33at θ=500
forces are balanced
mω2r+μmg cos(50)=mgsin(50)
9(0.45)+μ(9.8)(0.64)=(9.8)(0.766)
solving this
μ=0.55
q a rod oa rotates about a horizontal axis about o with a constant anticlockwise velocity w=3 radians per second. as it passes the position θ=0 a small block of mas m is placed on it at aradial distance of r=450mm if the block is found to slip atθ=50 the coefficient of static friction b/w block n rod is
(sin 50=.766 n cos 50=0.64)
a)0.2
b)0.55
c)0.8
d)1
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3 Answers
33
3wah!!! priyam bhai kya pic lagaya bhai ..... ek dam killer type .....:P
