Find the angular velocity of the particle as a function of time in the form omeg

Question

Find the angular velocity of the particle as a function of time in the form omega(t) = A + Bt. A 0.34 kg point mass moving on a frictionless horizontal surface is attached to a rubber band whose other end is fixed at point P. The rubber band exerts a force F = bx toward P, where x is the length of the rubber band and b is an unknown constant. The mass moves along the dotted line. When it passes point A, its velocity is 4.8 m/s directed as shown. The distance AP is 0.6 m and BP is 1.0 m Find the speed of the mass at points B and C. Find b. The figure below shows a thin, uniform bar of length D = 1.27 m and mass M = 0.68 kg pivoted at the top. The rod, which is initially at rest, is struck by a particle whose mass is m = 0.30 kg at a point x = 0.80d below the pivot. Assume that the particle sticks to the rod. If the maximum angle between the rod and the vertical following the collision is 60 degree, find the speed of the particle before impact.

Explanation / Answer

at point A :

Va = speed at A = 4.8 m/s

m = mass = 0.34 kg

force by rubber band = centripetal force

F = m Va2/AP

bx = m Va2/AP

b (0.6) = 0.34 (4.8)2 / (0.6)

b = 21.8 N/m

At B ::

bx = m Vb2/BP

21.8 (1) = 0.34 Vb2/ (1)

Vb = 8.01 m/s

Vc = 4.8 m/s

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