233_pdfsam_math 54 differential equation solutions odd

233_pdfsam_math 54 differential equation solutions odd -...

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Exercises 4.7 With C =0 , F ( y )= [ k/ (2 m )] y 2 , and the energy E ( t )= 1 2 [ y 0 ( t )] 2 F [ y ( t )] = 1 2 [ y 0 ( t )] 2 ± k 2 m y 2 ² = 1 2 [ y 0 ( t )] 2 + k 2 m y 2 . By the energy integral lemma, 1 2 [ y 0 ( t )] 2 + k 2 m y 2 =const . Multiplying both sides by 2 m , we get the stated equation. 7. (a) Since, for a point moving along a circle of radius ` , the magnitude v of its linear velocity ~v and the angular velocity ω = dθ/dt are connected by v = ω` =( dθ/dt ) ` , and the vector ~v is tangent to the circle (and so, perpendicular to the radius), we have angular momentum = ` · mv = ` · m · dt ` = m` 2 dt . (b) From Figure 4.18, we see that the component of the gravitational force, mg ,wh ichi s perpendicular to the level arm, has the magnitude | mg sin θ | and is directed towards decreasing θ .Thu s , torque = ` · ( mg sin θ )=
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Unformatted text preview: torque = d dt (angular momentum) `mg sin = d dt m` 2 d dt `mg sin = m` 2 d 2 dt 2 d 2 dt 2 + g ` sin = 0 . 9. According to Problem 8, with ` = g , the function ( t ) satises the identity ( ) 2 2 cos = C = const . (4.9) Our rst purpose is to determine the constant C . Let t a denote the moment when pendulum is in the apex point, i.e., ( t a ) = . Since it doesnt cross the apex over, we also have ( t a ) = 0. Substituting these two values into (4.9), we obtain 2 2 cos = C C = 1 . 229...
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