A vertical shaft rotates at a constant angular velocity of 0.8 rad/s while also

Question

A vertical shaft rotates at a constant angular velocity of 0.8 rad/s while also moving in the vertical direction according to z = 3cos theta metres. A frictionless horizontal bar is attached to the vertical shaft as shown, and has a mass of 20kg sliding on it. The mass is attached to a spring, and executes a motion according to r = (3sin theta + 5) metres from the intersection point of the horizontal bar and vertical shaft. If theta is the angular displacement of the vertical shaft around its axis of rotation, what is the force applied by the horizontal bar to the mass at theta 120 degrees? What is the force applied by the spring to the mass at this same time? The apparatus is located on earth.

Explanation / Answer

Given:
Vertical motion of the shaft, z= 3cos m

Rotary motion of the rod , r = (3 sin +5) m

Speed of shaft, d/dt= 0.8 rad/s

Solution:

The veritcal motion of shaft , z= 3cos

then, dz/dt= -3sin(d/dt) m/s

= -3sin(0.8)= -2.4 sin m/s

then d2z/dt2= -2.4cos(d/dt)= -2.4cos(0.8) = -1.92 cos

the force acting on the mass doe to vertical motion of the shaft is given by Newton's second law

F= mxdv/dt = 20 x(-1.92 cos) = - 38.4cos N

Therefore at = 1200 the force is given by F= -38.4cos 120 = 19.2 N

The horizontal motion of the rod is given by

  r = (3 sin +5) m

dr/dt= 3cos(d/dt) = 3cos(0.8) = 2.4 cos m/s

d2r/dt2 = -2.4 sin (d/dt) = -2.4 sin (0.8) = - 1.92sin m/s2

The spring force is given by Fs = m x d2r/dt2 = 20 x ( - 1.92sin) = - 38.4sin N

at = 1200 Fs = -38.4 sin1200 = -33.26 N (towards the shaft)

therefore,

F= 19.2 N

Fs = -33.26 N   

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