(II) A car drives in a spiral path of gradually decreasing rad...

Question

(II) A car drives in a spiral path of gradually decreasing radius in a large empty parking lot. The speed is held constant at 10.0 m/s, and the car starts at a distance of 50.0 m from the center of the spiral. If the coefficient of static friction is 0.80, how far is the car from the center when it first starts to skid?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice prom together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem, a motorcycle spirals inward on a large open track, maintaining a constant speed of 12.0 m per second from an initial position of 52.0 m from the center, given the friction coefficient between the tire and the track is 0.90. At what distance from the center will the motorcycle begin to slide? So that's our goal. Our goal, we're ultimately trying to figure out what the distance from the center will the motorcycle begin to slide? And that is our final answer that we're ultimately trying to solve for is what is this distance from the center of the track that the motorcycle will begin to slide? Awesome. So with that in mind, let's look at our multiple choice answers. Let us know that they're all in the same units of meters. So let's read them off to see what our final answer might be. A is 12 B is 13 C is 16 and D is 18. OK. So first off, let us draw a free body diagram to help us better visualize the forces at work in this particular problem. So I've gone and had and put together a fancy free body diagram for us to refer to the green rectangle represents our motorcycle and also has the word motorcycle written inside of it in case we forget and going up from our motorcycle pointing in the upwards direction, we have our FN our normal force. And of course, pointing in the downwards direction, we have our weight force of the motorcycle which is the mass of the motorcycle multiplied by the acceleration due to gravity, which is represented by G. And of course, mass is represented by M. And then we also have pointing to the left, we have our friction force which is denoted as capital F subscript fr and it is opposing the motion of our motorcycle. So it's going opposite to the direction that the motorcycle is going. Awesome. So with that in mind, now that we have a better visualization of what's happening in this problem. Let's take a moment to now write down all of our known and unknown variables. So let us note as we should be able to recall that G, the acceleration due to gravity is equal to 9.81 m per second squared. And let us also note that V the speed is equal to 12.0 m per second. And that is once again the speed of our motorcycle, we also know the coefficient of static friction, which we're gonna write as us. And this is given to us in the prom itself as 0.90. And finally, we do not know what the value of R is. Art is unknown. And this is the final answer that we're trying to solve for. And this is the distance from the center will the motorcycle begin to slide. So that's what R denotes. So at this point, as we should recall, the centripetal force is required in order to keep moving the observed object in a path of circular motion. And this value is given as a product of the object's mass and its centripetal acceleration which we can write as and let us call this equation one that FC, the centripetal force is equal to the mass multiplied by AC where M is the mass and AC is the centripetal acceleration. Let us note that since the centripetal acceleration is given as the quotient of the squared speed of the object and the radius of the circle circular path, we can rewrite equation one as and let's call this equation two, that FC is equal to M multiplied by V squared divided by R. So now we must note that we are dealing with static friction. So we must recall and use the equation for static friction, which is, and let us call this equation three, an equation three states that FFR is equal to MU s multiplied by FN Awesome. Where FN once again is the normal force acting on an object which is equal to the weight of the object. Thus, when we consider this in equation three, so if we consider this particular piece of information and apply it to equation three, equation three will become, and let's call this equation four, that SFR is equal to me s multiplied by M multiplied by G. So now at this point, if we set equations two and four equal to each other, we will get, and let's call this equation five parenthesis. M multiplied by V squared parenthesis divided by R is equal to us multiplied by M multiplied by G. So now at this point, we need to rearrange equation five to isolate and sulfur R which is our final answer. So I'm gonna draw a little line to separate it. So we can write equation six which once again, we're rearranging equation five to isolate and sulfur R. So when we use a little bit of algebra, we will find that R is equal to V squared divided by parenthesis us multiplied by G. So now at this point, we need to plug in all of our known variables into equation six to determine what the value of how far the R distance is from the center. And this is our final answer that we're solving for. So we will find that R is equal to which we know the value of V is 12.0 m per second. And don't forget that our speed is squared divided by are coefficient of static friction, which is 0.90. And we need to multiply this by the acceleration due to gravity which is 9.81 m per second squared. So when we plug that into our calculator, we will find that R is equal to 16.3 meters, which is approximately equal to 16 m when we round to the nearest hole number. And that's it we solve for this problem. Hooray. So looking at our multiple choice answers, the correct answer has to be C 16 m and that's it. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

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