How to Calculate Axle Diameter for Efficient Cement Lifting?

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To calculate the axle diameter for lifting 800N buckets of cement at a steady speed of 2.00 cm/s while rotating at 7.5 rpm, the relationship between linear speed, radius, and angular velocity is crucial. The axle radius was determined to be approximately 0.0255 m, resulting in a diameter of 0.051 m. For the upward acceleration of 0.400 m/s², the angular acceleration was calculated to be around 15.68 rad/s². The discussion emphasizes the importance of understanding tangential acceleration in relation to angular acceleration. Proper application of the relevant equations is essential for solving the problem effectively.
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Homework Statement


You are to design a rotating cylindrical axle to lift 800N buckets of cement from the ground to a rooftop 78.0m above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turns, the buckets will rise.

What should the diameter of the axle be in order to raise the buckets at a steady 2.00 cm cm/s when it is turning at 7.5rpm?

If instead the axle must give the buckets an upward acceleration of 0.400 m/s^2, what should the angular acceleration of the axle be?

Homework Equations



KE=0.5 I w^2
GPE = mgh or force x height
I=0.5 m r^2 (for a disc/cylinder)

The Attempt at a Solution



GPE=800x78=62400

Then i would have though KE=GPE but as there's no mass for the cylinder that can't be right

thanks in advance
 
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Don't worry about energy or moments of inertia, you won't need those for this problem.

Instead, how about the equation that relates radius, speed, and angular velocity?
 
For the steady speed question
1.What does 7.5 rpm translate to in revolutions per second?
2.For the rope to rise at 2cm per second, the 2cm traced out on the circumference of the axle must correspond to the answer of my first question. Now find the axle radius (and hence diameter) which will satisfy this.

For the acceleration question:
You keep the axle diameter you have just calculated then relate the linear acceleration of the bucket to the angular acceleration of the axle.
 
v=rw
0.02m/s=r x 0.785 rad/s
r = 0.0255 m
d=0.051 m
(correct)

a = v^2 / r = w^2 r

a = 0.016 m/s^2

but that's only that radial actually... is that what i want?
 
Last edited:
cantgetno said:
... but that's only that radial actually... is that what i want?

Nope. Try differentiating with respect to time:

v = rw
 
ummm

a=dw / dt ?

a(tan)=r (dw/dt) also
but how do i do dw/dt when i don't have an equation?
 
cantgetno said:
ummm

a=dw / dt ?
The left hand side is correct, but what happened to the "r" from the original equation?

a(tan)=r (dw/dt) also
This time you got the right hand side correct.

but how do i do dw/dt when i don't have an equation?
If you do the derivatives correctly, you'll have the equation. dw/dt will be the only unknown quantitiy.
 
ok using:
a(tan)=r (dw/dt)
0.4=0.255 (a)
a= 15.68 rad/s^2

Sorry i wasnt thinking properly

thanks lots
 
Looks good.

I didn't realize a(tan) meant tangential acceleration. I was trying to figure out how the tangent function got into this :blushing:
 
  • #10
ha sorry ^^

thanks for the help
 
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