Mechanics Lecture 6 PDF

# Note That ## Velocity - $v = \pm \sqrt{value}$ - Case 1: direction of motion is towards +ve - Case 2: direction of motion is towards -ve - x increasing with time, t - Y = +ve sign - x decreasing with time t - V = -ve sign ## Acceleration - $a$ =-ve sign If say in example one of t...

# Note That ## Velocity - $v = \pm \sqrt{value}$ - Case 1: direction of motion is towards +ve - Case 2: direction of motion is towards -ve - x increasing with time, t - Y = +ve sign - x decreasing with time t - V = -ve sign ## Acceleration - $a$ =-ve sign If say in example one of the following - The acceleration direction is towards O (origin) - Retardation [decreasing acceleration] # Example (1) x=0 A particle P is projected from O in the +ve direction of x-axis, with initial velocity $v_o$ at t=0. The particle moves with retardation of magnitude K$V^2$, where V is velocity and k is constant. Find the velocity and displacement at any time t. ## Solution - F = -KV^3/2 - dt = -KV^3/2 dv - $\int$ V dv = -k $\int$ t dt - V = -kt + C1, D-I-C t=0, Vs Vo C1=Vo - V = -kt + Vo - V = (KVo t + 2)^- 1/2 ## Find X(t) and V(t) - dx = (4Vo)/(KVo t + 2)^2 dt - x = (4Vo/K) (KVo t + 2) ^(-1) + C2 - x = (4Vo/K) (KVo t + 2)^(-1) +(4Vo/K) # Example (2) A particle moves in a straight line OX, with an acceleration which always direction to Oo, and varies inversely with the square of its distance from O, if initially the particle was at rest at the point A (2a, 0). Prove that The ratio between The times of motion from A to The point B (0, 0) and from B to O is equal to (π + 2)/(π - 2) ## Solution - F = -K/x^2 K is const. - dV/dx = -K/x^2 => $\int$ VdV = -K $\int$x^-2 dx - 1/2 V^2 + K/x + C1, I-C (x=2a, V=0) => C1 = - (K/2a) - 1/2 V^2 = K/x - K/2a= K(2a-x)/2ax - V = - $\sqrt{k/x}$ √2a - x/x - V= - $\sqrt{k/a}$ √2a - x/√x - -ve sign because x decreasing with time t - dx/ dt= - $\sqrt{k/a}$ √2a - x/√x - $\int$ √x/(2a-x) dx = - $\int$ dt - I = - √(k/a)t + C2 ## Now we calculate The integration I - I = $\int$ √x/(2a-x) dy by substituting - Let x=2 a sin^2 θ dx= 4a sin θ cos θ dθ - I = $\int$ √(2a sin^2 θ)/ (2a - 2a sin^2 θ) 4a sin θ cos θ dθ - I = 2a $\int$ (2 sin θ cos θ dθ / (cos^2 θ) - I = 2a $\int$ 2 sin θ dθ - I = 2a -[- cos 2θ] sin θ=√x/√2a cos θ= √2a - x/√2a - I = 2a [- √(2a - x) / √(2a) + √x/√(2a)] ## From ① into ① - 2a sin √(x / 2a) - √x √(2a-x) = - √(k/a) t + C2 - I-C t = 0, x=2a C2 = 2a sin π/2 - t = √(a/k) π + √x √(2a-x)/√(2a) - 2a sin √x/√2a ## At x = a => t = tAB - tAB = √(a/k) [-π + (√2a)/√2 - π)] = √(a/k) (- π + 1) ## At x = ∞ => t = tA - t = √(a/k) [π - 0] => t = √(a/k) π ## At x= 0 => t =t Bo - tBo = √(a/k) (π - π - 1) => tBo = √(a/k) (- 1 ) ## Ratio - tAB /tBo = (π + 1)/(-1) = (π + 2)/(π - 2) # Example (3) A particle moves in a straight line to wards a center. Which starts from rest at distance ‘a’ from o. If the acceleration always directed to o varies inversely as the cube of distance from o. Find the time of reaching a point at distance b and its velocity ## Solution - F = - k/x^3 k constant, x=a, V=0, t=0, C1 = K/2a^2 - dv/dx = - k/x^3 - ∫vdv = - ∫k/x^3 dx - 1/2 V^2 + k/2x^2 + C1 = -k/2x^2 + C1 - 1/2 V^2 = - k/x^2 + K/a^2 - V = ±√[K(a^2 - x^2)/a^2x^2] => V = ± √(k/a) √(a^2 - x^2)/x - V = - √k/a √(a^2 - x^2)/x -ve sien because x decreasing with time t - dx/dt = -√(K/a) √(a^2 - x^2)/x - ∫(x/√(a^2 - x^2)) dx = ∫ √(K/a) dt - ∫(a^2/√(a^2 - x^2)) dt = √(K/a) dt (t=0, x=a) C2=0 - C2 + 2√(a^2 - x^2) = √(K/a) t + C2 - 2√(a^2 - x^2) = √(K/a) t => t30 = √(a/K) 2 √(a^2 - b^2)