A disc rotates about its axis of symmetry in a horizontal plane at a study rate of 3.5 revolutions per second a coin placed at a distance of 1.25cm from the axis of rotation remains at rest on the disc the coefficient of friction between the coin and the disc : (g = 10m/sq )
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Here's your answer mate
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ashish55587:
what is f here
f is the frequency
can you explain me
what
how at place of omega it is 4π²f²
omega^2=(2πxfrequency)^2
=4π^2xf^2
is it formula
Of course
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see the attachment⬇⬇⬇
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