# What is the RATE of Rotation help

• 8parks11
In summary, an automatic tumble dryer with a basket of 0.65m diameter rotates at a rate of 37 revolutions per minute in order for the clothes to fall away from the basket's edge at a point 60 degrees from the vertical. The equation used to calculate this rate is 2.53 multiplied by 60 and then divided by twice the value of pi multiplied by the radius of the basket.
8parks11
What is the RATE of Rotation! help!

An automatic tumble dryer has a 0.65m diamter basket that rotates about a horizontal axis. As the basket turns, the clothes fall away from the basket's edge and tubmle over. IF the colthes fall away from the basket at a point 60 degrees from the vertical, what is the rate of rotation in units of revolutions per minute.

i thought this problem would be extremely simple.

I was wrong :)

First i get

V^2 / rg = tan theta
so i get

Vc^2 = rg tan theta
soif ifind Vc i get 1.785

Now this wouldn't make snese...
does someone else have a different Apporpach

ok i can't get mass from anything. what do i do?

it has tobe something like 0.65 / x = tan theta or something?

F=mv^2/r

At sixty degrees the vertical part of this centripetal force is provided solely by the weight of the clothes ie mg.

thus mv^2/r*cos60 = mg, am I wrong?

v= sqrt(r*g/cos60) = 2.53m/s? Then remember you have to make it revs/min

cannot see why you are using tan

the answer is 37revs/minute
i did get 2.53m/s and then multiplied it by 60/2pi
wrong wrong :(

Yeah I get 74 revs/minute =/

I did 2.53/r to get it in rads/sec and then /2pi * 60 = 74...

2 times 37, strange

maybe because the problem gave it in diameters and u used that.
i'm still trying to get it

wait i still don'tget how uget 2.53m/s to 74rev/min

2.53 / (2pi x 60) doesn't work nor (2.53/2pi) x 60

no i used radius and I told you 2.53/r to get radians per seconds.

2.53/2pi*60 doesn't mean anything. What you want is

2.53/r/2pi*60 to get 74

ok so any idea why it is 74 instead of 37?

Unfortunately no clue =(

can you give me the exact equation that gives 74.
ive tried so many.. it's so weird.

so 2.53m/s is in seconds so i multiply it by 60 to make it in minutes.
so it turns 2.53x60 m/minutes
now since its in meters i have to divide it by 2pi to make it by revolutions
this gives me 24... Not 74... :(

am i looking for r? what did i do wrong?

What is 2.53 it is velocity, IE m/s

why would 2.53/2pi give you revs/s? It does not. However 2.53/r gives you rads/s which divided by 2pi gives you revs/s

The exact equation as stated before is!

2.53/r/2pi*60 to get 74

(2.53 x 60) / (2 * pi * .65) = 37.1688006

I think you're forgetting r.

BSCS said:
(2.53 x 60) / (2 * pi * .65) = 37.1688006

I think you're forgetting r.

But r is .65/2 is it not?

"0.65m diamter", you are correct.

Last edited:

## What is the rate of rotation?

The rate of rotation is the speed at which an object rotates around its axis. It is typically measured in rotations per unit of time, such as degrees per second or revolutions per minute.

## How is the rate of rotation calculated?

The rate of rotation can be calculated by dividing the angle of rotation by the time it takes for the object to complete one full rotation. This can be expressed as angular velocity, which is measured in radians per second.

## Why is the rate of rotation important?

The rate of rotation is important because it affects the overall motion and behavior of an object. It can determine the stability and balance of an object, as well as its ability to generate forces and move certain distances.

## What factors can affect the rate of rotation?

The rate of rotation can be affected by a variety of factors, including the shape and size of the object, the distribution of mass, the applied forces, and external factors such as friction and air resistance.

## Can the rate of rotation change over time?

Yes, the rate of rotation can change over time. If the forces acting on an object change, its rate of rotation can also change. External factors such as air resistance can also cause the rate of rotation to decrease over time.

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