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# 2&10&18 - t is in seconds Required The integral of...

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TAM 212 Spring, 2004 1.2 Given: ( 29 ( 29 ( 29 k i 4 / cos 20 4 / sin 20 t t L p p - = v slug*ft/s, where i and j , are unit vectors fixed in the frame of reference. t is time measured in seconds. Required: ( 29 ) 3 s t L F = & v Solution: Note: Be careful here! You must take the derivative first and then evaluate it for t = 3s. Since we are only worried about one frame of reference, we suppress it in this problem. From the rules of vector derivatives, if k i z x L L L + = v , then k i z x L L L & & & v + = (since the unit vectors are not changing in time). Applying this to what’s given ( 29 ( 29 ( 29 2 / * 4 / cos 20 4 / sin 20 / s ft slug t t dt d L k i p p - = & v (Minor point: The extra factor of 1/s comes from the derivative and the definition of t as in s.) So… ( 29 ( 29 ( 29 lb t t L k i 4 / sin 5 4 / cos 5 p p p p + = & v (Unit check: The units are that of force, as the time derivative of momentum should be.) 1.10 Given: ( 29 ( 29 ( 29 2 / * 4 / cos 20 4 / sin 20 s m kg t t k i F p p - = , where t is in seconds
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Unformatted text preview: t is in seconds . Required: The integral of the force from t = 2s to t = 5s Solution: ( 29 ( 29 ( 29 s m kg dt t t dt s s t s s t / * 4 / cos 20 4 / sin 20 5 2 5 2 -= ∫ ∫ = = k i F p p Again, the extra factor of s comes from the integral and that t is in s. ( 29 ( 29 s m kg dt t dt t s s t s s t / * 4 / cos 20 4 / sin 20 5 2 5 2 -= ∫ ∫ = = p p k i ( 29 ( 29 ( 29 ( 29 [ ] s m kg t t t t / * 4 / sin / 4 20 4 / cos / 4 20 5 2 5 2 = =--= p p p p k i ( 29 ( 29 ( 29 ( 29 [ ] s m kg / * 1 2 / 2 / 4 20 2 / 2 / 4 20------= p p k i ( 29 ( 29 [ ] s m kg / * 1 2 / 2 2 / 2 / 80 k i + + = p 1.18 Done in class ( 29 ( 29 ( 29 lb lb s t L k i k i +-= +-= = p p 2 5 . 2 2 / 2 2 / 2 5 3 & v...
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