PHY218 SPRING 2015
RECITATIONS PROBLEMS: Week 8
Mechanical energy
These are the problems that you and a team of other 2-3 students will be asked to solve during the recitation
session next week. Your team can do better if you think about the approach and explanation for these
problems BEFORE coming to class. You may want to follow the solved example on the last pages of this form.
Roller coaster: A roller coaster car has a mass of m. It is launched horizontally from a giant spring, with spring
constant k into a frictionless vertical loop-the-loop track of radius R.
a) Draw free body diagrams for the car when it is:
i. Compressing the spring
ii. At the bottom of the loop,
iii. Half way to the top,
iv. At the top.
b) What is (are) the point (s) where the car would leave the loop? Justify your choices.
c) What is the minimum compression of the spring such that the cart will pass through the loop?
d) The car, launched at the speed calculated in part c), exits the loop and starts sliding up a frictionless ramp
set at 30o. How far does the car slide up the ramp before reversing its motion?
Bungee Jumpin': A thrill seeker, who has a mass of___ kg and is ___ m tall, will bungee jump from the middle of
a bridge ____m above a ____ m deep pool of Jello. The bungee cord has an unstretched length of ___m, and it
will behave like an ideal spring when it stretches during his jump. What should the elastic (spring) constant of the
bungee cord be if the thrill seekers head is just to graze the top of the pool of Jello without becoming immersed
in it?
1
PHY218 SPRING 2015
RECITATIONS PROBLEMS: Week 8
Approach: Under this tab, list the steps taken by your team for finding each solution. You answer here the
questions WHAT? and HOW?
Approach 1.
Approach 2.
Explanation: Under this tab, explain why your team has chosen those approaches. You answer here the questions
WHY? and WHEN?
Explanation 1.
Explanation 2.
2
PHY218 SPRING 2015
RECITATIONS PROBLEMS: Week 8
Mechanical Energy: Solved Example
______________________________________________________________________
Roller coaster: You are designing a new roller-coaster. The main feature of this particular design is to be a vertical
circular loop-the-loop where riders will feel like they are being squished into their seats even when they are in
fact upside-down (at the top of the loop). The coaster start at rest a height of 80m above the ground, speeds up
as it descends to ground level, and then enters the loop which has a radius of 20m. Suppose a rider is sitting on a
bathroom scale that initially reads W (when the coaster is horizontal and at rest). What will the scale read when
the coaster is moving past the top of the loop? (You can assume that the coaster rolls on the track without
friction).
Conceptual Analysis:
- The coaster begins with zero kinetic energy and with non-zero gravitational potential energy with respect to
the ground level. Because the surfaces are frictionless, the mechanical energy will stay the same during the
motion. Energy conservation dictates that the coaster cannot reach a position above its initial height.
- Two forces act on the rider: gravitational force and normal force. Their resultant is the centripetal force
maintaining the rider on the loop.
- The scale reads the normal reaction force. This equals the weight only if the rider is on a horizontal surface
and at rest.
Strategic Analysis:
- From energy conservation, find the speed at the top of the loop.
- Calculate the required centripetal force at the top of the loop.
-Calculate the normal force at the top of the loop. Use the given weight W as a unit of measure.
Quantitative Symbolic Analysis:
- Begin by labeling the given quantities
H height in the initial state
m mass of the person
R radius of the loop
g gravitational acceleration (constant)
W=Fg gravitational force
FN normal reaction force from the scale
Fnet net force
 From the energy conservation:
mgH=mg(2R)+mv2/2
 The speed at the top of the loop enters into
the calculation of the centripetal force
Fcp=mv2/R= 2mg(H/R-2)=2W(H/R-2)
Since the rider pushes against the scale while
moving upwards, the scale pushes him back:
the normal force is directed towards the center
of the loop (see on the diagram.)
At the top of the loop, both the weight and the
normal are directed vertically down. The net
force there is
Fnet=W+FN
By imposing that the net force gives the
necessary centripetal force, we get
W+FN=2W(H/R-2) or FN=W(2H/R-5)
Quantitative Numerical Analysis:
-Inserting the numerical values given by the problem: FN=W(2*80/20-5)=3W
You will feel like in a place where the gravitational acceleration is 3 times greater!
3