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Suppose a car is driving around a flat_horizontal; circular track at constantl speed: We know there is anet force on the car (a centripetal force) [Therelis also an...

Question

Suppose a car is driving around a flat_horizontal; circular track at constantl speed: We know there is anet force on the car (a centripetal force) [Therelis also an angular momentum for rotations about the center of the circular Ipath:How does the centripetal force change the angular momentum of the car fori rotations about the center of the circular path?KDi The centripetal force causes the angular momentum to increase with timeD The centripetal force causes the angular momentum to decrease wi

Suppose a car is driving around a flat_horizontal; circular track at constantl speed: We know there is anet force on the car (a centripetal force) [Therelis also an angular momentum for rotations about the center of the circular Ipath: How does the centripetal force change the angular momentum of the car fori rotations about the center of the circular path? KDi The centripetal force causes the angular momentum to increase with time D The centripetal force causes the angular momentum to decrease with time: The centripetal force causes the angular momentum to first point uPl thenchange [0 down: The angular momentum points parallel to the centripetallforce solas the carGoes around the circle the centripetal force changes the directiomofiche angulan momentum IThe centripetal force does not change the aneular momentum ef the GarFon rotalions about the centeciefnG Gixchkan Pakh becauise exarts na torquat



Answers

Determine the tangential and centripetal components of the
net force exerted on a car (by the ground) when its speed is
$27 \mathrm{m} / \mathrm{s},$ and it has accelerated to this speed from rest in 9.0 $\mathrm{s}$
on a curve of radius 450 $\mathrm{m} .$ The car's mass is 1150 $\mathrm{kg}$ .

In this question. Okay, you are given an automobile. Um Uh huh, Acceleration is 0.8. You guys 2nd square. And then the time during the acceleration is 20 seconds, and the radios of the tire is 0.33 m. So basically the diagram looks like this. Okay, so the a is 0.8 m/s squared and then it Accelerates for 20 seconds. So the car will travel some distance and there is no sleeping. If you want to find out angle to which the view has rotated. Okay, so to solve this problem. Okay, so we know that uh distance travel by the uh car is equal to that. Uh huh. Number of call revolutions. Oh, higher Times two Pi R into my eyes. The circumference. Okay. Um Oh God. Of the tire. Okay, so the distance traveled by car can be calculated using the kingdom attics equations. Eight other exit goes to, you know, t plus half a T square. So, uh, we're not is equal to zero is 0.8. You know, it's a close 2nd square. He's 20 seconds. Okay, so delta x. Yes. Half a T square. Okay. Your .8. Hence 20 square. So you calculate this, you get 160. Yes. Okay. So you can calculate the angular displacement. Yeah. To be using using you have the eggs goes to our data. Okay? So data is equal to X. R. Which is 116 um by 0.3 train. And this is equal to 485 regions. Yeah. Since the tire, it is rotating clockwise. Mhm. We use our convention that the clockwise is negative. So data is uh 485 begins. Okay, so this is the answer for this question and that's all.

Well, because the centripetal forces modifying your momentum along a particular factor you would, in theory, sense of centripetal force, continue along a straight line. That Harvey Word centripetal force modifies data factor by inducing an acceleration in a different direction. Well, within the frame of reference off a turning car, you feel a force pressing you two words on the outside wall of the car. This is the apparent force. Knowing is centrifugal force. Well, in this context harbor, it is our pictures. It is not. It's not really afford. It's the result of inertia and not a reactive force in opposition to the centripetal force. Right. So it is the result off, uh, energy.

Hi in this question we have there is cost starts from riffed on the fundamental radio struck And has an acceleration of constant rate of 0.5 minutes with second squid. So I want to find first of all the speed of the car where the centripetal in tangential explorations are equal. So that's for the centripetal acceleration. You see go to him go to v squared. Remember if you have lost the R. Is the radius of the truck. And since it's called to the tangential acceleration that is 0.5 the second square. So solving for V B squared recorder there by fire times are A movie called the skirt of 0.5. There are are is going to be funded with us. So Visa cold suit 14 .14 industries occurred. Yeah. Others the 2nd 1. Want to find that this nice to hold. Yes. So for me we can use the equations of motion for constant acceleration since the solution is constant and we can use fee square is go to U squared just to a Yes. So V is the final velocity at the point where the center product celebration and tangential acceleration is equal. Which is this. This 14.14 squared is a call to U squared. And I was going to be there because that's the rest is used in shell velocity. First to I was accelerations deserved by five times the distance unknown as And so the distance traveled is approximately called 200 m. Where was she? Mhm Want to find the laps laps time. So you can use to physical to v minus you whether by the acceleration, So the final velocity 14.1 for -10 travel. Lost his error divided by the exploration, which is german for, and this would be cool to me 28.28 seconds. Mhm. So access for emergency.

Institutions. We have to tell that what is the centripetal acceleration. So centripetal acceleration is the exhibition of a body when body is traveling in a circular path. So it's coming to picture when body is in circular motion and it acts towards the center. Its magnitude is given by that is a C. Z equals two. The square of the bodies is paid divided by the distance from the center of the circle to the moving body. That is we square by our. We can show it by diagram. Also let this is the circular motion and this is our body and its velocity is let this is a tangential velocity which is we and this is the distance. Are so and and so at this point the centripetal acceleration on the body is to ask the center that is a C. Now let after sometimes the body reached at this point. So the velocity will be into this direction. This is me. And now the centripetal acceleration is towards the center. This is our centripetal X solution. Grand body point reach at this point. Let so the velocity at this point is the Now the centripetal acceleration is at towards the center at this point. This is that wasn't capitalization. So centripetal acceleration is perpendicular to the tangential velocity. So the fut perpendicular to this tangential velocity here. Also it is the perpendicular here. Also it is perpendicular. Now in the second part of the solution we have mm tell that a car is moving around this carl and we have to tell the direction of this interpreter exhibition. So let this is the circular track and this is the center of the track. And we have told that the car is moving on this track. So this is our car. So on this card, the centripetal acceleration is towards the center, This is this interpreter expiration, This is the centripetal acceleration and this is also the centripetal exception and the velocity of the car. Let this is the velocity. So when the car is moving around the curve, the centripetal acceleration is worth the center of circular. Uh This is the answer for our second part that thank you.


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