Rolling grain mill pushing on floor with force equal to twice its weight

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a rolling grain mill having radius R is constrained to roll in a circle of radius r at a constat angular speed of W by a rigid shaft of negligible moment of inertia. For what value of W will the mill stone push on the floor with a force equal to twice its weight?

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my problem... i don't understand :cry: the question. does the question suggest that the faster the mill rolls, the larger the downward force? as in normal? pls offer your guidance and tell me what are the things i should look in order to solve this question. thanks in advance...
 

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As the mill spins is exert and inward and thus an outward force of mw^2r (I hope that right never did get the centrifugal/centripedal difference). Therefore at some speed w the wheel will exert and outwards force equal to its weight and thus push dowonwards with twice its weight (and no net force up).

Well that's my interpretation, but I'm a dope so I could be wrong ^__^
 


Hi there! Don't worry, I will try my best to explain the question and guide you through solving it.

First, let's break down the given information. We have a rolling grain mill with a radius R, which is constrained to roll in a circle of radius r. This means that the mill is moving in a circular motion with a constant angular speed, W. The mill is also connected to a rigid shaft with negligible moment of inertia, which means that the shaft will not affect the motion of the mill.

Now, the question is asking for the value of W that will make the mill stone push on the floor with a force equal to twice its weight. This means that the downward force exerted by the mill stone on the floor should be twice its weight, which is given by the formula F = mg, where m is the mass of the stone and g is the acceleration due to gravity.

To solve this problem, we need to use the concept of centripetal force. In circular motion, there is a constant inward force called centripetal force that keeps an object moving in a circular path. This force is given by the formula F = mv^2/r, where m is the mass of the object, v is the velocity and r is the radius of the circle.

Now, we can equate the centripetal force to the force exerted by the mill stone on the floor. Since we want this force to be twice the weight of the stone, we can write the equation as:

2mg = mv^2/r

We know that the velocity of the mill stone is given by v = Wr, where W is the angular speed and r is the radius of the circle. Substituting this in the equation above, we get:

2mg = m(Wr)^2/r

Simplifying, we get:

2g = W^2r

Finally, we can solve for W by dividing both sides by r and taking the square root:

W = √(2g/r)

Therefore, the value of W that will make the mill stone push on the floor with a force equal to twice its weight is √(2g/r). I hope this helps you understand the problem and how to solve it. Good luck!