- #1

Hoophy

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## Homework Statement

Question: A 600 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 5.01 N perpendicular to the tube. The maximum tension the tube can withstand without breaking is 40.0 N. Assume the coefficient of kinetic friction between steel block and steel table is 0.60.

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

## Homework Equations

My variables and constants:

Steel Block Mass = m = 0.600 kg

Radius = r = 1.2 m

Maximum Tension = T = 40.0 N

Thrust/Tangential Force = F = 5.01 N

g = 9.8 m/s

^{2}

Weight = mg

Coefficient of Kinetic Friction = μ = 0.60

Normal Force = N

Friction = z

Centripetal Acceleration = a

_{c}

Angular Acceleration = α

Maximum Angular Velocity = ω

Maximum Tangential Velocity = v

Tangential Acceleration = a

_{t}

## The Attempt at a Solution

a) I drew a partial free body diagram showing only the vertically acting forces Normal Force (up, positive y) and Weight (down, negative y)

b) I summed the forces in the vertical axis:

F

_{vertical}: 0 = N - mg

c) I isolated N:

N = mg

d) I drew another partial free body diagram showing only Friction (z), Thrust, and Tension. The situation I used is one in which the Steel Block is directly North of the center-point of the circular path such that Tension points South, Thrust points West, and Friction points East.

I defined my coordinate system as:

1) Positive x is West

2) Positive y is South

e) I summed the forces in the y axis:

F

_{y}: ma

_{c}= T

a

_{c}= T/m

f) I summed the forces in the x axis:

F

_{x}: ma

_{t}= F - z

ma

_{t}= F - μmg

a

_{t}= (F/m) - μg

g) I did circular motion shenanigans:

a

_{c}= v

^{2}/r = T/m

See (e)

v = sqrt(Tr/m)

ω = v/r

ω =

**( sqrt(Tr/m) )/r**= ( sqrt[ (40)(1.2)/(0.6) ] )/(1.2) =

**7.45356 rad/s**

α = a

_{t}/r

See (f)

α = ( (F/m) - μg )/r =

**F/(mr) - (μg)/r**= (5.01)/( (0.600)(1.2) ) - (0.60)(9.8)/(1.2) =

**2.05833 rad/s**

^{2}h) More circular motion shenanigans:

Kinematic Equation: ω

^{2}= 2αΔθ

Δθ = ω

^{2}/(2α) = (7.45356)

^{2}/(2(2.05833)) =

**13.4953 rad**

i) I computed the number of revolutions:

Δθ/2π = (13.4953)/2π =

**2.1478 revs**=

**2.1 revs**

**4. A note**

2.1 revs is not the correct answer, nor is 2.1478 revs. Also I do not know what the correct answer is...

I don't know where I went wrong.

Thanks for taking the time to read this!

**[EDIT] 2.1 revolutions is correct.**

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