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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.
I'm testing<ref name="Tester">''The Tester'', issue 12</ref> the new citation style<ref>Cite.php</ref> and can I do this twice?<ref>Cite.php</ref> Probably not; the test<ref name="Tester" /> will probably fail.


==F<sub>0</sub>: x = 0==
<references />
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>

Revision as of 05:47, 23 January 2006

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.

F0: x = 0

There will be no rotation.

F = m a {\displaystyle F=ma}

F 0 = M a {\displaystyle F_{0}=Ma}

a = F 0 M {\displaystyle a={\frac {F_{0}}{M}}}

v ( t ) = a t {\displaystyle v(t)=at}

K t ( t ) = 1 2 m [ v ( t ) ] 2 {\displaystyle K_{t}(t)={1 \over 2}m^{2}}

K t ( t ) = 1 2 M ( a t ) 2 = 1 2 M ( F 0 t M ) 2 = F 0 2 M t 2 2 M 2 = F 0 2 t 2 2 M {\displaystyle 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: F 0 2 t 2 2 M → ( k g ⋅ m s 2 ) 2 s 2 k g = k g 2 ⋅ m 2 ⋅ s 2 s 4 ⋅ k g = k g ⋅ m 2 s 2 {\displaystyle {\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 ) = F 0 2 t 2 2 M {\displaystyle K_{t}(t)={\frac {F_{0}^{2}t^{2}}{2M}}}

F1: x = L / 2

There will be some rotation.

τ i = F i l = F 1 L 2 {\displaystyle \tau _{i}=F_{i}l=F_{1}{\frac {L}{2}}}

τ = I α {\displaystyle \tau =I\alpha }

F 1 L 2 = ( 1 12 M L 2 ) α {\displaystyle F_{1}{\frac {L}{2}}=\left({\frac {1}{12}}ML^{2}\right)\alpha }

α = F 1 L 2 1 12 M L 2 = 12 F 1 L 2 M L 2 = 6 F 1 M L {\displaystyle \alpha ={\frac {F_{1}{\frac {L}{2}}}{{\frac {1}{12}}ML^{2}}}={\frac {12F_{1}L}{2ML^{2}}}={\frac {6F_{1}}{ML}}}

ω ( t ) = α t {\displaystyle \omega (t)=\alpha t}

K r ( t ) = 1 2 I [ ω ( t ) ] 2 = 1 2 ( 1 12 M L 2 ) ( α t ) 2 = 1 2 ( 1 12 M L 2 ) [ ( 6 F 1 M L ) t ] 2 = 36 M L 2 F 1 2 t 2 24 M 2 L 2 = 3 F 1 2 t 2 2 M {\displaystyle 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}}}