In an amusement park ride a cart of mass 300 kg and carrying four passengers each of mass 60 kg

First, we need to find the total mass of the cart and passengers. There are four passengers, eac

In an amusement park ride a cart of mass 300 kg and carrying...

In an amusement park ride a cart of mass $300 \mathrm{kg}$ and carrying four passengers each of mass $60 \mathrm{kg}$ is dropped from a vertical height of $120 \mathrm{m}$ along a frictionless path that leads into a loop-the-loop machine of radius $30 \mathrm{m}$ The cart then enters a straight stretch from $A$ to C where friction brings it to rest after a distance of $40 \mathrm{m} \text { . (See Figure } 8.19 .)$ (a) Find the velocity of the cart at A. (b) Find the reaction force from the seat of the cart onto a passenger at $\mathrm{B}$. (c) What is the acceleration experienced by the cart from A to C (assumed constant)?

Transcript

00:01 Hi, here in this given problem, first of all we draw the given track in the amusement park.

00:10 In that amusement park, there is a ride in which the cart is coming down, then going over a loop -the -loop, and then going straight.

00:27 Okay, and we assume this point to be given by a.

00:34 This point c and this upper most point of the loop the loop this is the cart and this point that is b height initial height from there the cart is coming down this height that is given as 120 meter radius of loop the loop that is 30 meter and the distance covered by the card before coming to rest this ac that is 40 meter mass of the cart along with the passenger four passengers in it that will be given by mass of the card only plus four times the mass of single passenger which is 60 kg so that is calculated to be equal to 540 kilogram.

01:52 In the first part of the problem we have to find the velocity of the card at point a for which simply we will be using energy conservation which says kinetic energy of the card along with the passengers in it half mv a square that should be equal to initial gravitational potential energy because no kinetic energy energy should be there at the top of the track because it was dropped from the vertical height, which is 120 meter.

02:36 So canceling this m, we get an expression for va.

02:39 This is given by square root of 2, gh, plugging in the known values here, two times of 9 .8, multiplied by h which is 120.

02:52 So this va, speed of the cart, at point a, calculated to be the value.

02:56 To 48 .5 meter per second answer for the first part of this given problem here.

03:06 Then in the second part of the problem, this was the first part, find the velocity of the cart at a.

03:13 Second part, find the reaction force from the seat of the cart onto a passenger at b.

03:20 So when the cart will be at point a, it will be under the influence of centrifugal force.

03:28 And for a single passenger, his weight acting vertically down.

03:33 So at point p, normal reaction exerted by the seat on a passenger, it will be given by fc centrifugal force acting readily out and means vertically up minus weight of the passenger vertically down.

03:56 So an expression for the centrifugal force m vb squared speed of the card at point b divided by radius r minus mg.

04:10 So first of all we should find this vb.

04:13 So to find vb using energy conservation at point a and b, sorry, this is b.

04:37 It says, kinetic energy of the cart at point a, half mv, a squared, that should be equal to kinetic energy of the cart at point b plus gravitational potential energy...

Question

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