The horizontal rod OA rotates about a vertical shaft according to the relation t
ID: 1438227 • Letter: T
Question
The horizontal rod OA rotates about a vertical shaft according to the relation theta = 2t^2, where theta and t are expressed in rad/s and seconds, respectively. A 300 g collar B is held by a cord with a breaking strength of 27 N. Neglecting friction, determine, immediately after the cord breaks: a. How long it takes for the cord to break b. The relative acceleration of the collar with respect to the rod. c. The magnitude of the horizontal force exerted on the collar by the rod. d. For the situation where the rod OA is rotating at omega = 16 rad/s when the collar is released from its initial position of 0.5 m and will hit the stop at A which is 0.8 m from point O, calculate the angular velocity [rad/s] at this state.Explanation / Answer
a)the chord breaks when radial force becomes 27N
mw2 r = 27
0.3*(2t2)2*0.5 =27
t4 =27/0.6 = 45
t = 2.59s
b)a =27/0.3 = 90 m/s2
c) Since just after chord breaks, the collor is free to move in radial direction,
therefore radial component of the force exerted by rod will be zero
tangential component = m*alpha *r = 0.3 * 4t*0.5 = 1.2t =0.6*2.59 = 1.554N
d) when w=16rad/s
to= sqrt(16/2) =2.828s
radial acceleration of collar = w2r= 16*16*0.5= 128 m/s2
s= 0.8-0.5=0.3m
s = 0.5at2
0.3=0.5*128t2
t2=0.3/64
t=0.068 s
total time= to +t =2.828s +0.068 s =2.896s
REQUIRED w= 2*2.8962= 16.784 rad/s
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