Centripetal Force and maximum force

AI Thread Summary
The discussion revolves around calculating the maximum force exerted by the wall of a front-loading washer on a wet Teddy bear during spin drying. The user initially calculated the centripetal force using the formula Fc = mass (V^2/r) but found their answer incorrect. It was clarified that the maximum force from the wall, denoted as N, is not equal to the centripetal force at all points, especially when the bear is at the bottom of the spin. The correct approach involves considering the forces acting on the bear, specifically the relationship N - mg = Fc to determine the maximum normal force. The user ultimately resolved their confusion with assistance from other forum members.
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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