Curves you can take to the BANK! - Banked Curves

Solutions from the sheet:

1.
a) 22.78 degrees
b) 15.6 m/s

2.
a) T_max = 23.65 N to break
b) T = mg/cos(theta) so theta = arc_cos(mg/T_max) = 78.0 degrees
c) v = (T*L sin(theta)*sin(theta)/m)^1/2 = 8.239 m/s

1995B5
a) acceleration is to the left, velocity is towards the top of the page.
b) .586 m/s
c) v_max = (mu*g*r)^(1/2) = .828 m/s
d) since mass divides out in deriving the result for speed in (c), the presence of the second coin will not change the answer.

Book problems - Chapter 5:
Question 2.
A sharp curve will have a smaller radius of curvature, so the centripetal acceleration will be greater as compared with a curve with a greater radius at the same speed.

Problems:
14.
For the occupants to feel weightless, the apparent weight must be zero. Thus mg = mv^2/R.

You should get that the speed of the riders is 8.57 m/s.

Now you need to convert this to revolutions/minute:

8.57 m/s * (1 revolution/(2*pi*7.5 m))*(60 s/1 min) = 10.91 revolutions per minute.

17.
Calculate the speed required if it is perfectly banked. (12.07 m/s)

Since the 90 km/h (25 m/s) is greater than this, you know the car will slide UP the incline, which means friction must be present and directed down the incline to keep this from happening.

From the x-direction, you should obtain that Fn = (mv^2/R - Fs*cos(theta))/sin(theta).

From the y-direction, you should obtain that Fn*cos(theta) - mg - Fs*sin(theta) = 0.

Substitute the first result into the second, and after some algebra, you can obtain that Fs = mv^2/R * cos(theta) - mg*sin(theta). Substituting the values, you get that Fs = 8035 N.

18.
Your FBD should have Fn towards the center of the circle, mg down, and static friction force up.

In the y-direction, Fnet = 0, so Fs = mg. Since we are looking for the minimum value of mu, Fs = Fsmax = mu*Fn. This means that mg/mu = mv^2/R.

SOlving, mu = gR/v^2 = .198.

There is no force pressing the riders against the wall - the riders feel the increased normal force as a greater apparent weight.

20.
Follow the same steps as for #1 on the worksheet today.
The angle of the incline is 25.75 degrees.
Following the same steps as with part (b) in #1, mu = .765