A pocket-sized introduction to dynamics (original) (raw)
Chapter 1 is devoted to problems based on one and two dimensions. The use of various kinematical formulae and the sign convention are pointed out. Problems in statics involve force and torque, centre of mass of various systems and equilibrium. 1.1 Basic Concepts and Formulae Motion in One Dimension The notation used is as follows: u = initial velocity, v = final velocity, a = acceleration , s = displacement, t = time (Table 1.1). (i) v = u + at X (ii) s = ut + 1/2at 2 X (iii) v 2 = u 2 + 2as X (iv) s = 1 2 (u + v)t X In each of the equations u is present. Out of the remaining four quantities only three are required. The initial direction of motion is taken as positive. Along this direction u and s and a are taken as positive, t is always positive, v can be positive or negative. As an example, an object is dropped from a rising balloon. Here, the parameters for the object will be as follows: u = initial velocity of the balloon (as seen from the ground) u = +ve, a = −g. t = +ve, v = +ve or −ve depending on the value of t, s = +ve or −ve, if s =−ve, then the object is found below the point it was released. Note that (ii) and (iii) are quadratic. Depending on the value of u, both the roots may be real or only one may be real or both may be imaginary and therefore unphysical.
A man kicks the 150-g ball such that it leaves the ground at an angle of 60° and strikes the ground at the same elevation a distance of 12 m away. Determine the impulse of his foot on the ball at A. Neglect the impulse caused by the ball's weight while it's being kicked. Solution Kinematics. Consider the vertical motion of the ball where (s 0) y = s y = 0, (v 0) y = v sin 60° c and a y = 9.81 m>s 2 T , (+ c) s y = (s 0) y + (v 0) y t + 1 2 a y t 2 ; 0 = 0 + v sin 60°t + 1 2 (-9.81)t 2 t(v sin 60°-4.905t) = 0 Since t 0, then v sin 60°-4.905t = 0 t = 0.1766 v (1) Then, consider the horizontal motion where (v 0) x = v cos 60°, and (s 0) x = 0, (S +) s x = (s 0) x + (v 0) x t ; 12 = 0 + v cos 60°t t = 24 v (2) Equating Eqs. (1) and (2) 0.1766 v = 24 v v = 11.66 m>s Principle of Impulse and Momentum. (+ Q) mv 1 + Σ L t 2 t 1 Fdt = mv 2 0 + I = 0.15 (11.66) I = 1.749 N # s = 1.75 N # s Ans.