Download ( ) 13.0m / s ( ( ) 8.0m / s ( ( ) 8m / s ( ( ) 7.2m / s (

Momentum
Problem 1
Two skaters Daniel (mass 65 kg) and Rebecca (mass 45 kg) are practicing. Rebecca
moving a 13m/s collides with a stationary Daniel. After colliding Rebecca moves at 8 m/s
at 53.1! w.r.t. her initial direction. What are the magnitude and direction Daniel’s velocity
after the collision? What is the change in kinetic energy as the result of the collision? Is
this an elastic collision?
Before
Rebecca
m R = 45kg
After
v R1 = 8.0m / s
Daniel
m D = 65kg
+y
α = 53.1°
β=?
v R 0 = 13.0m / s
v D0 = 0
+x
v D1 = 0 ?
(a) The simplest way to solve the problem is to assume that the +x direction is the same
as Rebecca’s initial velocity, as shown in the diagram. There must be conservation of
momentum for both the x and y components.
x-component mR vR0 = mR vR1 cos 53.1! + mD vD1 cos !
( 45kg ) (13.0m / s ) = ( 45kg ) ( 8.0m / s ) ( 0.6 ) + ( 65kg ) vD1 cos ! " ( 65kg ) vD1 cos ! = 369kg • m / s
vD1 cos ! = 5.68m / s (1)
y-component 0 = m R v R1 sin 53.1! " m D v D1 sin ! Take note of the negative sign!
0 = (45kg )(8.0m / s )(0.8) " (65kg )v D1 sin # ! (65kg )v D1 sin # = 288kg • m / s
v D1 sin ! = 4.43m / s (2)
v sin " 4.43m / s
v sin "
Divide equation (2) by (1) D1
=
! D1
= tan " = .78
v D1 cos " 5.68m / s
v D1 cos "
5.68m / s
" = tan !1 (0.78)= 38 ! . Substituting into equation (1) v D1 =
= 7.2m / s . Daniel
cos 38 !
velocity is 7.2 m/s, 38° below the horizontal as shown in the above diagram.
1
1
(b) The initial kinetic energy is K 0 = m R v R2 0 = (45kg )(13m / s )2 = 3800 J (only
2
2
Rebecca is moving). The final kinetic energy (Rebecca and Daniel are moving) is
K1 =
1
1
1
1
2
2
2
2
mR vR1
+ mD vD1
= ( 45kg ) ( 8m / s ) + ( 65kg ) ( 7.2m / s ) = 3125J . The change in
2
2
2
2
kinetic energy is "K = K1 ! K 0 = !675 J . Note that kinetic energy is not conserved,
and this is an inelastic collision.
Problem 2
In the diagram below a 8.00 g bullet strikes and embeds itself in a 0.922 kg block. The
block then slides on a frictionless floor, compressing the spring, till it stops after 15 cm.
Calibration of the spring shows that a force of 0.750 N is required to compressed the
spring 0.250 cm. a) Find the impact of the block’s velocity just after impact. B) What was
the initial speed of the bullet?
A bullet with a mass of 8.00 × 10-3 kg strikes and embeds itself in a block with mass
0.992 kg that rests on a frictionless surface. The impact compresses the spring 15.0 cm.
Experimental data shows that a force of F = 0.750 N compresses the spring
x = 2.50 " 10 !3 m . Using Hooke’s law we obtain a force constant of
F
0.750N
N
F = kx " k = =
= 300 .
-3
x 2.50 ! 10 m
m
(a) For this part, use conservation of energy. Initially the elastic potential energy is zero,
while at the maximum compression the kinetic energy is zero. Hence,
(KE of block + bullet before impact) = (elastic PE at max. compression)
1
(mbullet + M block )vtot2 = 1 kx 2
2
2
k
300 N / m
(0.15m )= 2.60m / s
vtot =
x=
mbullet + M block
0.008kg + 0.992kg
(b) This part uses the conservation of momentum. Before impact the block does not
have a momentum, only the bullet has a nonzero momentum. After impact the block +
bullet moves together with speed vtot = 2.60 m/s.
(momentum of bullet) = (momentum of bullet + block)
mbullet v = (mbullet + M block )vtot
m
+ M block
1.0kg
(2.60m / s )= 325m / s .
v = bullet
vtot =
mbullet
0.008kg
Problem 3
In the diagram below a 50 kg cart rolls to the left at 5 m/s and a 15 kg box slides down an
incline and leaves the edge at 3 m/s 4 m above the bottom of the cart. A) What is the
speed of the package just before it lands in the cart? B) What is the speed of the cartpackage after the collision?
37°
vBi = 3.0 m/s
vBi-y
vBi-x
4.00 m
initial
just before
collision
final
vTF
vCi = 5.0 m/s
vCi = 5.0 m/s
vBf
The leftmost sketches show the box just before it leaves the chute with speed
vBi = 3.0 m/s. First find its x and y components: v Bi ! x = (3.0m / s )cos 37 ! = 2.4m / s and
v Bi ! y = !(3.0m / s )cos 37 ! = !1.8m / s
(A) Just before the collision between the box and the cart
To find the speed of the box, vBf, just before it hits the box use conservation of
mechanical energy:
1
1
2
2
U 1 + K1 = U 2 + K 2 ! m B gh + m B v Bi
= 0 + m B v Bf
2
2
(
)
2
v Bf = 2 gh + v Bi
= 2 9.8m / s 2 (4m )+ (3m / s )2 = 9.35m / s
(B) Just after the collision between the box and the cart
To find the speed of the cart plus box after the collision, vTF, use conservation of
momentum in the horizontal component (can you see why the vertical component is not
relevant?). Just before the collision the x-component of the cart’s velocity is -vCi = -5.0
m/s and the x-component of the cart’s velocity is vBi-x = 2.4 m/s
! mC vCi + m B v Bi ! x
! mC vCi + m B v Bi ! x = ( mC + m B )vTF " vTF =
mc + m B
! (50.0kg )(5.0m / s )+ (15kg )(2.4m / s )
= !3.29m / s
65kg
The cart plus box move with a velocity of 3.29 m/s to the left.
vTF =
Rotation and Torque
Problem 4
A computer disk starts rotating from rest at constant angular acceleration. If it takes 0.750
s to complete its second revolution: a) How long does it take to complete the first
complete revolution; b) What is the angular acceleration?
(A) Let t be the time it takes to rotate through one revolution " = 2! . Then we obtain
1
1
# = # 0 + "0 z t + ! z t 2 with ! 0 = 0 and !0 z = 0 , which gives 2" = ! z t 2 (1). Now the
2
2
second revolution takes 0.75 s to complete, so it takes t + 0.75 s to rotate from ! 0 = 0 to
1
1
" = 4! , which using # = # 0 + "0 z t + ! z t 2 and !0 z = 0 gives 4" = ! z (t + 0.75s )2 (2).
2
2
1
2
2
t + 0.75s )
4! 2 " z ( t + 0.75s )
(
=
#2=
Dividing (2) by (1) gives
, which simplifies to
1
2!
t2
" zt 2
2
2t = ( t + 0.75s ) ! t " 1.5t " .5625 = 0 ! t =
2
2
2
1.5 ± 1.5 2 + 4 (1) ( 0.5625 )
2
or simply
t = 1.81s,!0.31s . Of course t = 1.81 s is the physical solution.
1
2
(b) Using (1) 2! = " z t 2 gives 4! = " z t 2 # " z =
4!
4!
rad
=
= 3.84 2 .
2
2
t
s
(1.81s )
Problem 5
In the figure calculate the net torque about an axis of rotation perpendicular to the origin.
F1 = 18.0N
F2 = 26.0N
F3 = 14.0N
F2
F1
l2
l1
l1 = l2 = 0.090 m
l3 = (0.090 m)/sin45°=(0.090m)√2
l3
45°
F3
Torque #1, " 1 = ! F1l1 = !(18.0 N )(0.090m ) = !1.62 N • m , clockwise (-)
Torque #2, ! 2 = F2l2 = (26.0 N )(0.090m ) = 2.34 N • m , counterclockwise (+)
Torque #3, ! 3 = F3l3 = (14.0 N )(0.090m ) 2 = 1.78 N • m , counterclockwise (+)
" net = " 1 + " 2 + " 3 = !1.62 N • m + 2.34 N • m + 1.78 N • m = 2.50 N • m , counterclockwise
(+), or the net torque is pointing out the page.
Newton’s second law for rotation and translation
Problem 6
A 50 kg grindstone of radius 0.260m is rotating at 850 rev/min, when an ax grinds against
the rim with a normal force of 160N (see below). It comes to rest after 7.50s. Find the
coefficient between the ax and the grindstone. Assume grindstone is a solid disk.
f
α
r
FN
ω
!0
rev
rad
1
, with ! 0 = 850
" 2#
"
= 89s $1
t
min
rev 60 sec/ min
89s "1
1
!=
= 11.86s "2 . Newton’s second law for rigid body is ! = I" = MR 2" , where
7.5s
2
we assume the gridstone is a solid cylinder. This gives for R = .260m, ! = 20Nim .
The Torque arise from the friction force between the ax and the stone. It should be clear
that the moment arm equals the radius, ! = f (.260m ) " f = 77N . The friction force is
77N
fk = µ k FN with the normal force FN = 160N . µ k =
= 0.48
160N
Problem 7
In the diagram below a 5 kg crate slides down an incline with coefficient of kinetic
friction of 0.25. A string is wrapped around a flywheel of mass 25 Kg, and moment of
inertia 0.5kg • m 2 with respect to the axis of rotation through O. The string pulls without
slipping at a distance 0.2 m from the axis. Find the tension in the string, and acceleration
of the crate.
Use ! = 0 = ! 0 + " t # " =
+y
T
T
n
T
ƒk
acm
36.9°
mg
α
R
T
+x
36.9°
(a) As mentioned in class, this problem is can be decomposed into translational and
rotational parts:
Translation: Refer to the middle diagram of the box
y-axis: n = mg cos 36.9 !
friction: Note here that the force of static friction ƒk must be up the incline (opposing the
translational motion), in order to prevent slipping: fk = nµk = mg µk cos 36.9! . [1]
x-axis: mg sin 36.9! ! fk ! T = macm . Using [1] we obtain
mg sin 36.9! ! mgµ k cos 36.9! ! T = macm . [2]
Rotational: Refer to the rightmost diagram on the flywheel.
! = TR = I" → T =
relation T =
I! R
. Using the no-slip condition (eq. 10.14) acm = R! , gives the
R2
Iacm
. [3]
R2
Substituting [3] into [2] mg sin 36.9! ! mgµ k cos 36.9! !
a cm =
a cm =
I
acm = macm , which gives
R2
mg sin 36.9! ! mgµ k cos 36.9!
m + I / R2
( 5.00kg ) 9.8m / s 2 ( 0.6 ) ! ( 5.00kg ) 9.8m / s 2 ( 0.25 ) ( 0.8 )
(
)
(
)
5.00kg + 0.500kg • m / ( 0.200m )
(b) Using [3] from part (a)
(
)
(
)(
2
2
= 1.12m / s 2
)
0.500kg • m 2 1.12m / s 2
Iacm
T= 2 =
= 14.0N
R
( 0.200m )2
Problem 8
In the diagram below, two weights are connected by a very light string, which is passed
over a 50 N solid-disk cylinder of radius 0.3 m. The 125 N accelerates downward,
without the rope slipping. What force does the ceiling exerts on the pulley?
75N
9.8m / s 2
= 7.65kg
m1 =
Clockwise
is positive
α
125N
m2 =
9.8m / s 2
= 12.75kg
+y
FW
r = .3m
a
T125
T75
a
125 N
75N
Fg1 = 75N
Fg2 = 125N
From FBD diagrams above:
75 kg box (Translation)
125N Box (Translation)
net
net
Fy = T75 ! 75N = m1a
Fy = T125 ! 125N = !m2 a
T75
!
net
Fp= 50N
T125
Pulley (rotation)
= T125 r " T75 r = I#
[3]
[1]
[2]
Subtract [2] by [1], T125 ! T75 = 50N ! ( m1 + m2 ) a [4]
1
1
Using [3] I = Mr 2 (Solid disk), T125 ! T75 = Mr" . Using the no-slip condition a = r! ,
2
2
50N
1
= 5.1kg Substituting into [4] gives
gives T125 ! T75 = Ma , M =
m
2
9.8 2
s
1
50N
50N
Ma = 200N ! ( m1 + m2 ) a " a =
=
1
2
m1 + m2 + M 7.65kg + 12.75kg + 2.55kg
2
m
m
a = 2.18 2 . Substituting into [1] gives T75 = 75N + 7.65kg ! 2.18 2 = 91.7N .
s
s
m
Substituting into 2 gives T125 = 125N + 12.75kg ! 2.18 2 = 153N .
s
Using FBD of the pulley, the y-component of the net forces must be zero, otherwise, the
pulley would translate. Fynet = Fw ! FP ! T75 ! T120 = 0
FW = 50N + 91.7N + 53N = 195N .
Conservation of energy for a rigid body rotating and translating
Problem 9
In the diagram below, s solid uniform ball roll without slipping with a linear speed of
25.0 m/s. It moves up the hill, and goes over a vertical cliff. A) How far from the foot of
the cliff does the ball land. B) Calculate the linear speed of the ball when it lands. Does
the fact that this speed is greater than the initial linear speed means that the ball gains
energy? C) Calculate the initial and final linear and rotational kinetic energy.
Solution
Solid Uniform Ball with
moment of inertia
I = (2/5)mR2
v2 = ?
ω2
Position 1
U1g = mgh
v1 = 25.0m/s
ω1
h
ω3
Position 3
U3g = 0
R
v3 = ?
Position 1
U1g = 0
x
a) First use conservation of total mechanical energy between position 1 and 2.
1
1
1
1
1
1
1
1
U1g + mv12 + I! 12 = U 2g + mv22 + I! 22 " mv12 + I! 12 = mgh + mv22 + I! 22
2
2
2
2
2
2
2
2
2
1
12
1
2
Since I = mR 2 ! I" 12 =
mR 2" 12 = m ( R" 1 ) . Using the condition for rolling
5
2
25
5
1 2 1 2
1
1
without slipping v1 = R! 1 gives I! 1 = mv1 , and similarly I! 22 = mv22 .
2
5
2
5
1 2 1 2
1 2 1 2
1 2 1 2
1 2 1 2
mv1 + I! 1 = mgh + mv2 + I! 2 " mv1 + mv1 = mgh + mv2 + mv2
2
2
2
2
2
5
2
5
1 2 1 2
1
1
7
7
mv1 + mv1 = mgh + mv22 + mv22 ! mv12 = mgh + mv22
2
5
2
5
10
10
(
)
10
10
2
gh = ( 25m / s ) !
9.8m / s 2 ( 28.0m ) = 15.26m / s
7
7
Hence the ball linear speed v2 decreases. Its angular speed ! 2 also decreases.
Now use conservation of total mechanical energy between position 2 and 3.
1
1
1
1
1
1
1
1
U 2g + mv22 + I! 22 = U 3g + mv32 + I! 32 =" mgh + mv22 + I! 22 = mv32 + I! 32
2
2
2
2
2
2
2
2
Now note that as the ball roll up the hill, from position 2 to 3, without slipping, the
friction between the ball and ground induces a torque that changes its angular velocity.
But as it leaves the cliff to go from position 2 to 3 there is no friction to change the
angular velocity: ! 2 = ! 3 . This gives
1
1
1
1
1
1
mgh + mv22 + I! 22 = mv32 + I! 32 " mgh + mv22 = mv32
2
2
2
2
2
2
v2 = v12 !
v3 = v22 + 2gh =
(15.26m / s )2 + 2 ( 9.8m / s 2 ) ( 28.0m ) = 27.9m / s
At the top of the hill (position 2) its velocity is v2 = 15.26m / s horizontally. It has no
vertical component of the velocity! Hence the time, t, it takes to fall h = 28.0 m to the
2 ( 28.0m )
1
2h
bottom of the cliff is gt 2 = h ! t =
=
= 2.39s . Note now that at
2
g
9.8m / s 2
position 3 the horizontal component of the velocity remains v2 = 15.26m / s . The
horizontal distance traveled is simply x = v2t = (15.26m / s ) ( 2.39s ) = 36.5m
b) It is noted that the linear speed at position 3 ( v3 = 27.9m / s ) is greater than at position
1 ( v2 = 25m / s ), even though they are at the same height. This is explained by the fact
that the kinetic energy includes rotational component. When the ball roll up the hill from
1 to 2 it gains PE of mgh and it loses mgh in KE, which since there is friction both the
linear and rotational component lose energy so that v1 > v2 and ! 1 > ! 2 . When it falls off
the cliff from 2 to 3 it regains mgh in KE, but only the rotational component cannot
change so that ! 2 = ! 3 , meaning that the linear component v2 < v3 regains the whole
mgh so that v1 < v3 , but also ! 2 = ! 3 < ! 1 , so the rotational energy at position 3 is less
than at 1. However it is noted that the total kinetic energy (rotational + linear) are the
same at position 1 and 3.
Problem 10
In the diagram below, a uniform spherical boulder starts from rest and rolls down a 50.0
m high hill. The top half of the hill is rough so that the boulder rolls without slipping, but
the bottom half is covered with ice, and so is frictionless. What is the translational speed
of the boulder when it reaches the bottom of the hill?
Point 1
U1
!1 = 0
v1 = 0
Solid sphere (table 9.2) I =
M – mass of sphere
R – radius of sphere
!2 = ?
Rough
25.0 m Roll without
slipping
Point 2
U2
Smooth
Sphere slips
25.0 m
2
MR 2
5
+
Clockwise
is positive
v2 = ?
!3 = ! 2
v3 = ?
Point 3
U3
Point 1 to 2 The total energy of the sphere is K + K rot + U , with K is the translational
kinetic energy, Krot is the rotational kinetic energy, and U is the gravitational energy.
There is conservation of energy from point 1 to 2: K 1 + K rot ,1 + U 1 = K 2 + K rot , 2 + U 2 .
Taking the zero gravitational energy at point 2, U 2 = 0 , and at point 1 U 1 = Mgh , with h
= 25.0 m. This gives
1
1
1
1
K 1 + K rot ,1 + U 1 = K 2 + K rot , 2 + U 2 " Mv12 + I!12 + Mgh = Mv 22 + I! 22
2
2
2
2
1
1
Using !1 = 0 and v1 = 0 , Mgh = Mv12 + I!12 .
2
2
From point 1 to 2 there is rolling without slipping: v = "R ! " 2 = v 2 / R . Using this and
2
I=
2
1
1'2
1
1
10
$' v $
MR 2 gives Mgh = Mv 22 + % MR 2 "% 2 " ! gh = v 22 + v 22 ! v 22 = gh
5
2
2&5
2
5
7
#& R #
(
)
10
9.8m / s 2 (25.0m ) = 18.7 m / s .
7
Point 2 to 3 Taking the gravitational energy to be zero, U 3 = 0 ,at point 3, and
U 2 = Mgh , with h = 25.0 m. This gives
1
1
1
1
K 2 + K rot , 2 + U 2 = K 3 + K rot ,3 + U 3 " Mv 22 + I! 22 + Mgh = Mv32 + I! 32
2
2
2
2
From point 2 to 3 there is no friction so the angular velocity remains constant: ! 2 = ! 3 .
The linear (translational) speed at point 2 is v 2 =
This gives
1
1
1
1
Mv22 + Mgh = Mv32 ! v32 = v22 + gh ! v32 = v22 + 2gh .
2
2
2
2
The linear (translational) speed at point 3 is
v3 =
(18.7m / s )2 + 2 ( 9.8m / s 2 ) ( 25.0m ) = 29m / s
Angular Momentum
Problem 11
In diagram below a 2kg rock is at point P traveling horizontally with a speed of 12 m/s.
At this instant what is the magnitude and direction of the angular momentum? If the only
force acting on the rock is its weight, what is the rate of change (magnitude and direction)
of the angular momentum?
SOLUTION TO BE PRESENTED Tomorrow
Problem 12
A diver comes off the board with straight arms and legs has a moment of inertia of
18kg • m 2 . She then tucks into a small ball decreasing her moment of inertia to 3.6kg • m 2 .
While tucked in she makes two complete revolution in 1.0s. If she hadn’t tucked in how
many revolutions would she have made in 1.5 seconds?
This is a conservation of angular momentum problem. Initially the diver is straight arm
with large moment of inertia, I1 = 18kg • m 2 , and low angular velocity !1 = ? . In the
final scenario she tucks reducing the moment of inertia to I 2 = 3.6kg • m 2 , and rotates 2
revolution in one seconds. Hence her final angular velocity is
rad %
" rev % "
!2 = $ 2
( $ 2)
'
' = 4) rad / s = 12.56 rad / s.
# s & #
rev &
initial angular momentum = final angular momentum
I1! 1 = I 2! 2 " ! 1 =
I2
3.6kg • m 2
!2 =
(12.56rad / s ) = 2.51 rad / s
I1
18kg • m 2
If she hadn’t tucked her angular velocity would remain ! 1 = 2.51rad / s . In 1.5 seconds she
would rotate ! = " t = ( 2.51rad / s ) (1.5s ) = 3.77rad. In revolution she would rotate
# 1 rev &
! = ( 3.77rad ) %
= 0.600rev.
$ 2" rad ('
Problem 13
A 40 kg child is standing at a center of a large rotating disk (i.e. a merry-go-round) of
moment of inertia I = 1200kg • m 2 . Initially the disk makes one revolution in six seconds.
The child then runs (without slipping) to a distance 2m from the center, and then remains
there. What is the final angular speed of the disk in rad/s?
Moment of Inertia of Disk I disk = 1200kg • m 2 , w.r.t center of disc
Child of mass m = 40 kg
Axis of rotation is perpendicular to page, passing through the center of disk
+
!f
!i
2m
I tot = I disk , child does not
contribute to moment of inertia
axis of rotation.
I tot = I disk + I child = I disk + mR 2 , with R
= 2.0 m. Child contributes to moment of
inertia axis of rotation.
Initial total moment of inertia I tot ,i = I disk = 1200kg • m 2
Final total moment of inertia
2
I tot , f = I disk + mR 2 = 1200kg • m 2 + (40kg )(2m ) = 1360kg • m 2
Conservation of Angular Momentum (Disk + Child is an isolated system)
initial angular momentum I tot ,i ! i = I tot , f ! f final angular momentum , with
1 rev.
rad
rad
2!
= 1.057
6s
rev
s
I tot ,i ! i 1200kg • m 2 (1.057 rad / s )
=
= 0.93rad / s or
.! f =
I tot , f
1360kg • m 2
"i =
(
" f = (0.93rad / s )
)
rev
rev
= 0.148
.
2! rad
s