04 - friction-forces and energy conservation: Ch. 6: P43, P49, P51, P57; Ch. 7: P8, P13, P16, P22, P25, P32. • chapter 6, problem 42 (|| 43): Suppose the coefficient of static friction between the road and the tires on a car is 0.60 and the car has no negative lift. What speed will put the car on the verge of sliding as it rounds a level curve of 30.5 m radius? Solution: Let the car be moving at speed v, be of mass m, and the curve-radius be R; the acceleration, and the consequent radial force, is computed as,
ar
v2 R
Fr mar
mv 2 R
(1.1)
The normal force, and the consequent friction force, is, F N Fg mg FFr S FN S mg ;
(1.2)
If the car has no negative ne gative lift, and does not slip, you are at a critical speed. We solve for the critical speed,
Fr FS m
v2 R
solve for v S mg v R S g 30.5 0.60 9.81
m s2
13
m s
48
km hr
;
(1.3)
•• chapter 6, problem 48 (|| 49): A roller-coaster car has a mass of 1200 kg when fully loaded with passengers. As the car passes over the top of a circular hill of radius 18 m, its speed is not changing. At the top of the hill, what are the (a) magnitude FN and (b) direction (up or down) of the normal force on the car from the track if the car’s speed is v = 11 m/s? What are (c) FN and (d) the direction if v = 14 m/s? Solution: Let the normal force (on the car from the rail) point in the +y direction, and gravity in the – the – y direction. Inwards towards the circle-center, then, is the negative direction. The force balance appears as,
v2 F F F F 0 F m ( g ) m F m g ; N g r N N r r v2
(1.4)
3
(a) When v = 11 m/s, we obtain F obtain F N = 3.7 10 N on solving (1.4) for the unknown F N . (b) F N points upward. 3
3
(c) When v = 14 m/s, we obtain F obtain F N = – = – 1.3 1.3 10 N, or | F N | = 1.3 10 N. (d) The fact that this answer is negative means that F N points opposite to what we had ha d assumed. Thus, the magnitude of F N is | F N | 1.3 kN and its direction is down. down. Afterword – Afterword – on on “constrained motion”: motion”: For the case of a roller coaster bolted b olted onto a track, the normal no rmal force exerted by the rails switches directions. This is difficult to anticipate, and is an artifact of constrained motion. motion. Interestingly, “constrained motion” is the topic of higher -level -level physics classes, and the techniques developed to analyze constrained motion are the foundation of some of the most modern tools of physics.
•• Chapter 6, Problem 70 (|| 51): The figure shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at constant speed. (The cord sweeps out a cone as the bob rotates.) The bob has a mass of 0.018 kg, the string has length L = 0.79 m and negligible mass, and the bob follows a circular path of circumference 1.3 m. What are (a) the tension in the string and (b) the period of the motion? T
Solution: Notice I drew an angle tan 1 T y sin 1 x
r L
T T x Ty (T sin ) Ty T 2
2
2
2
sin1 T T in the figure. x
T y 2
solve for T
1 sin 2
mg 1
(0.018 kg)(9.81 sm2 )
r 2 L2
1 (
1.3 m/(2 ) 2 0.79 m
0.18 N ; (1.5)
)
Maybe I shouldn’t have used the letter T for tension. Let “t” be the period. Then,
mar m
( 2 t r ) 2 r
4 2 rm t2
mg
Tx T sin
r
2 1 Lr 2 L
4 2 rm ( Lr ) 2 1 4 2 L2 r 2 (1.6) t mg g ( L ) 2 1 mg
solve for t 2
2
r
In (1.6), we solved for t in symbols. We note that mass has canceled out. As a sanity check, we remember that the period of uniform circular motion is independent of the mass. Plugging in numbers, t
4 2 L2 r 2 g
4 2 (0.79m )2 ( 1.3 m )2 2 9.81 sm2
3.1s ;
(1.7)
• chapter 7, problem 9 (|| 8): The only force acting on a 2.0 kg canister that is moving in an xy-plane has a magnitude of 5.0 N. The canister initially has a velocity of 4.0 m/s in the positive x-direction and, some time later, has a velocity of 6.0 m/s in the positive y-direction. How much work is done on the canister by the 5.0 N force during this time? Solution: By the work-kinetic energy theorem,
W K
1 2
mv f2
1 2
mvi2
1 2
(2.0kg) (6.0m/s)2 (4.0m/s)2 20 J.
We note that the directions of v f and vi play no role in the calculation.
Chapter 7, problem 15 (|| 13): Figure 7-27 shows three forces applied to a trunk that moves leftward by 3.00 m over a frictionless floor. The force magnitudes are F 1=5.00N, F 2=9.00N, and F 3=3.00N, and 60 . During the displacement, (a) what is the net work done on the trunk by the three forces and (b) does the kinetic energy of the trunk increase or decrease?
(1.8)
Resolving all of the abovementioned vectors into components, d 3.00 x ˆ
m;
F1
5.00x N ; ˆ
F2
9.00N cos60x sin60 y ; ˆ
ˆ
3.00Ny;
F3
(1.9)
ˆ
Thus, the work done by each force is,
W1 F1 d (3.00m)(5.00 N ) x x 15.0 J ; W3 (3.00m)(3.00 N ) x y 9 J 0 0 ; ˆ
ˆ
ˆ
ˆ
(1.10) 1 3 W2 F2 d (3.00m)(9.00 N )(cos 60 x x sin 60 x y ) 27.0 J ( 1 0) 13.5J ; 2 2 ˆ
ˆ
ˆ
ˆ
Net work can be computed as the sum of the contributions of (1.10); which can be shown to be the same as if we computed the vector-resultant (harder) of the sum of the forces
F1
W W1 W2 W3 F1 d F2 d F3 d F1 F2 F3 d
F2 F3 F ,
F d 15.0 13.5 0 J 1.5J (1.11)
Chapter 7, problem 12 (|| 16): A can of bolts and nuts is pushed 2.00 m along an x axis by a broom along the greasy (frictionless) floor of a car repair shop in a version of shuffleboard. Figure below gives the work W done on the can by the constant horizontal force from the broom, versus the can’s position x. The scale of the figure’s vertical axis is set by Ws = 9.00 J. a) What is the magnitude of that force? b) If the can was initially at rest and had a mass of 0.500 kg, what is its speed at the end of the 2.00 m? Solution: start with the differential definition of work, and realize the linear function plotted,
dW F dx F
W K
1 2
dW
linear function
dx
mv 2 v
2 K m
W 9.00J 4.5N ; x 2.00m 2(9.00 J ) 0.500kg
6.00
m s
(1.12)
;
(1.13)
•• chapter 7, problem 21 (|| 22): A cord is used to vertically lower an initially stationary block of mass M at a constant downward acceleration of g/4. When the block has fallen a distance d, find (a) the work done by the cord’s force on the block, (b) the work done by the gravitational force on the block, (c) the kinetic energy of the block, and (d) the speed of the block. Let down be negative, and up be positive,
Wcord Fcord d
F F d F F d M 4 g Mg d 4 1 Mgd 1
g
g
1
3 4
Mgd ; (1.14)
W grav Fg d M ( g )( d ) Mgd ; K Wnet Wcord Wgrav
2( 14 Mgd ) 1 1 3 2 1 Mgd Mgd Mv v 4 2 M 4
(1.15) gd 2
02 2(
1 4
g )(d ); (1.16)