[/tex]
Using the final [tex]\omega[/tex] I came up with [tex]T_f = \frac{46801.3 "sec"}{"rev"}[/tex] - isn't that a longer day? Yet the question states "But, thanks to the asteroid's angular momentum, the planet rotates faster after the imapact than it did before."