Centripetal Force and maximum force

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Homework Help Overview

The problem involves a front-loading clothes washer with a spinning drum, focusing on the centripetal force exerted on a wet Teddy bear inside the drum. The parameters include a rotation speed of 1 rotation per second, a radius of 42 cm, and a mass of 3.9 kg for the bear.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of centripetal force and its components, questioning how the normal force and gravitational force interact in different positions of the bear within the drum.

Discussion Status

Some participants have provided insights on the relationship between centripetal force and the forces acting on the bear, particularly at the bottom of the spin. There is ongoing exploration of how to correctly interpret the problem's requirements regarding the maximum force exerted by the wall.

Contextual Notes

There is a mention of the implications of the drum's perforations on the bear's motion, suggesting that the bear maintains a constant speed during rotation. Participants are also navigating the distinction between centripetal force and the force exerted by the wall on the bear.

uno
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Please help with the following problem:

1. A front-loading clothes washer has a horizontal drum that is thoroughly perforated with small holes. Assuming it to spin dry at 1 rotation per second, have a radius of 42 cm, and contain a 3.9 kg wet Teddy bear, what maximum force is exerted by the wall on the bear?




2. The first step I took was change 42 cm to .42m

The main formula I used was Fc= mass (V^2/r)

R=.42m
T= 1 second
Mass = 3.9 kg

I figured V = 2pi(r)/t = 2 (3.14)(.42)/1 = 2.6389

Fc = 3.9 x (2.6389^2 / .42m = 64.664

According to the online hwk, this answer is incorrect. Please let me know where I am going wrong. Thanks.
 
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Here the centripetal force has two components: the normal force from the wall and the weight.
The way the problem is formulated (with the perforations on the walls) seems to imply a constant speed for the bear during the rotation (the perforations will keep it from sliding). Then the centripetal force should be the same in every position.
However, you may have Fc=N+mg (top), Fc=N-mg (bottom) and so on.
Which one gives the maximum N?
 
uno said:
Please help with the following problem:

1. A front-loading clothes washer has a horizontal drum that is thoroughly perforated with small holes. Assuming it to spin dry at 1 rotation per second, have a radius of 42 cm, and contain a 3.9 kg wet Teddy bear, what maximum force is exerted by the wall on the bear?




2. The first step I took was change 42 cm to .42m

The main formula I used was Fc= mass (V^2/r)

R=.42m
T= 1 second
Mass = 3.9 kg

I figured V = 2pi(r)/t = 2 (3.14)(.42)/1 = 2.6389

Fc = 3.9 x (2.6389^2 / .42m = 64.664

According to the online hwk, this answer is incorrect. Please let me know where I am going wrong. Thanks.
This is a front loading machine, and the problem asks for the max force of the wall on Teddy bear. What are the forces on Teddy when he's at the bottom of the spin?
 
I am still a little confused, could you please be more specific.

Thanks for your help.

Brad
 
I figured it out. Thanks.
 
You did calculate the centripetal force (Fc).
The problem asks for the force exerted by the wall. This is not equal to Fc (excepting the points where bear is at 90 deg from the bottom).

PhantomJay gave you the solution already: calculate the force when the bear is on the bottom. Here you have N-mg=Fc where N is the force from the wall.
How much is N?
 

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