# KINEMATICS IN TWO-DIMENSIONS

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```					                Chapter 5: NEWTON’S LAWS II Hints (Updated 3/26/12)

When solving problems with Newton’s 2nd Law the correct free-body-diagrams are
essential. Keep in mind the following:
1. Draw a “free-body diagram” for each object of interest in the problem.
2. Only real interactions (gravity, surface contact, tension & springs) can apply forces
on objects.
3. Choose a coordinate system that is parallel and perpendicular to the direction of
acceleration.
4. Write “Fnet=ma” equations in each direction for each “fbd”. These equations will
solve the problem.

1. There are some common misconceptions about the normal force that deserve consideration.
Answer TRUE or FALSE to the following statements and justify your decision.

a) The normal force is not always equal to the weight.
b) The normal force is always vertical.
c) The normal force can accelerate an object.

(a) True, on an incline for example. (b) False. N is perpendicular to any pair of surfaces,
horizontal or vertical. (c) True, in an elevator for example.

2. The force of friction can be difficult to deal with. Answer TRUE or FALSE to the
following statements and justify your decision.

a)   Kinetic friction is generally less than static friction.
b)   Kinetic friction is always equal to the coefficient of kinetic friction times the normal force.
c)   Static friction is always equal to the coefficient of static friction times the normal force.
d)   Friction always slows down the speed of an object so it cannot accelerate an object.
e)   Friction can make an object turn around.
f)   Friction cannot exist on a vertical surface.
g)   The coefficients of friction are measured in Newtons.

a) F. Kinetic friction is always less than the max static, but not generally, since static
friction is often less than the max. b) T. c) F. Static friction can be less than the maximum.
d) F. Friction accelerates your car forward or around a curve. e) T. f) F. g) F.  is unitless.

3. In order to solve problems in circular motion it is important to understand the centripetal
force. Answer TRUE or FALSE to the following statements and justify your decision.

a) The centripetal force is an extra force on a body generated by the circular motion.
b) The centripetal force is simply another name for the net force on a rotating body directed
radially toward the center the rotation.
c) In a rotating frame of reference an object in circular motion experiences a sense of being
pulled outward by a (so-called centrifugal) force, but this force does not exist in an
unaccelerated (inertial) frame of reference.
d) The centripetal force can be any real force (such as weight, tension, friction, etc) or
combination of real forces.
e) Static friction cannot be a centripetal force because the object is in circular motion.
f) One should never draw the centripetal force in a free-body diagram.
a) F. See (b). b)T. c) T. You feel pulled out because your body want to go in a straight line and
must be pulled toward the curved path. d) T. e) F. Several example of static friction as Fc were
given in class. f) T. You should only draw real forces from interactions on your diagram.

4. In the following cases draw the “fbd” for the highlighted object paying special attention to the
force of friction and especially to its direction. For each case determine whether the force of
friction is accelerating, decelerating or balancing the other forces. Determine a valid expression
for the force of friction on each. You may assume that the coefficients of friction are known.
Remember that i & j are direction unit vectors.

Illustration              Free-body diagram            Acceleration,   Force of friction        Net force
deceleration   and its direction    on object and its
or neither?                            direction.
Box of mass M at rest relative      Forces are mg, N,            Accelerating    fs = Ma (i)         Fnet = fs = Ma (i)
to an accelerating cart.            & fs                         forward with
N
a                                                      the cart.       [fs < sN ]
fs

mg

Book of mass M is held at rest                                   No              fs = mg (j)         Fnet = 0
fs
on a wall by a                                                   acceleration
horizontal force F                    N                     F

mg

A car rounding a curve on the                                    Acceleration    fs = Fc= Mac (-r)   Fnet = fs = Mac (-r)
N
toward the      fs = Mv2/R
center of
fs
curvature.
mg

Block of mass M slides up a                 N                    Decelerating    Calling + upward, Fnet = [mgcos
rough ramp of angle                                        x
fk = N (-i)     +mgsin (-i)
vo
fk = (mgcos)
fk

mg

5. A student sets up an experiment to determine the coefficient of friction between a block and a
long board. He places the block on the board, then he slowly tilts the board up increasing the
angle until the block just begins to slide over the board.

a) What does the student need to measure to determine the coefficient of
friction between the block and the surface? Does this method give the                            

coefficient of static friction or of kinetic friction?
b) Would the result in (a) change if: (i) the block was heavier? or if (ii) the contact surfaces
were larger?
c) When the block begins to slide it will actually accelerate downward (if no change is
made on the angle). Explain why. Determine the acceleration.
d) How could the student determine the coefficient of kinetic friction?

a) Student needs to measure only the angle of slip because tanslip=s (recall that fs=mgsin). This
gives the static coefficient of friction.
b) Neither would make any difference since they do not affect the normal force or the weight.
c) Once sliding starts friction becomes kinetic, but since the incline is at the angle of slide, the
downward force still equals the maximum static friction. Since fs >fk , acceleration results. Fnet here
will be equal to [(s  mgcos]. Divide out the mass and you have the acceleration.
d) He should reduce the angle until the block barely slides with constant speed. The new angle
gives the kinetic coefficient of friction.

6. Consider the problem of a mass on an incline connected to another mass over a pulley. We have
done some versions of this problem in class, but there are many variations. We used an angle of 30o
for the incline, and the masses were 0.3 kg on the incline and 0.2 kg hanging. In class we showed that,
with no friction, the acceleration of the system was 1 m/s2 with the hanging mass moving downward.

a) Suppose you replace the hanging mass with a different                     a
2
one and, as a result, the acceleration changes to 1 m/s in
the opposite direction. What is the value of the mass and

tension in the rope if we assume that no friction is present?
b) Determine the force of friction that would keep the original system at rest. What
minimum coefficient of static friction does this require?
c) Assume the system at rest in (b) is jolted temporarily and, as a result, the masses begin to
accelerate. The tension drops from 2.0 N to 1.9 N in that case. Determine the acceleration and
the coefficient of kinetic friction.
d) Determine the angle that would keep the system in (b) at rest with no help from friction.
e) Determine the acceleration in two extreme cases: (i) the angle of the incline is zero, and (ii) the
angle of the incline is 90o.

Review the problem done in class and use the same basic techniques.
a) m=0.11 kg; T=1.2 N; b) fs=0.5 N; minimum s=0.19
c) The jolt causes the friction to change from static to kinetic which is less, therefore the mass
accelerates. a =0.5 m/s2; k =0.096
d) Here the Fnet=ma equations combine to give: (0.3)g sin =(0.2)g  sin =2/3;  =42º
e) (i) This is the similar to the Newton’s 2nd Law lab: With friction as in (c), T=1.3 N, a=3.4 m/s2;
(ii) The two masses would be hanging on either side of a pulley and there would be not friction, so
T=2.4 N and a=2 m/ s2.

7. Two masses 2-kg each and two identical springs are arranged in three different ways. In Fig. 1 below
the lower spring is stretched 5 cm. Assume the springs have negligible mass. Remember that the spring
constant is the force/stretch of the spring (k=F/s = ∆F/∆s)

a) Show that the spring constant is 400 N/m and that the upper spring in Fig.1 is stretched 10 cm.
b) Determine the amount of stretch in Fig.1 if the entire system accelerates downward at 3 m/s2.
c) Now consider the other two arrangements (Fig.2 & Fig.3) of these same elements. Determine the
stretches in each spring in each case assuming no acceleration.
d) A spring is considered “stiffer” than another spring if it exerts more force with less deformation (this
means the spring constant “k” is larger). Combining two (or more) springs can increase or decrease
the overall stiffness of the combination. In Fig. 2 the springs are acting “in parallel”, while in Fig. 3
the springs are acting “in series”. In which case is the overall stiffness greater than that of a single
Fig.                                         Fig. 2                                   Fig. 3

a) Force in lower spring=20 newtons=k (0.05). Force in the upper spring = 40 newtons. Use these to
derive the given values of k and the stretch of the upper spring.
b) Draw the fbds and remember that the masses are accelerating. For the lower mass Fnet=mg -
ks1=mas1=0.035 m. Similarly, for upper mass, Fnet=mg + ks1 – ks2 =ma s2=0.07 m.
c) In 2nd illustration: The two springs share the total weight of 40 N, so the force exerted by each spring
is 20 newtons. Each spring therefore stretches 5 cm.
In the 3rd illustration: Each spring has to support the entire 40N weight, so the force exerted by each
spring is 40 newtons. Each spring stretches 10 cm.
d) Fig. 2 with parallel springs is the “stiffer” system. This is a general principle for multiple springs:
spring “in parallel” produce a larger overall “k”; spring “in series” produce a smaller overall “k”.
For two spring, k1 and k2, the combined spring constant k12, is given by the following expressions: In
“parallel”(as in Fig.2), k12=k1+k2; in “series”(as in Fig.3), 1/k12=(1/k1)+(1/k2).

8. Double pulley systems can give you what is called “advantage”. For the following cases, determine
the relationship between the tension in the difference strings, and the relationship between acceleration
of the masses involved.

Illustration            Tension Relationships                     Acceleration          Acceleration in terms of the
Relationship               masses and “g”
The string that wraps around the          a1 = 2 (a2)           a1=(2m1 - m2)g/(2m1 +½ m2)
pulleys has the same tension (T)
throughout. The tension of the
m1                      string supporting m2 is 2T and
m2
that of the string supporting the
large pulley is 3T.
The string that wraps around the          aM = 2 (am)           aM = mg/(2M + ½ m)
M                       pulleys has the same tension (T)
throughout. The tension of the
string supporting the hanging
m        mass is 2T.

9. Consider three versions of a problem with three block of masses m, 4m, and 3m. The small mass is
always on top of the large mass, and the 3m mass is accelerating downwards (but at different rates).
There is friction between the m and the 4m-block but NOT anywhere else. The coefficients of kinetic
and static friction are 0.6 and 0.9 respectively. Pulleys and strings are massless.

m                                   T2                                            T2
fs                 fs    T1                     fk      fk                                    fk      fk
4m                                                 T1                                            T1
T2

a1
a2                                      a3
3m
a) Draw the forces of friction in each case. Which are static and which are kinetic?
b) In which case is the acceleration the greatest? …the least? Justify your choice. How would
that affect the tension on the string connecting to the “4m” and the “3m” mass?
c) Determine the accelerations in each case and check your answers in (b).

a) In all cases there is a pair of action-reaction friction forces between the m and 4m blocks that depend
on the normal force between these two blocks, N=mg, (there is no friction between the table and the
4m-block). In the first case the friction is static (fs<µsmg) because there is no relative motion between
the two blocks, but in the other two cases the friction is kinetic (fk=µkmg) because the blocks move
relative to each other.
b) The best way to make a comparison is to look at the forces on the 4m-block. This block is always
pulled by the string connecting it to the “3m” hanging mass. In the first case the opposing force is
static friction, in the second case the opposing force is kinetic friction, in the third case the opposing
force is kinetic friction plus the tension from the other string connecting the blocks over the pulley
attached to the wall. Without some calculations, it’s unclear which of a1 and a2 is larger since that
depends on knowing which friction force, static or kinetic (remember fs ≤ fmax), is larger in these
cases. But a3 is clearly less than a2 since there is an extra tension force opposing the motion of the
4m-block in the 3rd case. The tension will be the least when the acceleration of 3m is the greatest.
c) Draw “fbd’s” (I’ve drawn the horizontal forces on the illustrations) and use Newton’s 2nd Law:
Case 1: You can treat the two blocks as one “5m” mass. All objects accelerate at a1 = 0.375g.
Case 2: Block m is at rest, and the forces on it are its weight, normal, fk, and tension. Forces on the
4m block are its weight, 2 normals (from below and above), fk, and tension. The 4m-mass and
hanging mass accelerate at a2=0.343g.
Case 3: Here the 4m-mass has an additional tension force from the connection over the pulley next to
the wall. All masses accelerate at the rate of a3= 0.225g, but “m” is moving toward the wall.

10. A 2-kg box sits on a long platform of mass 3-kg and length 80 cm. There is friction between
the box and the platform (k=0.3 and s= 0.4). There
is no friction between the platform and the ground.                               f                      F
f

a) A horizontal force F is applied to the platform so that both objects accelerate together and the
box stays at rest relative to the platform. Determine the maximum possible acceleration and
the value of F in this situation.
b) Draw the “fbd” for the box. What is the accelerating force on the box? What function does
the force of friction serve in this situation? Is it static or kinetic friction that is involved here?
c) Draw the “fbd” for the platform. What is the accelerating force on the platform? Compare
the force of friction on the platform to the force of friction on the box.
d) A temporary jolt causes the box to break the static contact and to begin to slide relative to the
platform. Determine the acceleration of the box relative to the ground, and the acceleration
of the box relative to the platform. Make sure to give directions as well as magnitudes.

a) The “a” is a max when the friction force between the box and the platform is the max static friction,
fmax=(0.4)(2g), which is also the force that accelerates the box a=4 m/s2; F=20 N
b) You should have already done the “fbd’s” to solve for (a). There are 3 forces on the box: mg, normal
from platform, and static friction to the right. The static friction is the accelerating force.
c) There are five forces on the platform: Mg, a normal force up from the ground and a normal down from
the box, a static friction to the left, and the force F=20N to the right. The accelerating force on the
platform is (F-fs). Friction on box and platform are an action-reaction pair of equal and opposite forces.
d) Here friction drops to kinetic. This decreases the acceleration of the box and increases the acceleration
of the platform relative to the ground. The box will appear to accelerate backwards but this is only
relative to the platform: abox/ground =3 m/s2 (i); aplatform/ground =4.67 m/s2(i); abox/platform=-1.67 m/s2(i).
11. Here is another version of the problem above. A 2-kg box sits on a long platform of mass 3-
kg and length 80 cm. There is friction between the box and the platform (k=0.3 and s= 0.4).
There is no friction between the platform and the ground. In this                       F
version the horizontal force F is applied directly to the box.        f
f

a) Assume that both objects accelerate together and that the box stays at rest relative to the
platform. Determine the maximum possible acceleration and the value of the force F in
this situation.
b) Is the resultant acceleration smaller or larger than in the previous problem? Describe the
differences between these two problems.
c) The force F is increased to 20 N and the box then slides relative to the platform.
Determine the acceleration of the box relative to the ground, and the acceleration of the
box relative to the platform. Make sure to give directions as well as magnitudes.
d) In (c) the box is dragged over the full length of the platform. How far did it actually move
relative to the ground? (Hint: you need to find the time it took the box to slide from one
end of the platform to the other.)

a) As in previous problem, “a” is max when the friction force between the box and the platform is the
max static friction, fmax=(0.4)2g. Static friction accelerates the platform a=2.67 m/s2; F=13.3 N
b) The acceleration is less. In this case the static friction is accelerating the platform forward and
fighting the acceleration of the box. Since the platform has more mass, the result is less max
acceleration overall (so F is less). In this case it is more difficult to keep the masses from
separating.
c) The acceleration of the box (relative to the ground) increases to 7 m/s2. The acceleration of the
platform (relative to the ground) decreases to 2 m/s2. Finally, abox/platform = 5 m/s2 (i).
d) Using the kinematics formulas in the frame of reference of the platform, it can be determined that
the time for the box to move 80 cm with an acceleration of 5 m/s2 will be 0.566 s. Now change to
the ground frame of reference to get 1.12 m of distance traveled with respect to the ground.

12. In the different cases below determine the force F needed to move a mass M at constant
speed (either up or down) on an incline of angle with a coefficient of kinetic friction k.
(a)                            (b)                               (c)                          (d)

V constant
V constant                     V constant                          V constant

F                                                              F                                   F
                                    F
                                                                

Draw “fbd” and note the differences: the kinetic friction has different directions depending on
which way the block moves. When F is horizontal (as in c & d) it has to be broken into
components that change the value of the normal force and of the friction.
a) F= mg(sin + k cos); b) F= mg(sin - k cos)
c) F= mg(tan + k)/(1 - k tan); d) F= mg(tan - k)/(1+k tan)

13. A box is given an initial speed vo up an incline of angle .
The coefficients of kinetic and static friction are k and s respectively.                             vo
Express your answers to the questions below in terms of the given terms.

a) Show that the ratio of the acceleration of the box going up the ramp to
the acceleration of the box going down the ramp is:
aup/adown = (tan+ k)/(tan - k).
b) Assume that the box goes up and that it immediately comes down. Determine the final
speed when it returns to the original position.
c) Draw a velocity vs. time graph for (b).
d) Explain why it takes longer for the box to come down than it takes for it to go up.
Determine the ratio of the time up to the ratio of the time down.
e) If the angle is sufficiently small the box will reach the top but it will not come down.
Determine that maximum angle.

a) The accelerations are directed downward in both cases (whether the block moves up or down). If you
call down “+”, then when the block moves up, Fnet =mg(sin +k cos)=ma1. When the block moves
down, Fnet =mg(sin - k cos) =ma2. The ratio of these two should give you the result. Beware that
“up” and “down” here refer to the direction of the velocity, not the acceleration, which is down the
incline in both cases.
b) The speed changes from vo to 0 on the way up with a1 as the downward acceleration. The speed
changes from 0 to vfinal on the way down with a2 as the downward acceleration. Since the block slides
up the same distance that it slides down, you can use the appropriate kinematics formula (2a∆x=v2-
vo2) to relate the speeds: vfinal =vo[(tan - k)/(tan + k)] 1/2.
c) v
The areas under the two lines must be equal…

d) There is simply greater deceleration on the way up and less acceleration coming down, so the block
slows down faster going up than it will speed up coming down: [tup/tdown]=[(tan-k)/ (tan +k)] 1/2.
e) The angle of slip arctan(s). Once the box stops static friction would hold it in place.

14. A person is pulling on a crate of mass M at an angle. The crate is dragged at constant
speed over the floor where the coefficient of kinetic friction is k.                                F

a) Determine the value of F if the angle were zero.
b) If the angle is increased from zero, does it get easier or
c) At what angle does it take the least effort to drag the crate? (This problem requires
calculus to find the minimum value of a function.)                      F
d) Instead of pulling the crate the person decides to try to push                 
on it from above at an angle . What is the critical angle
above which no amount of pushing will move the crate?
e) What is the value of k that would make it impossible to push the crate forward at that angle?

a) F= kMg for =0. In general: F= kMg/(cos + ksin).
b) The amount of force necessary to move the crate at constant speed actually decreases at first
because the component of F upwards reduces the amount of normal force and hence the amount of
kinetic friction opposing the motion. After a critical angle the needed force will increase, because
the horizontal component of the force is getting too small to accelerate the crate.
c) First write a function of the force F with respect to the angle of application F= kMg/(cos +
ksin). Take the derivative of this function and set it equal to zero to determine extreme values. The
solution turns out to be [optimum = arctan k].
d) Here the normal force is increased by the push so it gets harder to push the crate as the angle
increases: F=kMg/(cos - ksin). For F to be less than infinity the denominator must be greater
than zero: cos >ksintherefore cot > k.
e) It should occur to you that this basically the same problem as (d), but stated differently. The
answer is based on the same relationships and k > cot.

15. A box of mass m sits on a wedge of mass M and angle . A horizontal force F is applied to the
wedge. Assume all surfaces are frictionless. The wedge and box are accelerating horizontally such
that the box remains at rest relative to the wedge (the box doesn’t move up or down the incline). Use
the given quantities to determine the following. (Hint: Note the direction of the acceleration!)
a
a) Draw a “fbd” for the box. What’s the accelerating force?                     F
b) What is the acceleration “a” in this situation? What happens if the                                        
acceleration of the wedge is greater (or lesser) than the “a” derived above?
c) What is the value of the force F required to maintain this acceleration?
d) If there is some friction between the wedge and the box (s for example), there is a range
of accelerations for which the box can remain at rest relative to the wedge with the help                       a
of static friction. Determine the maximum and minimum accelerations of this range.
e) Consider this problem in the extreme case of the angle going to 90o. What minimum F
acceleration is required to keep the box at rest relative to the surface?
N
Check you class notes. Pay attention to the direction of acceleration when analyzing forces.

a) Note “fbd” on the right. In this case fs=0, so net Fbox=Nsin=ma. Also mg=Ncos. fs
b) Combining the equations above, a=gtan. With more acceleration the box will
slide up the incline, with less acceleration the box will slide down the incline.             mg
c) The force “F” accelerates both the wedge and box together, so F= (M +m) gtan.
d) Static friction will be directed down the incline when a >gtan or up the incline when a<gtanto
keep the box from sliding). To determine the max and min “a”, assume static friction has its maximum
value (fmax=µN) either up or down. The extra friction force changes the normal and the accelerating
force so: mg=Ncos + µNsin, and ma=NsinµNcosCombining these give the answer: amax/min
=g[(tan ± s)/ (1 + s tan)]
fs

e) On the small block, the force of friction is equal the weight and the normal
force is the accelerating force. Fbd is shown on the right. Answer: amin=g/s                       N

16. In the following cases of circular motion determine a valid expression for the centripetal mg
force and for the tangential force (if any) in terms of the actual forces on the highlighted body.
Remember that centripetal and tangential forces are net forces on a rotating body.

Illustration                     Free-body diagram                  Centripetal and tangential forces
Conical pendulum of                                                          Fc=mg tan
T
length L, mass m, and                                      
Ft=0
angle 
Satellite in orbit                                   mg                      Fc=mg
mg
near the earth surface                                                       Ft=0

Swinging pendulum at the                                   T          Fc= T- mg cos
bottom and at the ends                                               Ft= mg sin
mg
A man standing on a                             N                            Fc < fs =sm
rotating turntable                                                           Ft=0
fs

mg
N
A jogger running at                                   fs               Fc = N sin± fscos
constant speed around                                                        Ft=0
a banked track                                           mg
The main point of this exercise is for you to learn that the centripetal force is NOT an “extra” force
on a body, but that it represents the sum total of all the radial forces (from gravity, tensions,
normals, friction, etc.) on a body in circular motion. Necessarily the net radial force in circular
motion MUST point toward the center of the circle of motion (must be centripetal) because an object
in circular motion is accelerating toward the center. Fc always equals mvt2/r, which is Newton’s 2nd
law for circular motion. In addition, there can be tangential forces if the speed is changing.

17. A bug is crawling toward the rim of a turntable along a radius. The turntable is rotating at a
rate of 0.5 rps (revolutions/sec) and it has a radius of 15 cm.

a) When the bug is 10 cm from the center, it begins to slip on the surface of the turntable.
What is the coefficient of static friction between the turntable and the bug?
b) As the bug crawls radially outward, does the time to go around one revolution increase,
decrease, or stay the same? Does its speed increase, decrease, or stay the same? Does its
c) How slow would the turntable have to rotate so that the bug can reach the rim of the
turntable without slipping?

The way to do this problem is to first determine the period of the rotation T (time for one
revolution), then use this to determine the tangential velocity (v=2r/T) and the centripetal
acceleration (a=v2/r) wherever the bug is on the turntable. Static friction is the centripetal force
(fs=mv2/r) here, and the bug starts to slip when static friction is at its max=sN.
a) µs=0.10
b) Period is the same because turntable rotates all at once; velocity and acceleration increase as
the radius increases because the circumference of the circle of motion increases. Check out
the definitions of these concepts.
c) Work the problem backwards: max fs =N=mv2/r=m(2r/T)2/r. Solve for T. Remember that th
frequency is the inverse of the period: freq =(1/T) =0.41 rps

18. Consider a car going around a corner on a level road. Assume that the mass of the car is M,
the coefficient of static friction is  and the radius of curvature is R.

a) Draw the “fbd” on the car and discuss the role of the force of friction on the car. Is the force
of friction static or kinetic? What would happen if the car hit a slippery patch on the road?
b) Determine the maximum speed with which the car can safely round the corner. Is there a
minimum speed? Explain.
c) Suppose that the radius of curvature of the road is 25 m and that the posted speed limit is 20
mph (about 10 m/s). What minimum coefficient of friction is implied by that information?
d) What would be the difference if the car were loaded with people instead of just carrying the
driver?

a) This problem was used as a class example. The centripetal force is static friction, and N=mg.
b) vmax= (sRg)1/2;   c) 0.4
d) Nothing would be different. The mass drops out of the equations.

19. Consider a car going around a corner on
a road banked at an angle . Assume that the
Cross-section of
mass of the car is M, the coefficient of static  the banked road                  
friction is s and the radius of curvature is R.
Some typical values would be R=90 m, = 35o, and s=0.5.                               R
a) Show that if the car’s speed is equal to (Rgtan)1/2 the car can safely round the curve
without generating any friction. (Hint: Draw the “fbd” and pay attention to the direction
of acceleration of the car).
b) What would be the range of safe speeds (vmax and vmin) in the banked road if friction were
taken into consideration? (Hint: Add the force of friction to the “fbd” above and note that
its direction can be either up or down the incline.)
c) Consider the extreme case of a 90o-banked curve.
This amounts to a vertical cylindrical wall. Can a car
drive around the inside of a cylindrical wall? How fast
would it have to travel to keep from sliding down the wall?

This problem has many similarities with problem 15 (the fbd and equations are the same), except
that here the acceleration “a” is centripetal and ac=vt2/R. So check out the hints in that problem.
Draw fbd with vertical and horizontal coordinate system. With no friction, the Nsin component
of the normal force is the centripetal force. With static friction present the normal force is
affected (Ncos=mg±µsNsin), and the centripetal force includes a component of the friction
(Fc=Nsin±µsNcos).
Because static friction can act in either direction (down the incline for vmax                    N
and up for vmin), a range of velocities are possible for the car to round the                 
banked road without sliding off. The fbd for (a) and (b) is shown on the right:        fs
a) vbank= (Rgtan)1/2= 25.1 m/s
b) vmax= [Rg(tan + s)/ (1 - s tan)] 1/2= 47.4 m/s                                          mg
b) vmin= [Rg(tan - s)/ (1 +s tan)] = 11.5 m/s
1/2

c) Yes, the car’s motion would generate a normal force. Similar to 15e, vmin= [Rg/s] 1/2=42 m/s

20. Amusement parks have many examples of circular motion. Consider the “loop-the-loop”. This is
a vertical track that allows a passenger to ride a cart upside down. Assume that the track is
frictionless and that the radius of the track is R. Consider the cart of mass M at different locations on
the track. Determine the centripetal force in terms of the real forces on the cart. (Hint: This problem
is nearly identical to a pendulum making a full vertical swing.)

a)
v                                    v
Illustration

v                                      
g                g                                                                g
g                g    v
v

Centripetal      Fc=N-mg          Fc=N              Fc                    Fc=N+mg          Fc =[N- mgcos
force          Ft=0             Ft=mg             =[N+mgcos          Ft=0             Ft = mgsin
Ft = mgsin
Does         Heavier,         Neither           Lighter,              Lighter,         Heavier,
passenger       N >mg                              Typica                Typically,       N >mg
feel lighter or                                      lly,N                 N <mg
heavier?                                          <mg

b) Determine the minimum speed that the cart can have at the top of the ride.                                     N=0
ø
c) Determine the speed of the cart if it loses contact (N =0) with the track before it
reaches the top. Assume the cart makes angle ø with the vertical at that point.
d) For safety’s sake the “loop-the-loop” is actually made up two circular
circular tracks where the cart’s wheels ride. Explain how a double track
keeps the cart from leaving the circle regardless of speed.
a) In general, for a position angle of with the vertical: Fc =[N ± mgcosand Ft = mgsin. Passengers
feel heavier at the bottom and near the bottom, and typically feel lighter at the top and near the top
(unless the centripetal acceleration is larger than 2g). Remember that it is the normal force that gives
you the “feeling of weight”.
b) This was done in class, vmin at top = [Rg ] 1/2
c) Since N=0, only a component of the weight pulls toward the center, so Fc = mgcosø =mv2/R
v=[Rgcosø ] 1/2
d) The secod track means that the wheels can receive a force from either track (up or down). This means
that the cart can have any speed and still get the required net force to remain in the track since the
normal force can come from the top or the bottom. For example, at the top the weight is down but the
normal force can be up from the bottom track (if the cart is moving slower than √Rg) or it can be down
from the top track (if the cart is moving faster than √Rg). So the cart can have just about any speed.

21. There are many other interesting amusement park rides. Determine the centripetal force in
each of the following.

a) Determine centripetal forces in terms of the actual forces on the rider. Let down be (-).
Roller coaster                   Ferris wheel                Spinout
Illustration            N
g                                                             g
g        fs
N
mg
N

mg                        mg

Fc at the top            Ftop= (mg – N) (-j)            Ftop= (mg – N) (-j)            x
Fc at the bottom          Fbot= (N – mg )(+j)           Fbot= (N – mg)(+j)              x
Fc at the middle               x                        Fmid= N sin(±i)              Fc = N(±i)

b) Why is it unsafe to go too fast over the top of a roller coaster? Determine the maximum
safe speed in terms of radius of curvature on the track and “g”.
c) Describe how a person “feels” riding the Ferris wheel at different locations.
d) What is the minimum rotating speed in the Spinout that would allow a person to “stick to
the wall” while the floor is lowered?

I have added “fbds” to the drawings.
b) It is unsafe because the faster you go the less normal force (contact with the seat of the cart)
you have and you can fly off the cart altogether: vmax= [Rg ] 1/2
c) Passenger feels heavier at the bottom and near bottom, and feels lighter at the top and near top.
Remember that it is the normal force that gives you the “feeling of weight”. In the Ferris wheel the
seats can swing back and forth and they tilt so the normal force can provide a centripetal force, as
in the banked track. At the sides the acceleration is horizontal and the normal tilts toward the
center of the rotation so that Fc= Nsinwhere  is the angle with the vertical.
d) Class example, so check your notes. Similar solution to 15e and 19c, vmin= [Rg /s] 1/2

22. A 5-kg mass is attached to a 50 cm long string. A rigid spring of constant 200 N/m is also
attached to the mass. The other end of the string and the spring are attached to a stick and set 40
cm apart. The stick is rotated at different rates about its length making the spring compress or
stretch. This forces the string to take on different positions and tensions. Three situations are
illustrated below. In the first illustration the spring is compressed 10 cm (remember that
compressed springs push), in the second the spring is relaxed (exerting no force), and in the third
the spring is stretched 10 cm (spring pulls). For each situation determine the tension in the string,
the speed of the mass, and the rate of rotation or frequency (freq=1/T, in rps) of the system.

50 cm                                    50 cm                     66.4º        cm
40 cm                                                                                                  T
40 cm   51.3º     T               40 cm     r= 46 cm
T
r= 39 cm                                             Fs=ks =20 N
Fs=ks=20 N
mg=50 N                         cm
30 cm                                    cm                                               mg=50 N
mg=50 N

I’ve added fbd’s and the needed angles and circular motion radii to the drawings (these are obtained
using the geometry of the triangle formed by the string, spring and stick). Recall that compressed
springs push, while stretched springs pull, so the spring force is outward in the first case. Use fbds to
write equations relating forces in the x and y directions (note that centripetal acceleration is always
horizontal). Combine equations to find velocities, and from them find the periods and frequencies.
a) Here mg=Tcos37º and Fc= Tsin37º- ks v=1.02 m/s; freq=0.54 rps
b) Here mg=Tcos51.3º and Fc= Tsin51.3º v=2.2 m/s; freq=0.90 rps
c) Here mg=Tcos66.4º- (ks)cos66.4º and Fc= Tsin66.4º +(ks)sin66.4º  v=3.7 m/s; freq=1.3 rps

23. A small ball with a mass of 400 g is attached to a stick with two strings A and B of equal
length of 50 cm. The strings are attached 80 cm apart on the stick. The stick rotates at a rate of 2
rps (revolutions per sec) keeping the strings extended and taut.

a) Determine the tension in each string.
b) Slow down the rotation until the lower string is barely taut.
How slow is this rotation?                                                                    
c) Repeat the problem for the case where the stick is rotating
about a horizontal axis. Determine the tensions when the
ball is at its highest point and when it is at its lowest point.



a) Draw “fbds”! The centripetal force consists of the two horizontal components of the string
tensions. The vertical components of the tensions balance the weight of the mass. Combining
the two equations to solve for the tensions give the answers. I got: TA =18.4 N; TB= 13.3 N
b) If you keep slowing down the rotation, the tension in the strings will keep decreasing, but
the tension in the upper string will always be larger than the lower string. At some speed the
tension in the lower string will be ~0 but the string will still be fully extended. At that point
the only forces are the weight and the top string tension. The answer I got is 0.796 rps.
c) Ttop= 12.5 N; Tbot=19.1 N

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