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| Given a long, thin rod of uniform density, in space, at rest relative to our reference frame. Length ''L'', mass ''M''. A very small, light rocket can be attached to the rod at anywhere along its length, providing a constant force ''F''. ''F<sub>0</sub>'' is applied at the center (''x''=0); ''F<sub>1</sub>'' is along the end (''x''=''L''/2). ''K'' is the kinetic energy; ''K<sub>t</sub>'' is the translational kinetic energy, and ''K<sub>r</sub>'' is the rotational kinetic energy. | |||
| ==F<sub>0</sub>: x = 0== | |||
| There will be no rotation. | |||
| <math>F = ma</math> | |||
| <math>F_0 = Ma</math> | |||
| <math>a = \frac{F_0}{M}</math> | |||
| <math>v(t) = at</math> | |||
| <math>K_t(t) = {1 \over 2}m^2</math> | |||
| <math> | |||
| K_t(t) = {1 \over 2}M(at)^2 | |||
| = {1 \over 2}M{\left( \frac{F_0 t}{M} \right)}^2 | |||
| = \frac{F_0^2 Mt^2}{2M^2} | |||
| = \frac{F_0^2 t^2}{2M} | |||
| </math> | |||
| Test: | |||
| <math> | |||
| \frac{F_0^2 t^2}{2M} \rightarrow \frac{\left( \frac{kg \cdot m}{s^2} \right)^2 s^2}{kg} | |||
| = \frac{kg^2 \cdot m^2 \cdot s^2}{s^4 \cdot kg} | |||
| = \frac{kg \cdot m^2}{s^2} | |||
| </math> | |||
| <math>K_t(t) = \frac{F_0^2 t^2}{2M}</math> | |||
| ==F<sub>1</sub>: x = L / 2== | |||
| There will be some rotation. | |||
| <math>\tau_i | |||
| = F_i l | |||
| = F_1 \frac{L}{2} | |||
| </math> | |||
| <math> \tau = I \alpha</math> | |||
| <math>F_1 \frac{L}{2} = \left( \frac{1}{12} ML^2 \right) \alpha | |||
| </math> | |||
| <math> | |||
| \alpha = \frac{F_1 \frac{L}{2}}{\frac{1}{12} ML^2} | |||
| = \frac{12F_1 L}{2ML^2} | |||
| = \frac{6F_1}{ML} | |||
| </math> | |||
| <math>\omega (t) = \alpha t</math> | |||
| <math> | |||
| K_r(t) = \frac{1}{2}I^2 | |||
| =\frac{1}{2} \left( \frac{1}{12} ML^2 \right) \left( \alpha t \right)^2 | |||
| =\frac{1}{2} \left( \frac{1}{12} ML^2 \right) \left^2 | |||
| =\frac{36ML^2 F_1^2 t^2}{24M^2 L^2} | |||
| =\frac{3F_1^2 t^2}{2M} | |||
| </math> | |||