Newton_Second_Law_of_Motion

Newton_Second_Law_of_Motion - Newtons 2nd Law in one Space...

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1 Newton’s 2 nd Law in one Space Dimension d dt m V = F or m d V + V d m = F If the body is rigid, then m d V = F because d m = 0 Since ! = d V Then we obtain the popular form of Newton’s Second Law : F = m where: F force external to the body ( Nt) m the mass of the solid body ( ρ solid ( π d 3 /6), kg ) V the velocity of the solid body ( m/s ) t the time ( s ) α acceleration of the body ( m/ s 2 Newton’s 2 nd law applied to free falling bodies in vacuum s s = 0 h m W D m d V d t = - m g ! d V d t = - g where: m the mass of the solid body ( solid ( d 3 /6), kg ) V the velocity of the solid body ( m/s ) t the time ( s ) g acceleration of gravity (9.81 m/s 2 ) W weight of sphere ( N ) h original height ( m ) d diameter of the solid body ( m ) s vertical distance ( m )
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2 Since the mass has been eliminated the body’s acceleration and consequently its falling velocity is not a function of the size and makeup of the solid. The free-fall velocity is found by the solution of the simple differential equation by separation of variables: d V d t = - g ! V = - g t + c 1 Initial condition: t = 0, V = 0, c 1 = 0 , or V = - g t . The velocity is related to distance by: ds/dt = V , or d s d t = - g t ! a second integration yields s = - 1 2 g t 2 + c 2 Application of the second initial condition: t = 0, s = h, c 2 = h, gives: s = - 1 2 g t 2 + h Newton’s 2 nd law for free falling bodies when atmospheric air is present Let us now consider the example illustrating Newton’s 2 nd law, see figure below, when atmospheric air is present. Assume the solid to be spherical with a diameter ( d ). Newton’s 2 nd law m d V d t = - m g + 1 2 ! air C D A V 2 (1) Where: ρ air the density of air (1.1774 kg/m 3 ) solid the density of solid (7,849 kg/m 3 ) C D the drag coefficient of the solid body (for a sphere C D = 0.4 (found in tables of fluid mechanics books), no dimensions, dimensionless number, or clear number,) A projected area of the solid on a plane perpendicular to the body’s motion ( π d 2 /4, m 2 ) F D drag force ( N ) D diameter of the chute (m) s s = 0 h m W D F Dividing Eq. (1) by m we have: d V d t = - g + K m V 2 where K = 1 2 air C D A ` (2) This equation is of the Riccati type and it can be solved by the separation of variables method. d V - g + K m V 2 = dt ! d V - g + K m V 2 = dt + c 1 ________________________________________________________________________ From integral tables
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3 d x - b + a x 2 = 1 2 ab ln a x - b a x + b ________________________________________________________________________ 1 2 g K m ln K m V - g K m V + g = t + c 1 ! ln K m V - g K m V + g = 2 g K m t + c 1 ' , c 1 ' = 2 g K m c 1 or K m V - g K m V + g = c 1 '' exp 2 g K m t , c 1 = exp c 1 ' K m V - g = c 1 exp 2 g t K m V + g K m V - = K m V c 1 exp 2 g K m t + c 1 g exp 2 g K m t K m V - K m 1 exp 2 g K m t = c 1 g exp 2 g K m t + g K m V 1 - c 1 exp 2 g K m t = g 1 + c 1 exp 2 g K m t V = g m K 1 + c 1 exp 2 g K m t 1 - c 1 exp 2 g K m t Initial condition: t = 0, V = 0 0 = g m K 1 + c 1 exp 2 g K m 0 1 - c 1 exp 2 g K m 0 ! 0 = 1 + c 1 1 - c 1 Therefore, c 1 ’’ = -1, or V = g m K 1 - exp 2 g K m t 1 + exp 2 g K m t For large times (mathematically when t ) the velocity is approaching an asymptotic value: V lim t ! " = V T = - g m K V T i s known as the terminal velocity or V = V T 1 - exp 2 g K m t 1 + exp 2 g K m t
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This note was uploaded on 12/07/2011 for the course ENGR 213 taught by Professor Mr.ram during the Spring '10 term at Concordia Canada.

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Newton_Second_Law_of_Motion - Newtons 2nd Law in one Space...

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