Problem Set 2 Solution

# 3 the equation of motion for the upward motion is

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Unformatted text preview: p; 2k ' g # kv ( (1) (2) where the origin is taken to be the point at which v \$ v0 so that the initial condition is x ) v \$ v0 * \$ 0 . Thus, the distance from the point v \$ v0 to the point v \$ v1 is s ) v0 + v1 * \$ 2-12. (3) The equation of motion for the upward motion is 2 ! g # kv0 " 1 log % 2& 2k ' g # kv1 ( m d2 x \$ # mkv 2 # mg dt 2 (1) Using the relation d 2 x dv dv dx dv \$ \$ \$v 2 dt dt dx dt dx we can rewrite (1) as (2) dmax % 2 v0 g "1 ' sin \$ # 2-15. mg sin θ θ mg The equation of motion along the plane is m dv % mg sin 6 & kmv 2 dt (1) Rewriting this equation in the form 1 dv % dt k g sin 6 & v 2 k 42 (2) CHAPTER 2 We know that the velocity of the particle continues to increase with time (i.e., dv dt ! 0 ), so that " g k # sin \$ ! v2 . Therefore, we must use Eq. (E.5a), Appendix E, to perform the integration. We find 1 k 1 v & ' tanh %1 * +(t)C g g * sin \$ sin \$ + k ,k - (3) The initial condition v(t = 0) = 0 implies C = 0. Therefore, v( g sin \$ tanh k " # gk sin \$ t ( dx dt (4) We can integrate this equation to obtain the displacement x as a function of time: x( g sin \$ . tanh k " # gk sin \$ t dt Using Eq. (E.17a), Appendix E, we obtain x( " # ln cosh gk sin \$ t g sin \$ ) C/ k gk sin \$ (5) The initial condition x(t = 0) = 0 implies C/ = 0. Therefore, the relation between d and t is d( 1 ln cosh k " gk sin \$ t # (6) From this equation, we can easily find t( "# cosh %1 e dk gk sin \$ (7) 2-16. The only force which is applied to the article is the component of the gravitational force along the slope: mg sin 0. So the acceleration is g sin 0. Therefore the velocity and displacement along the slope for upward motion are described by: v ( v0 % " g sin 0 # t x ( v0 t % 1 " g sin 0 # t 2 2 where the initial conditions v " t ( 0 # ( v0 and x " t ( 0 # ( 0 have been used. At the highest position the velocity becomes zero, so the time required to reach the highest (1) (2) "2! gs # "VB2 VB # 2! gs VB # 15.6 m/sec 2-25. a) At A, the forces on the ball are: N mg The track counters the gravitational force and provides centripetal acceleration N " mg # mv 2 R Get v by conservation of energy:...
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