Rate of revolution for a dryer?

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Homework Help Overview

The problem involves determining the rate of revolution for a cylindrical dryer tub that rotates wet clothes. The setup includes a specific angle at which a piece of cloth loses contact with the tub, and the radius of the tub is provided. The objective is to find the required rate of revolution in revolutions per minute.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the derivation of the equation involving the cosine of the angle and its relation to the velocity and gravitational force. There are attempts to clarify the components of forces acting on the cloth and the correct interpretation of the angle in relation to the radial direction.

Discussion Status

Some participants have provided alternative approaches to finding the velocity and questioned the initial assumptions made regarding the forces. There is an acknowledgment of differing interpretations of the angle's role in the calculations, and some guidance has been offered regarding the correct equations to use.

Contextual Notes

Participants note confusion regarding the application of the angle in the context of centripetal acceleration and the components of weight. There is also mention of a potential error in the professor's explanation, leading to further exploration of the problem's setup.

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Rate of revolution for a dryer?!?

Homework Statement



In a home laundry dryer, a cylindrical tub containing wet clothes is rotated steadily about a horizontal axis as shown in the figure below. So that the clothes will dry uniformly, they are made to tumble. The rate of rotation of the smooth-walled tub is chosen so that a small piece of cloth will lose contact with the tub when the cloth is at an angle of 62.5° above the horizontal. If the radius of the tub is 0.385 m, what rate of revolution is needed? (revs/min)



2. The attempt at a solution

First I found the velocity of the dryer tub:

Cos(62.5) = (v^2)/gR
Cos(62.5) = (v^2)/9.8(.385)
v = 1.32m/s

Then I used that velocity to find the rate of revolution:

(v/2(pi)(.385)) * 60sec = 32.74 Revolutions/min

I submitted this answer and it said it was wrong. I can't seem to see what I did wrong. Any help or suggestions would be appreciated. Thanks.
 

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pcandrepair said:
Cos(62.5) = (v^2)/gR
Where does the Cos(62.5) come from?
 
mgCos(62.5) = m(v^2)/R
mgCos(angle) is the horizontal component of mg

The masses would cancel and divide g by both sides to get:

Cos(62.5) = (v^2)/gR
 
Since mg acts downward, it has no horizontal component. Find the component of the weight in the radial direction--that's where the centripetal acceleration points.
 
Cos(62.5) = (v^2)/gR
That was the equation that our professor lead us up to in class to find velocity. Wouldn't the component of the weight in the radial direction just equal mgCos(62.5)?
 
I believe your professor made an error. 62.5 degrees is the angle between the radial and the horizontal, not between the radial and the weight.
 
so to find the velocity i would take the square root of g*R?

v = sqrt(9.8*.385)
v = 1.9424 m/s

then,
(v/(2pi*R)) * 60 = revs/min

(1.9424/(2pi*.385))*60 = 48.178 revs/min
 
pcandrepair said:
so to find the velocity i would take the square root of g*R?
No. Follow my advice in post #4.
 
sin(angle) = v^2 / gR ?
 
  • #10
pcandrepair said:
sin(angle) = v^2 / gR ?

Yes.
 
  • #11
Ok. The answer was correct. Thanks Doc Al!
 
  • #12
Good! (Mark this problem "solved".)
 
  • #13
There is no option in my "thread tools" to mark problems as solved. That's where the option should be right...?
 
  • #14
pcandrepair said:
There is no option in my "thread tools" to mark problems as solved. That's where the option should be right...?
Yes, that's where it should be. You are not the first to state that can't find that option. (Question: Was this thread moved from another forum into this one? That might explain it.)
 

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