Roller coaster

$I$ have used the fact that mechanical energy remains conserved in abscence of external forces. Here, the coaster car is released from height $H$ . Thereore, velocity at the $lowermost$ point (A) = $\sqrt{2gH}$ ( $\frac {1}{2}$ m $v_A^2$ = mg $H$ , g = accelaration due to gravity).

Meanwhile, for the coaster car to go around the circle velocity at the $lowermost$ point must be = $\sqrt{5gR}$ ( $R$ = radius of vertical loop)

{ Let velocity at the $lowermost$ point be $v_A$ corresponding to point A and at the highest point be $v_B$ at point B.

At point A, total mechanical energy is $\frac {1}{2}$ m $v_A^2$ and at B, it is equal to $\frac {1}{2}$ m $v_B^2$ + 2mg $R$ .

Also at B, force equation gives $N_B$ + mg = m $\frac {v_B^2}{R}$ . Since the car has to just pass and not fall off at the highest point, $N_B$ = 0 $;$ therefore, $v_B^2$ = g $R$ .

So, $\frac {1}{2}m v^2_A$ = $\frac {1}{2}m v^2_B$ + 2mg $R$ = $\frac {5}{2}$ mg $R$ .

$\Rightarrow$ $v^A$ = $\sqrt {5gR}$ }

$\Rightarrow$ $\sqrt {5gR}$ = $\sqrt {2gH}$ . Squaring both sides and putting values, we get $H$ = $\frac {5}{2}R$ = $\boxed{50}$ meters

the centripetal force acted on the roller must be equal to the weight so as not to fall from the loop (mv^2)/r = mg ==> 1 therefor v^2= rg and since the total energy of the roller at the highest point is K+U = zero + mgh and the total energy of the roller at the lowest point is (0.5)mv^2 + zero and since the energy is always constant "there is no friction and air resistance can be neglected" therefor (0.5)mv^2 = mgh ==> 2 from 1 and 2 : we find a relation between height and the radius of the loop where v^2= 2gh and v^2 = rg rg=2gh r=2h again the potential energy at the highest point (total energy) is equal to the total energy at any point mgH=mgh+(1/2)mv^2 and substituting h=2r and v^2=rg mgH=2mgr+(1/2)mrg h=2r+(1/2)r=2.5r then the minimum height of the loop for roller not to fall is 2.5 of the radius of the loop

It should be intuitively obvious that the cart needs a certain minimum velocity to stay on the track while moving through the top half of the loop. This minimum velocity $v$ is greatest at the very top of the loop, where the gravitational force $F_g$ is perpendicular to and points away from the track. To find $v$ , we set $F_g$ equal to the centripetal force $F_c$ :

$F_g = F_c \Rightarrow mg = m \frac{v^2}{r} \Rightarrow v = \sqrt{gr}$ .

Where $r$ denotes the loop's radius. The cart's kinetic energy at the top of the loop is therefore:

$K = \frac{1}{2}mv^2 = \frac{1}{2}mgr$ .

The gravitational potential energy at the same point is:

$U=mg \cdot 2r = 2mgr$ .

From energy conservation, the sum of $K$ and $U$ must equal the cart's initial potential energy at the top of the hill of height H:

$mgH = K + U = \frac{1}{2}mgr + 2mgr = \frac{5}{2}mgr$

Or,

$H= \frac{5}{2} r = \frac{5}{2} \cdot 20 m = 50 m$

The roller coaster is at ground it must have a minimum velocity of (5gr)^(1/2) (explained below) by conservation of energy.potential energy at highest point is mgH (we hv to find H) which should be equal to kinetic energy of coaster at ground and min k.e. can be obtained by keeping the value of v min. That is (5gr)^(1/2). MgH=0.5m×5gr =》H=0.5×5×20 _ _ _ × _ _ _ __ Explaination of min velocity Using energy conservation we can say that. 0.5mu^2=0.5mv^2+mg.2r U is vel at ground .v is velocity at highest level I.e. 2r
For completing the circle weight of coaster must be balanced my centrifugal force.thus Mg=mv^2/r from here keep the value of v^2 in the eqn if energy cons. And u will get min u.

At first we calculate the minimum speed that the roller coaster must have at the top of the loop so that it stays in touch with the track. At the top of the loop, the only forces acting on the rollercoaster are normal force (centrifugal force) and the weight of the rollercoaster. If the rollercoaster is about to fall off, then the normal force (centrifugal force) must equal zero. Therefore, the only force acting on the rollercoaster will be its weight, here centripetal force. Therefore, $centrifugal force \geq weight$ for it not to fall off,
i.e, $mv^2/r \geq mg$ , and hence $v \geq 14m/s$ Now at the top of the loop the roller coaster has a potential energy 2rgm in addition to its kinetic energy(=1/2 m $v^2$ /r).

The minimum height of the hill is such that the potential energy they have at the hill top, i.e. mgh, where h is the height of the hill, is at least 2rgm + 1/2m $v^2$ to allow the car to climb up the loop and still have sufficient velocity at the top of the loop. Hence, $mgh \geq$ 2rgm +1/2 rgm or $h \geq 2.5r$ Therefore, $h \geq 50$ , hence, minimum height is 50m.

Remark: For any radius, minimum height of the hill so that the roller coaster doesn’t fall off from the loop is: $2.5 \times (radius of loop)$

To start off with, we must get a few formulas straight. Centripetal acceleration = velocity (v) squared divided by radius (r) Kinetic energy = mass (m) times velocity divided by 2 Potential energy = mass times velocity times height (h)

The question is asking how fast we have to go for centripetal acceleration to overcome gravity. That is, we are looking for (v^2)/r=g. If we then multiply both sides by r we get v^2=rg.

If you look at a carts energy relative to the top of the loop, and use that the carts energy remains constant at the start and at the top of the loop, we get that mgh=(mv^2)/2. We can use simple algebra to then get 2gh=v^2.

As shown above, v^2=rg, and 2gh=v^2, which means rg=2gh. This simplifies to r=2h. Since r is 20, h must equal 10. Since this value was calculated at the height of the top of the loop, we must add the height of the loop to our answer to get the final answer. The height of the loop is 2r, which is 40, so the answer to the problem is 40+10=50.

If the coaster car is to stay on the loop, then at the top of the loop the acceleration of gravity must be equal to the centripetal acceleration, i.e.

$\frac{v^2}{r}=g$ .

We also know from conservation of energy that

$mgH=mg(2r)+\frac{1}{2}mv^2$

at the top of the loop, where $r=20~m$ . Solving the first equation for $v^2$ and plugging the result into the second gives

$H=40+10=50~\mbox{m}$ .

50 meters