The drawing shows a version of the loop-the-loop trick for a small car. If the car is given an initial speed?

Question

The drawing shows a version of the loop-the-loop trick for a small car. If the car is given an initial speed of v = 3.9 m/s, what is the largest value that the radius r can have if the car is to remain in contact with the circular track at all times?

I tried doing

PE = KE
mgh = 1/2mv^2
2gh = v^2
2g(2r) = v^2
4gr = v^2
4(9.8) r = (3.9)^2

r = 0.388

but it was wrong. anybody helP?

Here's the picture

http://www.webassign.net/CJ/p6-44alt.gif

Answer 1

Yep, that's wrong. The reason it's wrong is, if you use the PE=KE equation, then ALL of the car's KE will be exhausted (will go to zero) by the time the car reaches the top of the loop. This means that, at the top of the loop, the car will come to a dead halt, and will then flop straight down.

What you really need is to make sure that the normal force (perpendicular force) exerted by the track on the car, never drops to zero.

To do that, make an equation for the normal force, as a function of the car's initial speed vi, and its position on the track.

Suppose the car has traveled through an angle θ around the track. At that point:

Its height is: h = r - rcosθ.

its speed can be determined by conservation of energy:
ΔKE = ΔPE
m(vi)²/2 − mv²/2 = mgh
(vi)² – v² = 2gh
(vi)² – v² = 2gr(1–cosθ)
v² = (vi)²–2gr(1–cosθ)

At that point, centripetal component of the car's acceleration is:
a_c = v²/r
a_c = [(vi)²–2gr(1–cosθ)] / r
a_c = (vi)²/r – 2g(1–cosθ)

This acceleration is produced by the combination of two forces acting on the car:
1) The normal force F_n (the track pushng inward);
2) The radial component of the gravitational force. Its magnitude is: mgcosθ. Sometimes (depending on θ) this is positive (pushing toward loop's center); sometimes negative (pushing away from center); sometimes zero.

So, the net force pushing cenripetally inward, as a function of θ, is:

F_net_c = F_n – mgcosθ
(the minus sign reflects the fact that the gravitational component points _outward_ when θ = 0).

Now, by F=ma, we have:

F_net_c = m(a_c)
F_n – mgcosθ = m[(vi)²/r – 2g(1–cosθ)]

or:
F_n = m[(vi)²/r – 2g(1–cosθ) + gcosθ]
F_n = m[(vi)²/r – 2g + 3gcosθ]

Now, we said previously that we want F_n to be greater than or equal to zero for all θ. Therefore the following has to be true for all θ:

m[(vi)²/r – 2g + 3gcosθ] >= 0
(vi)²/r – 2g + 3gcosθ >= 0
(vi)² >= rg(2 – 3cosθ)

The right side is a maximum when cosθ = –1 (namely when the car is at the top of the loop). In that case, we need:

(vi)² >= rg(2 – 3(–1))
(vi)² >= 5rg

Therefore, the car's inital speed vi needs to be at least sqrt(5rg).

Answer 2

If the car remains in contact with the track at the highest position, it will do at all times. So we only need to discuss the question there.
At the highest position, 2 forces act on the car, namely the gravity mg and the normal force N, both vertically downward. Their net force make the car centripetal acceleration mv^2/r. Frm newton's 2nd law, we have
mg + N = mv^2/r
N = m(v^2/r - g)
N must greater than 0, at least equal 0, the maintain the contact.
so v^2/r - g > = 0
r < = v^2/g
therefore the largest value of r is v^2/g = 3.9^2/9.8 = 1.6 m

Answer 3

Small Car Drawing