5 an f1 car of mass 730 kg is traveling around a curve...

5. An F1 car of mass 730 kg is traveling around a curve, increasing its speed at a constant rate of 2 m/s². At a given instant, a speed radar reads that the speed of the car on a dangerous turn is 160 km/hr. Given that at this instant, the net force on the car has reached its maximum allowable value of 6000 N, what is the radius of curvature of the turn? (A) 318 m (B) 248 m m = 730 (C) 2840 m (D) 244 m (E) 240 m 6000 = 730

Transcript

00:01 So in this question, you have a car which weighs, which has a mass 750 kilograms, and it travels as a at a speed, which is v equals, sorry, 90 kilometer per hour, right? and you ask lecture to convert this bit into meters per second.

00:25 So you can just rewrite kilometers as 10 to the power of 3 meter divided by 1 .0, which is 3 ,600 seconds, right? and then you'll find this to be 90 divided by 3 .6 meter per second, right? so that would be the velocity.

00:50 That would be the speed in meter per second.

00:54 And you ask if this, well, this guy is actually going about a curve with radius of curvature equal to 160 meters.

01:05 You ask, what should the banking angle of the curve be so that the only force between pavement and the tass of the car is a normal reaction first.

01:12 So let's just try to sketch it.

01:15 Imagine this car is making a curve, going around the curve like this, and the curve has a radius.

01:27 Of curvature, other equals, according to the question is 1 to 60 meters.

01:35 And imagine you have a bank, right? so the bank goes up, the bank goes of this, really, right? so this is an angle theta, right? so the car actually is, let's say the car is actually somewhere around here, and it's moving around this curve, right? this, okay? and there's an angle theta here.

01:56 Now you look at the car, there's some forces that i can on it.

01:59 And there's a gravity for us, of course.

02:02 And of course, there's a normal force.

02:04 Now, according to the questions, suppose at this critical angle, the normal force is just perpendicular, of course, to the slope, to the bank slope.

02:14 That would be it.

02:15 So this is the angle seta we're looking at.

02:18 And of course, it's making the term.

02:20 So you must also, you know, if you decompose this normal force, if you decompose this normal force into two directions, once there's, there's this normal force.

02:32 This, then the horizontal force will actually provide, we actually provide, so this is actual cta, right? the horizontal component of this first, of this normal force, we provide the acceleration that occurs when the car makes the turn, right? so you can see that that will be v squared, that's the speed of the car, divided by r, the radius of the curvature.

03:00 This is acceleration, right? and that must be given by the horizontal component of this force, right? i'll call this force n, right? so that would be n times sin sita, basically.

03:14 So this is a normal reaction force, and it's perpendicular to slab...

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