University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

132

Pages:

5
Author:

Jabin F.

#Acceleration of object sliding down an inclined plane: equilibrium (normal) N = mg cosθ ΣF = ma Fnet = ma mg sinθ - μN = ma ma = mg sinθ - μmg cosθ a = g(sinθ - μcosθ) for sliding down: a > 0 sinθ > μcosθ μ < tanθ Q1. A projectile is thrown with speed u at an angle with horizontal. Calculate its angular momentum about point of projection after time t. Solution: → R = xi + yj = ut cosθ i + (ut sinθ - (1/2)gt^2) j where x = ut cosθ y = ut sinθ - (1/2)gt^2 v = dx/dt = u cosθ i + (u sinθ - gt) j l = r x p = (xi + yj) x m(vx i + vy j) = m[(xv_y - yv_x)k] = m[(ut cosθ)(u sinθ - gt) - (ut sinθ - (1/2)gt^2)(u cosθ)]k = m[u^2t sinθ cosθ - ugt^2 cosθ - u^2t sinθ cosθ + (1/2)ugt^3 cosθ]k = (1/2)mgut^3 cosθ k Q2. In figure shown, the block is thrown up the inclined plane with some speed. If t1 is time for upward motion and t2 is time for downward motion, then find t1/t2. Solution: Let the block is thrown with speed u. After traveling a distance 's', it comes to rest and then return back and after traveling a distance 's' (same distance) it will get a speed 'u'. Let t1 be upward time and t2 be downward time. Then, for upward motion: 0 = u^2 - 2g(sinθ + μcosθ)s for downward motion: 0 = u^2 - 2g(sinθ - μcosθ)s t1 = sqrt(s/g(sinθ + μcosθ)) t2 = sqrt(s/g(sinθ - μcosθ)) t1/t2 = sqrt((sinθ - μcosθ)/(sinθ + μcosθ)) For downward motion: v^2 = u^2 + 2as => v^2 = 0 + 2g(sinθ - μcosθ)s => v = sqrt(2g(sinθ - μcosθ)s) v^2 = u^2 + 2as => 0^2 = 0 + 2g(sinθ - μcosθ)s => s = v^2/(2g(sinθ - μcosθ)) t2 = v/a = sqrt(2g(sinθ - μcosθ)s)/(g(sinθ - μcosθ)) = sqrt(s/(g(sinθ - μcosθ))) t1/t2 = sqrt(s/(g(sinθ + μcosθ)))/sqrt(s/(g(sinθ - μcosθ))) = sqrt((sinθ - μcosθ)/(sinθ + μcosθ)) => t1 < t2, i.e., upward time is less than downward time.