The 2-M g race car has a projected front area of 1.35 m^2 and a drag coefficient of (CD)c=0.28.

step 1 In this problem, we're considering a race car with a parachute. And I've drawn the diagram right

The 2-M g race car has a projected front area of 1.35 m^2...

Key Concepts

Drag Coefficient

The drag coefficient is a dimensionless number that characterizes the drag per unit area an object experiences as it moves through a fluid. It encapsulates the effects of shape, surface roughness, and flow characteristics. Lower values indicate more streamlined shapes, while higher values typically correspond to blunt or flat surfaces, which significantly influence the magnitude of the drag force.

Parachute Aerodynamics

Parachute aerodynamics involves designing a system that maximizes drag force to slow down objects, especially during deceleration or descent. The effectiveness of a parachute is determined by its surface area, shape, and drag coefficient. By manipulating these parameters, engineers can ensure that a sufficiently high drag force is generated to achieve desired deceleration rates.

Deceleration and Force Analysis

Deceleration in physics is directly related to the net force acting on an object according to Newton's second law of motion. By equating the force required to decelerate an object over a specific time interval with the aerodynamic drag force produced (in this context, by a parachute), one can derive necessary design parameters to achieve a targeted reduction in speed.

Air Density

Air density is the mass per unit volume of air and significantly influences drag forces. Under standard conditions, variations in air density—due to temperature, altitude, or humidity—alter the magnitude of aerodynamic drag. Accurate knowledge of air density is essential for precise calculations in vehicle dynamics and parachute design to ensure that models predict performance reliably.

Aerodynamic Drag

Aerodynamic drag is the resistive force exerted by air as an object moves through it. It is calculated using the equation F_drag = 0.5 * air density * velocity^2 * drag coefficient * reference area. This force is crucial in vehicle dynamics because it opposes motion and must be overcome by propulsion to maintain speed or can be used advantageously to decelerate or stabilize the object.

Transcript

00:02 In this problem, we're considering a race car with a parachute.

00:05 And i've drawn the diagram right here.

00:10 There's a car with a parachute.

00:12 We are asked to find the diameter of that parachute that's needed to reduce the car speed to 20 meters per second in four seconds.

00:20 So our initial speed is given as 20 meters per second in our finals, excuse me, 60 meters per second.

00:26 We're also given the mass of the car as, let me see, a two megagram.

00:36 Which is two times two times 10 to the third kilograms.

00:43 We're told that it has a front area of 1 .35 meters squared.

00:49 This should be a meter squared.

00:51 Good thing i checked this.

00:54 We're given our coefficient of drag for the car and the course should drag for the parachute, initial final speeds, air temp, time zero, time, and i looked up the density of air.

01:07 Okay, so i've written down one, two, three equations and i've done a free body diagram and we're ready to start solving.

01:23 Okay, so let's see.

01:31 Okay, so let's figure out the acceleration.

01:35 It's a derivative of velocity.

01:37 So my acceleration will equal derivative of velocity over dt.

01:46 So we know that vx equals v so a x equals dv over d t because we only have horizontal and for a drag force on the car that will be equal to cd on the car times ap of the car times one half density u squared let's call this equations.

02:32 Do i need other equations here? let's call this four.

02:40 Let's call this five.

02:42 Let's call this six.

02:43 These are my equation numbering.

02:45 And then for my parachute, that will equal cd of my pair, whoops, times ap of my parachute times one half density u squared.

03:07 Okay.

03:08 And that'll be equation seven there we go so this the parachute is a hollow hemisphere so um in the diameter we will set equal to d so we will have api equals pi d squared divided by four and let's call this equation eight and then we're going to combine seven and eight and this will give us f d p equals c d p times now we'll have pi d squared over four times one half density u squared so let's solve for p fd p and that will equal pi over eight times c d p times density t times density times d squared times u squared and let's make this equation now.

04:33 Okay.

04:37 So if you were traveling with either one, we know u equals v and the velocity of the car, the velocity of the parachute.

04:54 In replacing u, we will get fd and my car, whoops, and my car will equal cd of my car times a, of my car times one half density v squared.

05:26 Okay, so fdc will equal one half times c -d -c times c -d -c times a -p -c times density times v squared.

05:44 Let's call this 11.

05:49 Let's call this 10.

05:50 Okay.

05:54 And if we substitute 10 into 9, you can go back and look at that...