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Chapter 8: Q. 70 (page 203)

A 500 g steel block rotates on a steel table while attached to a 1.2-m-long hollow tube as shown in FIGURE CP8.70. Compressed air fed through the tube and ejected from a nozzle on the back of the block exerts a thrust force of 4.0 N perpendicular to the tube.

The maximum tension the tube can withstand without breaking is 50 N. If the block starts from rest, how many revolutions does it make before the tube breaks?

Short Answer

Expert verified

The number of revolution before is breaks is : 0.995

Step by step solution

01

Given information

Mass of block = 500 g = 0.5 kg

Radius of rotation = 1.2-m-long

Thrust force of 4.0 N perpendicular to the tube.

The maximum tension the tube can withstand without breaking is 50 N.

02

Explanation

As the thrust is perpendicular to the tube. so , the tangential component of the thrust force is Ft=4N

and radial component Fr=0

Lets find the angular acceleration of the block

α=Ftmr

Substitute the values

α=4N(0.5kg)×(1.2m)=6.667rad/s2

We can find the angular velocity when it breaks by equating radial force with maximum withstand force

mrω2=50Nω2=50Nmr∣ω2=50N0.5kg×1.2mω=9.129rad/s

Use this angular speed and acceleration to find the rotations made.

ω2=ω02+2αθθ=ω2-ω022aθ=(9.129rad/s)2-(0rad/s)22×(6.667rad/s2)θ=6.25rad=0.995rotation

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