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Horizontal Motion with Linear Resistance Suppose a block is projected with initial velocity t, on a smooth horizontal surface and that there is air resistance such that the linear term dominates. Then, in the direction of the motion, Fo = 0 in Equations 2.4.1 and 2.4.2, and F(0) = -co. The differential equation of motion is then do G0 = m dt

Horizontal Motion with Linear Resistance Suppose a block is projected with initial velocity t, on a smooth horizontal surface and that there is air resistance such that the linear term dominates. Then, in the direction of the motion, Fo = 0 in Equations 2.4.1 and 2.4.2, and F(0) = -co. The differential equation of motion is then do G0 = m dt

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Solve step by step (4.3.17) to get (4.3.18) Thank you!

I have included the example 2.4.1 please read all the information well

70 CHAPTER 2 Newtonian Mechanics: Rectilinear Motion of a Particle EXAMPLE 2.4.1 Horizontal Motion with Linear Resistance Suppose a block is projected with initial velocity o, on a smooth horizontal surface and that there is air resistance such that the linear term dominates. Then, in the direction of the motion, Fo = 0 in Equations 2.4.1 and 2.4.2, and F(v) =-c,p. The differential equation of motion is then do -C0 = m dt which gives, upon integrating, mdo m t = Solution: We can easily solve for v as a function of t by multiplying by -cy/m and taking the expo- nential of both sides. The result is v = vge Thus, the velocity decreases exponentially with time. A second integration gives -, tim "dt -(1 showing that the block approaches a limiting position given by X m = muolc).
Transcribed Image Text:70 CHAPTER 2 Newtonian Mechanics: Rectilinear Motion of a Particle EXAMPLE 2.4.1 Horizontal Motion with Linear Resistance Suppose a block is projected with initial velocity o, on a smooth horizontal surface and that there is air resistance such that the linear term dominates. Then, in the direction of the motion, Fo = 0 in Equations 2.4.1 and 2.4.2, and F(v) =-c,p. The differential equation of motion is then do -C0 = m dt which gives, upon integrating, mdo m t = Solution: We can easily solve for v as a function of t by multiplying by -cy/m and taking the expo- nential of both sides. The result is v = vge Thus, the velocity decreases exponentially with time. A second integration gives -, tim "dt -(1 showing that the block approaches a limiting position given by X m = muolc).
Linear Air Resistance We now consider the motion of a projectile subject to the force of air resistance. In this case, the motion does not conserve total energy, which continually diminishes during the flight of the projectile. To solve the problem analytically, we assume that the resisting force varies linearly with the velocity. To simplify the resulting equation of motions, we take the constant of proportionality to be my where m is the mass of the projectile. The equation of motion is then =-myv -kmg (4.3.15) m Upon canceling m's, the equation simplifies to d'r = -yv -kg (4.3.16) Before integrating, we write Equation 4.3.16 in component form * = -yi ÿ =-rÿ 2 =-yż-g (4.3.17) We see that the equations are separated; therefore, each can be solved individually by the methods of Chapter 2. Using the results from Example 2.4.1, we can write down the solu- tions immediately, noting that here y = c,Im, c, being the linear drag coefficient. The results are (4.3.18)
Transcribed Image Text:Linear Air Resistance We now consider the motion of a projectile subject to the force of air resistance. In this case, the motion does not conserve total energy, which continually diminishes during the flight of the projectile. To solve the problem analytically, we assume that the resisting force varies linearly with the velocity. To simplify the resulting equation of motions, we take the constant of proportionality to be my where m is the mass of the projectile. The equation of motion is then =-myv -kmg (4.3.15) m Upon canceling m's, the equation simplifies to d'r = -yv -kg (4.3.16) Before integrating, we write Equation 4.3.16 in component form * = -yi ÿ =-rÿ 2 =-yż-g (4.3.17) We see that the equations are separated; therefore, each can be solved individually by the methods of Chapter 2. Using the results from Example 2.4.1, we can write down the solu- tions immediately, noting that here y = c,Im, c, being the linear drag coefficient. The results are (4.3.18)
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