A bug walks outward from the center of a turntable that is rot...

Question

A bug walks outward from the center of a turntable that is rotating at 65 rpm. If the coefficient of static friction between the bug and the turntable is 0.42, how far out from the center does the bug get before it starts to slip on the turntable? What happens to the bug after it moves farther out past that point?

Accepted Answer

Welcome back. Everyone in this problem. An insect crawls outward from the center of a spinning record player which turns at 66 revolutions per minute, given a 0.45 static friction coefficient between the insect and the record player. How far does the insect move before sleeping? How does an insect behave after moving farther out? A says, the insect moves 4.093 m before slipping and it continues to crawl outward without any change. B says 3.093 m and it slips towards the center of the record player. C says it's 0.093 m and it slips outward along a radial line away from the center due to the centrifugal force. And the D says it's 0.003 m and that it stops moving entirely. Now, let's try to draw a free body diagram to understand what's going on here with our insect. OK. So we have our insect and if we think about our insect, the force is acting on it. Ok, then we know that the insect's wit is acting downwards and it's being opposed by the normal force. Likewise, we know that a centripetal force is acting on the insect where the force of friction is acting against the centripetal force. Now, so far from the information we have, OK. We know we know that the angular speed omega, OK is 66 revolutions per minute. That's one of the things that we know so far. OK. And now if we wanted to change that to a line, this is its rotational speed. But if we wanted to change it to a linear speed, OK, then we could multiply it by 0.10472 radiance per second. OK. And when we do that, we get omega to also be a speed of 6.9 radiance per second. OK. So we know that that's the value of omega. We also know that our coefficient of static friction mu S equals 0.45. And we can take the acceleration, oops my bad. We can take the acceleration due to gravity G as 9.8 meters per second squared. Now, this distance that we're talking about how far the insect moves before sleeping must be calculated from the center of the spinning player. Now let's take a good look at our diagram notice that based on the diagram, the friction force must equal the centripetal force. So from the diagram, then we can say that are frictional affairs on a mirror. It beside it, our frictional force F fr is equal to the centripetal force which would be equal to Mr Omega squared. Now we know that when two objects are in contact and at rest with respect to each other, static friction is a force acting between them. And a frictional force opposes the relative motion or tendency between two surfaces that are in contact. In this case, our insect and the record player and it occurs only if the external force applied to one object is not sufficient to overcome the frictional force. A static friction force is proportional to the normal force, which is what the surface exerts perpendicular to the objects. Now, the maximum force of static friction is given by the product of the coefficient of static friction and the normal force. OK. So we also know that our frictional force, OK. In this case, our frictional force, our frictional force for a static friction. So let's call that FFRS. OK. And we know that it's the product of the coefficient of static friction mu S and the normal force. OK. So here since the normal force based on our diagram is equal to the weight of our object. OK. Notice that it's acting opposite to the weight of our object, we can really say then that, that frictional, that static frictional force is equal to MU S multiplied by MG since N equals MG. Now, since we have two equations here for our frictional force, we can equate the both of them. OK? Because no, this implies then that MS MG. OK. Or rather sorry, Mr Omega squared, it was MS MG. Notice that both of our equations have m so we can cancel M from both sides. And now if we're trying to figure out that distance, how far our insect gets. OK, then we can devote, divide both sides by omega squared. So that means I is actually equal to the merited below it, I is equal to MGK divided by omega squared. Now, do we have all this information? Well, yes, we do. We know we know what mu is MU S is are coefficient of static friction. We know what our angular speed is and we already have it in terms of its linear speed, which is what we are concerned with here and we know the acceleration due to gravity. So if we substitute that, then it's going to be 0.45 multiplied by G 9.8 m per second squared. And all of that is going to be divided by omega squared. Remember we said it was 6.9 radiance per second. So now we can get the distance R from our formula. And if you were to put that expression into your calculator, you should find that you get R to be equal to 0.09263 m. If we write that to three significant figures, then it's really 0.093 m. OK. So that's how far our insect would get Now, remember the second part of our question says, how does it behave after moving further out? Well, let's think about it. A centripetal force acts on an object moving in a circular path, causing it to change directions continuously but not the speed. That's what the word centripetal means. It means center seeking. So that tells us the force is always pointing towards the center in order to for in order for the object to move in a circular path. This force is essential. Now, based on Newton's law of motion, which says that the force acting on an object is proportional to its mass and acceleration. In the case of circular motion, the acceleration is always directed towards the center of the circle and has a magnitude equal to V squared divided by R where V is the speed of the object and R is the radius of the circle. So that means the force needed to keep the object moving in the circular path is given by the equation F equals ma which equals MV squared divided by R. So in this scenario, that means it's going to slip outward along a radial line away from the center due to the centrifugal force. Therefore, that means C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Centripetal Force

Static Friction

Coefficient of Static Friction

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