The particle has a mass of 0.3 kg and is confined to move along the smooth verti

Question

The particle has a mass of 0.3 kg and is confined to move along the smooth vertical slot due to the rotation of the arm OA. The rod is rotating with a constant angular velocity theta = 2 rad/s. Assume the particle contacts only one side of the slot at any instant. (Figure 1) Determine the magnitude of the force of the rod on the particle when theta = 30 degree. Express your answer to three significant figures and include the appropriate units. Determine the magnitude of the normal force of the slot on the particle when theta = 30 degree. Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

Given

   angular velocity W = 2 rad/s

   theta = 30 degrees
and mass m = 0.3 kg

from the fig , r can be calculated as cos theta = 0.5/r ==> r = 0.577m

the normal acceleration of the rod is a = w^2 *r

               a = 2^2*0.577 m/s3 = 2.308m/s2

and the vertical (normal ) acceleration a_y = a/cos 60 = 0.577/cos60 m/s2 = 1.154 m/s2

from free body diagram of the ball

PArtA

   the magnitude of the force on rod when theta = 30 degrees is

       Fx = F_R cos 60 - F_N =0 ===> F_N = F_R cos 60

       Fy = F_R cos 30 - mg = ma_y

   F_R = m(a_y+g)/ cos 30

   F_R = 0.3(1.154+9.8)/cos30 N = 3.795 N
Part B

   normal force of the slot on the particle is

   F_N = 3.795*cos 60 N

   F_N = 1.8975 N

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