A 400 g steel block rotates on a steel table while attached to a 1.20 m -long ho

Question

A 400 g steel block rotates on a steel table while attached to a 1.20 m -long hollow tube. Compressed air fed through the tube and ejected from a nozzle on the back of the block exerts a thrust force of 5.21 N perpendicular to the tube. The maximum tension the tube can withstand without breaking is 60.0 N . Assume the coefficient of kinetic friction between steel block and steel table is 0.60. (Figure 1)

Part A: If the block starts from rest, how many revolutions does it make before the tube breaks?

Explanation / Answer

here

Fnet = F1 - u * m * g

Fnet = 5.21 - 0.6 * 0.4 * 9.8

Fnet = 2.858 N

then

a = Fnet / m

a = 2.858 /0.4 = 7.145 m/s^2

then

alpha = a/r = 7.145 / 1.2 = 5.95 rad/sec^2

Breaking force = Fb = 50 = m * r * w^2 where w^2 = 2 * alpha * theta

theta = 60 / (2m * r * alpha)

theta = 60 / ( 2 * 0.4 * 1.2 * 5.95 )

= 10.5 radians = 1.67 rev

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