User:Knowledge Seeker/sandbox

From Wikipedia, the free encyclopedia

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. F0 is applied at the center (x=0); F1 is along the end (x=L/2). K is the kinetic energy; Kt is the translational kinetic energy, and Kr is the rotational kinetic energy.

[edit] F0: x = 0

There will be no rotation.

F = ma

F0 = Ma

a = \frac{F_0}{M}

v(t) = at

K_t(t) = {1 \over 2}m[v(t)]^2

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}

Test: \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}

K_t(t) = \frac{F_0^2 t^2}{2M}

[edit] F1: x = L / 2

There will be some rotation.

\tau_i = F_i l = F_1 \frac{L}{2}

τ = Iα

F_1 \frac{L}{2} = \left( \frac{1}{12} ML^2 \right) \alpha

\alpha = \frac{F_1 \frac{L}{2}}{\frac{1}{12} ML^2} = \frac{12F_1 L}{2ML^2} = \frac{6F_1}{ML}

ω(t) = αt

K_r(t) = \frac{1}{2}I[ \omega (t)]^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[ \left( \frac{6F_1}{ML} \right) t \right]^2 =\frac{36ML^2 F_1^2 t^2}{24M^2 L^2} =\frac{3F_1^2 t^2}{2M}