Circular Motion With Constant Angular Acceleration

In summary, the conversation discusses a scenario where a 600 g steel block is attached to a 1.20 m-long hollow tube and rotates on a steel table. Compressed air is fed through the tube and ejected from a nozzle on the back of the block, exerting a thrust force of 5.01 N perpendicular to the tube. The maximum tension the tube can withstand without breaking is 40.0 N, and the coefficient of kinetic friction between the steel block and steel table is 0.60. The question asks how many revolutions the block will make before the tube breaks, assuming it starts from rest. The solution involves analyzing forces and circular motion equations to find the angular velocity and number of revolutions.
  • #1
Hoophy
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Homework Statement


upload_2019-2-20_21-14-56.png

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/s2
Weight = mg
Coefficient of Kinetic Friction = μ = 0.60
Normal Force = N
Friction = z
Centripetal Acceleration = ac
Angular Acceleration = α
Maximum Angular Velocity = ω
Maximum Tangential Velocity = v
Tangential Acceleration = at

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:
Fvertical: 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:
Fy: mac = T
ac = T/m
f) I summed the forces in the x axis:
Fx: mat = F - z
mat = F - μmg
at = (F/m) - μg
g) I did circular motion shenanigans:
ac = v2/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 = at/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/s2
h) More circular motion shenanigans:
Kinematic Equation: ω2 = 2αΔθ
Δθ = ω2/(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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  • #2
Hoophy said:
2.1 revs is not the correct answer, nor is 2.1478 revs.
Using your analysis I get the same answer, but I wonder if we need to consider the flow of air in the tube. As that flows out from the centre it gains tangential velocity, and the work for that comes from the thrust.
 
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  • #3
haruspex said:
Using your analysis I get the same answer, but I wonder if we need to consider the flow of air in the tube. As that flows out from the centre it gains tangential velocity, and the work for that comes from the thrust.
My class has not progressed that far yet (I don't even know if this is within the scope of that class) so I doubt they would expect us to model this. Of course, you had no way of knowing that. :oldsmile:

I just checked again, apparently my calculated answer is correct and I simply typed in the wrong number for my submission...
...I guess I'll mark this question as solved. :bugeye:

Thank you for responding!
 
  • #4
Hoophy said:
My class has not progressed that far yet (I don't even know if this is within the scope of that class) so I doubt they would expect us to model this. Of course, you had no way of knowing that. :oldsmile:

I just checked again, apparently my calculated answer is correct and I simply typed in the wrong number for my submission...
...I guess I'll mark this question as solved. :bugeye:

Thank you for responding!
Very good.
My thought doesn't help anyway. We would need more information about the flow. E,g. by making the air at very high pressure the pipe could be kept very thin, so hardly any air mass.
 
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Related to Circular Motion With Constant Angular Acceleration

1. What is circular motion with constant angular acceleration?

Circular motion with constant angular acceleration refers to the movement of an object along a circular path at a constant rate. This means that the object's angular velocity, or the rate at which it rotates, is changing at a constant rate. This type of motion is commonly seen in objects such as planets orbiting around the sun or a ball rolling around in a circular track.

2. How is angular acceleration different from linear acceleration?

Angular acceleration is the rate of change of angular velocity, while linear acceleration is the rate of change of linear velocity. In simpler terms, angular acceleration measures the change in an object's rotational speed, while linear acceleration measures the change in an object's straight-line speed.

3. What is the formula for calculating angular acceleration?

The formula for angular acceleration is α = (ω₂ - ω₁) / t, where α stands for angular acceleration, ω₂ is the final angular velocity, ω₁ is the initial angular velocity, and t is the time it takes to change from ω₁ to ω₂.

4. How does increasing angular acceleration affect circular motion?

Increasing angular acceleration will cause the object to rotate at a faster rate, leading to a shorter period of rotation and a smaller radius of the circular path. This means that the object will cover a larger distance in a shorter amount of time, resulting in a higher linear speed.

5. Can angular acceleration be negative?

Yes, angular acceleration can be negative. This means that the object's angular velocity is decreasing, causing it to slow down and possibly even change direction. Negative angular acceleration is commonly seen in situations where an object is decelerating or changing direction, such as a car turning around a corner.

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