A car can just barely turn a corner on an unbanked road at 45 km/h on a dry sunny day. What is the car's maximum cornering speed on a rainy day when the coefficient of static friction has been reduced by 50%?

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Hey, everyone in this problem, we have a race car driver that's able to make a sharp turn on a dry unbanked road at a maximum speed of 75 kilometers per hour. However, due to unexpected rainfall, the coefficient of static friction is reduced by 70%. We're asked to find the maximum speed at which the race car can take that same turn on a rainy day without skidding off the road. And we're told to assume that the conditions of the road are identical except for that change in the coefficient of static friction. We're given four answer choices, option, a 75.1 kilometers per hour. Option B 62.7 kilometers per hour, option C 35.4 kilometers per hour and option D 48.4 kilometers per hour. Now, we have this unbanked road. So we have a flat curve. So we need to think about our centripetal acceleration and we're looking for a maximum speed. So let's recall that our speed is related to this centripetal acceleration through the following. We have ac the centripetal acceleration is equal to V squared divided by R and or V is the speed and R is the radius of that curve. Now, we wanna calculate V, we wanna calculate this speed. We don't actually know this value ac we don't know R so we need to at least figure out this acceleration may now recall that acceleration can be related to our force. OK. So our centripetal acceleration is related to our centripetal force through Newton's loss. So let's go ahead and draw a free body diagram and think about the forces we have acting in the system. We have a race car. There's gonna be a normal force acting upwards, the force of gravity acting downwards. We're gonna assume that this race car has a centripetal acceleration pointing to the right, right. And that means that the force of static friction is going to be acting to the right as well. And this is gonna be the force of static friction max F S max. OK? Because we're talking about that maximum speed that they can take. OK. So looking at our forces, we know that the sum of the centripetal forces is going to be equal to the mass multiplied by the centripetal acceleration. And in the X direction, the only force we have a acting is this force of static friction. So we have the F S max is going to be equal to the mass multiplied by the centripetal acceleration, which we know is equal to V squared over R now because we have this maximum static friction force, we're gonna have the maximum speed as well. So we have F S max is equal to M multiplied by V max squared divided by R. OK. So now we have this V max in our equation that maximum speed we're looking for, we don't know this friction force. OK? We know that it's reduced by 70%. So we know the relationship between the friction force when there's rain versus when there's no rain. But we don't know the absolute value of that number. And so let's, let's look at our two cases when there is no rain. What do we have? Well, we know that the maximum speed is 75 kilometers per hour. So what we get is we get that FX Max, hey, we're gonna start with just writing at that same equation is equal to M V max squared divided by R. I recall that our static friction force is given by us multiplied by N near the coefficient of static friction multiplied by the normal force. And this is gonna be equal to the mass M multiplied by 75 kilometers per hour squared divided by R. Now, we don't know any of this additional information. We don't know the mass, we don't know the radius of this curve. So let's go ahead and move all of the unknowns to the left hand side and have just the numbers on the right hand side. And we get the, the coefficient of static friction us multiplied by the normal force and multiplied by the radius R divided by the mass M is equal to 75 kilometers per hour squared, which is equal to kilometers squared per hour squared. And that's all we can do for the no rain case. We've used all of the information we've been told that's as far as we can go with this situation. So let's switch over to when it's raining. OK? And this is the V max, we wanna find, OK? That maximum speed that can take this corner on a rainy day. We have the same equation. F S max is equal to M V max squared divided by R. Now F S max, we're gonna say that we have us R here. So this is gonna be a coefficient of static friction in the rain multiplied by that normal force. The normal force doesn't change whether there's rain or no rain. And this is equal to the mass multiplied by V max squared divided by the radius R. Now this coefficient of static friction in the rain. Us R again, we don't know the value but we know that it's related to the coefficient when it, we're not in the rank. OK? It's 70% of that original value. So us R is gonna be 0. multiplied by us. OK? 70% of that original coefficient of static friction. When were not in the rape So we have 0.7 us multiplied by N which is equal to M V max squared divided by R. Now, we're gonna do the same thing. We wanna isolate for V max here. So we're gonna move everything else to the left hand side like we did with the no rain case. This is gonna give us 0.7 multiplied by us multiplied by N multiplied by R divided by M and this is equal to V max square. And you'll notice that this us N R divided by M is exactly what we have on the left hand side. For no rain. We had M us multiplied by N multiplied by R divided by M is equal to 5625. OK. So we can actually substitute that value in here. Hey, the N is the same, we have the same normal force. The rain doesn't change that. It's the same corner. So the radius of that corner is going to be the same and we have the same car. So the mass is the same. So we can write that V max squared is going to be equal to 0.7 multiplied by 5625 kilometers squared per hour squared. Ok. If we take the square route and we simplify, we're gonna take the positive route, ok? When we take the square root, we're gonna get both the positive and the negative value. We're talking about speed. So what we care is that magnitude, we don't care about the direction. And so we're gonna take the positive route when we get that V max is equal to 62. kilometers per hour. That is at maximum speed that we can take the corner in the rain that we were looking for. And so this was a really neat problem. We weren't given any information about the mass or the radius of the curve or even the coefficient of static friction. All we were given was the original maximum speed and the relationship between the coefficient of static friction in the rain versus not in the rain. And we were able to solve this problem. We compare what we found to the answer traces and we round to three significant digits. We can see that the correct answer is option B 62.7 kilometers per hour. Thanks everyone for watching. I hope this video helped see you in the next one.

A car can just barely turn a corner on an unbanked road at 45 km/h on a dry sunny day. What is the car's maximum cornering speed on a rainy day when the coefficient of static friction has been reduced by 50%?

Verified Solution

Key Concepts

Coefficient of Friction

Centripetal Force

Maximum Cornering Speed