Question

Please answer the following. Thank you, I'll rate.

A 500 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 4.91 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.

Part a.) If the block starts from rest, how many revoloutions does it make before the tube breaks?

Answer 1

mass of the steel block, m=500 g

m=500*10^-3 kg

length of the tube, L=1.2 m

thrust force, F_thrust=4.91 N

Tension, T=60 N

uk=0.6

a)

Torque=I*alpa

(F_thrust - f)*L=m*L^2*alpa

(F_thrust - uk*m*g)*L=m*L^2*alpa

(4.91-0.6*0.5*9.8)*1.2=0.5*1.2^2*alpa

===> alpa=3.283 rad/sec^2

and

Tension , T=m*w^2*L

60=0.5*w^2*1.2

===> w=10 rad/sec

use,

wf^2 -wi^2= 2*alpa*theta

10^2 - 0 = 2*3.283*theta

===> theta=15.23 rad

no of revolution, n= theta/2pi

n=15.23/(2pi)

n=2.434