Upside-Down antics - Vertical Circles

Remember the advice I gave you in class!

You MUST have one positive axis directed into the center of the circle, so before you start drawing vectors on your FBD, figure out where positive needs to be first.

Centripetal force is the net force in the radial direction of something moving in a circle - use this with Newton's 2nd law to solve the various problems ahead.

Also remember the condition for the minimum speed for something in a vertical circle - the only force acting at this minimum speed is gravity! This is true whrether something is being swung around by a string, riding on a roller coaster, or riding over a hill.

Homework solutions:
6. This is a horizontal circle problem, which we will discuss in class more tomorrow.

8. 3.139 N for the top (a), 9.019 N for the bottom (b)

10. Again, this is a horizontal circle. The key here is that the normal force (the force 'felt' by the trainee' is 7.75*mg, and this is the centripetal force.

Fn = mv^2/R so v = (Fn*R/m)^1/2 = (7.75*mg*R/m)^1/2 = 27.559 m/s

Since one revolution is 2*pi*R meters, this can be converted into revolutions per second using a unit conversion. The correct answer is .439 rev/s.

12. v = (g*R)^1/2 = 9.18 m/s

13.
a) newton's 2nd: mg - Fn = mv^2/R, so Fn = mg - mv^2/R = 5800 N
b) For driver, the equation is the same except that the mass m in the equation is of the driver, not the car. Fn = 406 N.
c) At the minimum speed, Fn = 0. Therefore mg = mv^2/R, and v = (g*R)^1/2 = 31.3 m/s


21. The pilot's path is a circle during the evasive maneuver - in order to NOT hit the ocean during the circular dive, the initial altitude of the pilot MUST be the radius of the path. This is then what the question is asking you to find.

The acceleration must NOT exceed 9g. The radius for this is R = v^2/a_c = 1090 meters.

63. You know that the maximum tension is 1400 N. If you draw a FBD of Tarzan, your equation from Newton's 2nd becomes:

T - mg = mv^2/R

Solving for v, v = (R(T - mg)/m)^1/2 = 6.08 m/s