a.- Determine the angular velocity ofthe bar immediately after the collision. b.

Question

a.- Determine the angular velocity ofthe bar immediately after the collision.

b.- Determine the speed v of the1-kilogram object immediately after the collision.

c.- Determine the magnitude of theangular momentum of the object about the pivot just before thecollision.

d.- Determine the angle?.

A 1.0 kilogram object is moving horizontally with a velocity of 10 meters pre second, as shown above, when it makes a glancing collision with the lower end of bar that was ganging vertically at rest before the collision. For the system consisting of the lower end of the hanging vertically at rest before the but kinetic energy is conserved. The bar, which has a length l of 1.2 meters and amass m of 3.0 kilograms, is pivoted about the upper end. Immediately after the collision the object moves with speed v at an angle ? relative to its original direction. The bar swing freely, and after the collision reaches a maximum angle of 90 with respect to the vertical. The moment of internal of the bar about the pivot is I bar ml2

Explanation / Answer

a) (1/2)*I*W^2 = m*g*(L/2)

(1/2)*(1/3)*3*1.2^2*W^2 = 3*9.8*0.6

W = 4.95 rad/s

b) 0.5*m1*u1^2 = 0.5*I*W^2 + 0.5*m1*v^2

1*10^2 = 3*9.8*0.6 + 1*v^2

v = 9.08 m/s

c) L =I*W = (1/3)*m*L^2*W = (1/3)*3*1.2^2*5.94 = 8.55 kg m^2/s

d) m1*u1*L = I*W + m1*v*cos(theta)

1*10*1.2 = 8.55 + 1*9.08*cos(theta)

theta = 67.67 degres

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