oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 3 2007, 3:00 pm
Question 1, chap 12, sect 2.
part 1 of 1
10 points
The tub of a washer goes into its spindry cycle, starting from rest and reaching an
angular speed of 2.3 rev/s in 12 s . At this
point the person doing the laundry opens the
lid, and a safety switch turns off the washer.
The tub slows to rest in 10.7 s .
Through how many revolutions does the
tub turn?
Assume Constant angular acceleration
while it is starting and stopping.
Correct answer: 26.105 rev (tolerance ± 1
%).
Explanation:
We will break the motion into two stages:
(1) an acceleration period and
(2) a deceleration period.
While speeding up,
θ1 = ω t
0+ω
t1
=
2
1
= ω t1
2
1
= (2.3 rev/s) (12 s)
2
= 13.8 rev .
While slowing down,
θ2 = ω t
ω+0
t2
=
2
1
= ω t2
2
1
= (2.3 rev/s) (10.7 s)
2
= 12.305 rev .
Assume: When the disk lands on the surface it does not bounce.
The disk has mass m and outer radius R
with a radial mass distribution (which may
not be uniform) so that its moment of inertia
3
is m R2 .
4
The disk is rotating at angular speed ω0
around its axis when it touches the surface,
as shown in the figure below. The disk is
carefully lowered onto a horizontal surface
and released at time t0 with zero initial linear
velocity along the surface. The coefficient of
friction between the disk and the surface is µ .
The kinetic friction force between the surface and the disk slows down the rotation of
the disk and at the same time gives it a horizontal acceleration. Eventually, the disk’s
linear motion catches up with its rotation,
and the disk begins to roll (at time trolling )
without slipping on the surface.
The acceleration of gravity is g .
m
θ = θ1 + θ2
= (13.8 rev) + (12.305 rev)
= 26.105 rev .
Question 2, chap 13, sect 2.
part 1 of 1
10 points
R , radius
ω0
3
I= m R2
4
µ
Once the disk rolls without slipping, what
is its angular speed?
1. ωrot =
2. ωrot =
3. ωrot =
4. ωrot =
So
1
5. ωrot =
6. ωrot =
7. ωrot =
7
ω0
16
2
ω0
9
2
ω0
5
3
ω0 correct
7
4
ω0
13
5
ω0
12
3
ω0
8
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 3 2007, 3:00 pm
1
8. ωrot = ω0
5
5
9. ωrot =
ω0
11
4
10. ωrot = ω0
9
Explanation:
From the perspective of the surface, let the
speed of the center of the disk be vsurf ace .
Using the frictional force f , we can determine
the acceleration
f = µmg,
X
and
Fsurf ace = m a , or
m a = µ m g , so
a = µ g , and
µg
α=
.
R
(1)
of the center of the disk; that is, there will be
no slipping. Setting the velocity ωdisk from
Eq. 4 equal to ωsurf ace from Eq. 2 gives
ωdisk = ωsurf ace
4 µg
µg
ω0 −
t=
t , or
3 R
R
7
µ g t = R ω0 , so
3
3 R ω0
t=
.
7 µg
ωrot = α t
µg
t
=
R µ g 3 ω0 R
=
T
7 µg
=
3
ω0 .
7
(6)
or using Eqs. 4 and 5, we have
(2)
After pure rolling begins at trolling there is no
longer any frictional force and consequently
no acceleration. From the perspective of the
center of the disk, let the tangential velocity
of the rim of the disk be vdisk and the angular
velocity be ω ; the angular acceleration is
X
τ = I α , so
τ
α=
I
µmgR
=
3
m R2
4
4 µg
=
.
(3)
3 R
4 µg
t
3 R 4 µg
3 R
= ω0 −
ω0
3 R
7 µg
ωrot = ω0 −
=
3
ω0 .
7
Four particles with masses 7 kg, 5 kg, 7 kg,
and 9 kg are connected by rigid rods of negligible mass as shown.
y
9 kg
7 kg
O
x
ω
(4)
When the disk reaches pure rolling, the velocity from the perspective of the surface will be
the same as the velocity from the perspective
(6)
Question 3, chap 12, sect 5.
part 1 of 1
10 points
6m
The time dependence of ω is
ω = ω0 − α t
4 µg
t.
= ω0 −
3 R
(5)
Using Eqs. 1 and 5, we have
Since
ωsurf ace = α t , we have
µg
t.
=
R
2
5 kg
4m
7 kg
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 3 2007, 3:00 pm
The origin is at the center of the rectangle,
which is 4 m wide and 6 m long.
If the system rotates in the xy plane about
the z axis (origin, O) with an angular speed
of 2 rad/s, calculate the moment of inertia of
the system about the z axis.
Correct answer: 364 kg m2 (tolerance ± 1 %).
Explanation:
Let : m1 = 7 kg ,
m2 = 5 kg ,
m3 = 7 kg ,
m4 = 9 kg ,
w = 4 m,
ℓ = 6 m,
From
I=
X
top left
bottom left
bottom right
top right
and
mj rj2 ,
j
where in this case all distances are equal to
s
h w i2 ℓ 2
r=
+
2
2
s
2 2
(4 m)
(6 m)
=
+
2
2
= 3.60555 m ,
we obtain
Iz = [m1 + m2 + m3 + m4 ] r 2
= [(7 kg) + (5 kg) + (7 kg) + (9 kg)]
× (3.60555 m)2
= 364 kg m2 .
Question 4, chap 12, sect 5.
part 1 of 1
10 points
A massless rod of length L has a mass 2 m
fastened at its center and another mass m
fastened at one end. On the opposite end
from the mass m, the rod is hinged with a
frictionless hinge. The rod is released from
rest from an initial horizontal position; then
it swings down.
3
What is the angular velocity ωmax as the
rod swings through its lowest (vertical) position?
r
6g
1. ωmax =
5L
r
g
2. ωmax =
4L
r
8g
correct
3. ωmax =
3L
r
g
4. ωmax =
L
r
4g
5. ωmax =
5L
r
3g
6. ωmax =
2L
r
2g
7. ωmax =
L
r
5g
8. ωmax =
6L
r
12 g
9. ωmax =
5L
r
g
10. ωmax =
2L
Explanation:
The mechanical energy of the system is conserved. Measuring heights from the point at
the bottom of the rod when it is vertical,
the initial potential energy of the system is
Ui = (3 m) g L. The potential energy at the
bottom of the swing, Uf = 2 m g (L/2) =
m g L . To calculate the final kinetic energy,
we need the moment of inertia of the system,
I = 2 m (L/2)2 + m L2 = 3 m L2 /2 . Therefore,
1 3
3mgL = mg +
m L2 ω 2 ,
2 2
r
8g
ωmax =
.
3L
Question 5, chap 12, sect 5.
part 1 of 1
10 points
An 856 kg car is held in place by an unrealiable winch. The gearbox of the winch
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 3 2007, 3:00 pm
4
v
u 2 m g ∆h
breaks and at that moment the car-drum sysu
v
=
u
tem free-falls from rest, shown in the figure.
Id Ip
t
m+ 2 + 2
During the car’s fall, there is no slipping berd rp
tween the (massless) rope, the pulley, and the
v
u
winch drum.
2 (856 kg) (9.81 m/s2 ) (9 m)
u
=
u
The moment of inertia of the winch drum
t
(33 kg m2 ) (7 kg m2 )
+
(856 kg) +
is 33 kg m2 and that of the pulley is 7 kg m2 .
(0.77 m)2
(0.06 m)2
The radius of the drum is 77 cm and the
radius of the pulley is 6 cm .
= 7.27479 m/s .
The acceleration due to gravity is
9.81 m/s2 .
Question 6, chap 12, sect 5.
7 kg m2
part 1 of 1
10 points
A circular disk, a ring, and a square, have
the same mass M and width 2 r.
77 cm
33 kg m2
856 kg
ring
disk
square
2r
2r
2r
9m
Find the speed of the car as it hits the
ground.
Correct answer: 7.27479 m/s (tolerance ± 1
%).
Explanation:
Let : Id = 33 kg m2 ,
Ip = 7 kg m2 ,
rd = 77 cm = 0.77 m ,
rp = 6 cm = 0.06 m ,
mb = 856 kg , and
∆h = 9 m .
For the moment of inertia about their center of mass about an axis perpendicular to
the plane of the paper, which of the following statements concerning their moments of
inertia is true?
1. Idisk > Iring > Isquare
2. Idisk > Isquare > Iring
3. Iring > Idisk > Isquare
4. Iring > Isquare > Idisk
5. Isquare > Iring > Idisk correct
6. Isquare > Idisk > Iring
Applying conservation of mechanical energy (Ki = Uf = 0) and v = r ω ,
1
1
1
m v 2 + Id ωd2 + Ip ωp2 − m g ∆h = 0
2
2
2
2
1
1
1
v
v2
m v 2 + Id 2 + Ip 2 − m g ∆h = 0
2
2
2
rp
rd
Explanation:
Basic Concept: Moment of inertia, rotational kinetic energy. Solution: In the ring,
the same mass of the disk is concentrated at
the maximum distance from the axis. Therefore
Iring > Idisk
In the square, the same mass of the ring lies
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 3 2007, 3:00 pm
5
at distances which are at least the radius of
the ring. Therefore
5.9 cm
Isquare > Iring
ω
4 kg
thus
Isquare > Iring > Idisk
29 kg
Alternative Part 1: Using moments of inertia as found in the book and the parallel axis
theorem,
1
1
m r2 = M r2
2
2
= m r2 = M r2
Idisk =
Iring
1
m ℓ2
12
1
2
2
Isquare = 4
mℓ + md
12
M
m=
4
ℓ = 2r
d=r M
1 M
2
(2 r) +
r2
Isquare = 4
12 4
4
4
= M r2 ,
3
rod
Icm
=
2.5 m
18 kg
Determine the speeds of the two masses as
they pass each other.
Correct answer: 2.34521 m/s (tolerance ± 1
%).
Explanation:
Let : M = 4 kg ,
R = 5.9 cm ,
m1 = 29 kg ,
m2 = 18 kg ,
h = 2.5 m ,
v = ωR,
1
I = M R2 , and
2
1
1
Kdisk = I ω 2 = M v 2 .
2
4
From conservation of energy
since
4
1
M r2 > M r2 > M r2
3
2
then
Isquare > Iring > Idisk .
Question 7, chap 13, sect 1.
part 1 of 1
10 points
A 29 kg mass and a 18 kg mass are suspended by a pulley that has a radius of 5.9 cm
and a mass of 4 kg. The cord has a negligible
mass and causes the pulley to rotate without
slipping. The pulley rotates without friction.
The masses start from rest 2.5 m apart. Treat
the pulley as a uniform disk.
The acceleration of gravity is 9.8 m/s2 .
K1 + K2 + Kdisk = −∆U
or
h
m1 v 2 m2 v 2 M v 2
+
+
= (m1 − m2 ) g
2
2
2
4
M
m1 + m2 +
v 2 = (m1 − m2 ) g h ,
2
h
where is the height. Taking no slipping into
2
account, we can solve for v
s
2 (m1 − m2 ) g h
v=
(1)
2 (m1 + m2 ) + M
s
269.5 J
=
49 kg
= 2.34521 m/s .
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 3 2007, 3:00 pm
Alternative Solution: The forces in the
vertical directions for m1 , m2 , and the torque
on the pulley give us
1
M a = m1 g − m1 a − m2 a − m2 g
2
2 (m1 − m2 ) g
.
a=
M + 2 (m1 + m2 )
11 g/s at 737 m/s
4.8 m
4 rev/min
m1 g − T1 = +m1 a
m2 g − T2 = −m2 a , so
T1 = m1 (g − a)
T2 = m2 (g + a) , and
I α = [T1 − T2 ] R
a h
1
= m1 (g − a)] R
M R2
2
R
i
− [m2 (g + a) R
6
1020 kg m2
11 g/s at 737 m/s
Figure: Rear view of spacecraft.
For how long must they turn on these jets
to stop the rotation?
Correct answer: 5.48981 s (tolerance ± 1 %).
Explanation:
Let : R = 4.8 m ,
v = 737 m/s ,
∆m
= 11 g/s ,
∆t′
ω = 4 rev/min ,
The velocity is
and
2
I = 1020 kg m .
vf2 = vi2 + 2 a ∆y
r
h
vf = 2 a
2
s
=
2 (m1 − m2 ) g h
,
2 (m1 + m2 ) + M
The angular momentum is
∆L = I ∆ω ,
and the force exerted by the gas on the spaceship is
F = ma =
which is the same as Eq. 1.
The magnitude of the net torque exerted by
the jet is
Question 8, chap 13, sect 4.
part 1 of 1
10 points
The figure shows the rear view of a space
capsule that is rotating about its longitudinal
axis at 4 rev/min. The occupants want to stop
this rotation. They have small jets mounted
tangentially at a distance 4.8 m from the axis,
as indicated, and can eject 11 g/s of gas from
each jet with a nozzle velocity of 737 m/s.
The moment of inertia of the ship about its
axis (assumed to be constant) is 1020 kg m2 .
m ∆v
∆m
=
∆v .
′
∆t
∆t′
∆L
∆t
2 ∆m
I ∆ω
vR =
.
∆t′
∆t
τnet = 2 F R =
Thus
I ∆ω
∆m
2
vR
∆t′
(1020 kg m2 )(4 rev/min)
=
2 (0.011 kg/s) (737 m/s) (4.8 m)
∆t =
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 3 2007, 3:00 pm
7
3 2
2π rad
1 min
v = g ∆h , so
×
×
4
rev
60 s
r
4 g ∆h
= 5.48981 s .
v=
r 3
4 (9.8 m/s2 ) (2 m)
=
Question 9, chap 13, sect 2.
3
part 1 of 1
10 points
= 5.11208 m/s .
A string is wound around a uniform disc of
radius 0.66 m and mass 1.6 kg. The disc is
released from rest with the string vertical and
Question 10, chap 13, sect 2.
its top end tied to a fixed support.
2
part 1 of 1
10 points
The acceleration of gravity is 9.8 m/s .
0.66 m
h
ω
1.6 kg
A 150 g basketball has a 32.4 cm diameter
and may be approximated as a thin spherical
shell.
Given: The moment of inertia of a thin
spherical shell of radius R and mass m is
2
I = m R2 , g = 9.8 m/s2 , and µ = 0.21 .
3
32
.4
cm
Calculate the speed of the center of mass
when, after starting from rest, the center of
mass has fallen 2 m.
Correct answer: 5.11208 m/s (tolerance ± 1
%).
From conservation of mechanical energy we
have
∆U + ∆Krot + ∆Ktrans = 0
1 1
1
2
−M g ∆h +
M R ω2 + M v2 = 0 .
2 2
2
21
0.
µ
=
4.
Let : R = 0.66 m ,
M = 1.6 kg , and
g = 9.8 m/s2 .
150 g
17
m
Explanation:
47◦
Starting from rest, how long will it take a
basketball to roll, without slipping, 4.17 m
down an incline that makes an angle of 47 ◦
with the horizontal?
Correct answer: 1.39261 s (tolerance ± 1 %).
When there is no slipping
Explanation:
v = Rω,
so
1 2 2
R ω +
4
1 2
v +
4
1 2
v = g ∆h
2
1 2
v = g ∆h
2
Given :
ℓ = 4.17 m ,
θ = 47◦ , and
32.4 cm
d
= 0.162 m .
R= =
2
2
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 3 2007, 3:00 pm
8
as the required time.
Question 11, chap 14, sect 99.
part 1 of 1
10 points
N
P
m
gs
in
θ
θ
mg
A uniform bar of length L and weight W is
attached to a wall with a hinge that exerts a
horizontal force Hx and a vertical force Hy on
the bar. The bar is held by a cord that makes
a 90◦ angle with respect to the bar; the bar
makes an angle θ with respect to wall.
The acceleration of gravity is 9.8 m/s2 .
The moment of inertia of the ball about
the point of contact between the ball and the
inclined plane is
IP = Icm + m d2
2
= m R2 + m R2
3
5
= m R2 .
3
90◦
L
W
θ
The net torque about the point of contact
between the ball and the inclined plane is
m g R sin θ = IP α
5
= m R2 α so
3
m g R2 sin θ
a=
5
m R2
3
3
= g sin θ .
5
If the sphere starts from rest, then its center
of mass moves a distance
ℓ=
1 2
at ,
2
which gives
What is the magnitude of the horizontal
force Hx on the pivot?
1. Hx =
2. Hx =
3. Hx =
4. Hx =
5. Hx =
t=
=
r
2ℓ
sa
6. Hx =
1
2
1
2
1
2
1
2
1
2
1
2
W tan θ
W sin θ cos θ correct
W cos2 θ
W sin θ
W cos θ
W sin2 θ
2 (4.17 m)
(4.30036 m/s2 )
= 1.39261 s .
Explanation:
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 3 2007, 3:00 pm
Analyzing the torques on the bar with the
hinge as the axis of rotation,
X
τ = LT −
L
sin θ W = 0
2
1
W sin θ .
2
T =
Analyzing the forces on the bar,
X
Fx = Hx − T cos θ = 0
1
W sin θ cos θ = 0
Hx −
2
1
W sin θ cos θ .
Hx =
2
Question 12, chap 13, sect 3.
part 1 of 2
10 points
A student sits on a rotating stool holding
two 2.7 kg masses. When his arms are extended horizontally, the masses are 0.89 m
from the axis of rotation, and he rotates with
an angular velocity of 1.2 rad/sec. The student then pulls the weights horizontally to a
shorter distance 0.35 m from the rotation axis
and his angular velocity increases to ω2 .
ωi
ωf
Let : M
R1
ω1
R2
ω2
9
= 2.7 kg ,
= 0.89 m ,
= 1.2 rad/sec ,
= 0.35 m , and
= 2.41831 rad/sec ,
As the student moves his arms, his moment
of inertia changes from
I1 = Is + 2 m R12
= (2.9 kg m2 ) + 2 (2.7 kg) (0.89 m)2
= 7.17734 kg m2
to
2
I2 = (2.9 kg m ) + 2 (2.7 kg) (0.35 m)2
= 3.5615 kg m2 ,
but his angular momentum is conserved,
L = I 1 ω1 = I 2 ω2 .
Consequently, his angular velocity increases
from
ω1 = 1.2 rad/sec , to
I1
ω2 =
ω1
I2
(7.17734 kg m2 )
(1.2 rad/sec)
=
(3.5615 kg m2 )
= 2.41831 rad/sec .
Question 13, chap 13, sect 3.
part 2 of 2
10 points
When the student pulls the weights in, he
performs mechanical work — which increases
the kinetic energy of the rotating system.
Calculate the increase in the kinetic energy.
Correct answer: 5.24653 J (tolerance ± 1 %).
For simplicity, assume the student himself
plus the stool he sits on have constant combined moment of inertia Is = 2.9 kg m2 .
Find the new angular velocity ω2 of the
student after he has pulled in the weights.
Correct answer: 2.41831 rad/s (tolerance ±
1 %).
Explanation:
Explanation:
The rotational kinetic energy is given by
L2
Lω
I ω2
=
=
.
K=
2
2I
2
For the system in question — the student
and the weights — the angular momentum
L stays constant while the moment of inertia
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 3 2007, 3:00 pm
decreases and the angular velocity increases;
hence K increases by
L ω2 L ω1
−
2
2
I 1 ω1
[ω2 − ω1 ]
=
2
(7.17734 kg m2 ) (1.2 rad/sec)
=
2
× [(2.41831 rad/sec) − (1.2 rad/sec)]
∆K =
= 5.24653 J .
Question 14, chap 13, sect 3.
part 1 of 3
10 points
Consider an Earth-like planet hit by an
asteroid.
The planet has mass Mp = 1.16 × 1023 kg
and radius Rp = 4.18 × 106 m, and you may
approximate it as a solid ball of uniform
density. It rotates on its axis once every
T = 26 hr. The asteroid has mass Ma =
4.86 × 1016 kg and speed va = 32900 m/s
(relative to the planet’s center); its velocity
vector points θ = 73◦ below the Eastward horizontal. The impact happens at an equatorial
location. The picture below shows the view
from above the planet’s North pole:
vm
θ
ω
R
First, calculate the planet’s angular momentum (relative to its spin axis) before the
impact.
Correct answer: 5.4422 × 1031 kg m2 /s (tolerance ± 1 %).
Explanation:
Basic Concept: A rigid body rotating
around a fixed axis has angular momentum
L =I ×ω
where I is the body’s moment of inertia about
the axis of rotation.
10
Approximating the planet as a solid ball of
uniform density, its moment of inertia is
2
Ip = M R 2
5
= 8.10719 × 1035 kg m2 .
Its angular velocity before the impact is
2π
ω=
T
= 6.7128 × 10−5 rad/s , so
L=I ×ω
=I ×ω
= (8.10719 × 1035 kg m2 )
×(6.7128 × 10−5 rad/s)
= 5.4422 × 1031 kg m2 /s .
Question 15, chap 13, sect 3.
part 2 of 3
10 points
Calculate the asteroid’s angular momentum relative to the planetary axis.
Correct answer: 1.95409 × 1027 kg m2 /s (tolerance ± 1 %).
Explanation:
Approximating the asteroid as a pointlike
particle, its angular momentum is
~a = R
~ × Ma ~va
L
~ is the asteroid’s radius vector and
where R
Ma ~va is its linear momentum vector.
At the moment of impact, both the radius
~ and the momentum vector Ma~va
vector R
of the asteroid lie in the planet’s equatorial
plane. Consequently, their vector product
is perpendicular to the equatorial plane and
parallel to the planet axis. Because the horizontal component of the asteroid’s velocity is
directed Eastward — the same as the planet’s
rotation — the asteroid’s angular momentum
~ a has the same direction as the planet’s anL
~ p.
gular momentum L
In magnitude,
La = Rp Ma × horizontal component of va
= Rp Ma × va cos θ
= (4.18 × 106 m)
×(4.86 × 1016 kg) (32900 m/s) cos(73◦ )
= 1.95409 × 1027 kg m2 /s .
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 3 2007, 3:00 pm
Question 16, chap 13, sect 3.
part 3 of 3
10 points
11
In terms of the planetary day, this change
means
The impact is totally inelastic — the asteroid is stuck in the planet’s crust. But
thanks to the asteroid’s angular momentum,
the planet rotates faster after the impact than
it did before. By how many seconds has the
collision shortened the planetary day?
For simplicity, ignore the effect of the asteroid’s mass on the planet’s moment of inertia
and assume
−∆T = T − T ′
2π 2π
=
− ′
ω
ω
2 π(ω ′ − ω)
=
ω ω′
2 π ∆ω 1
=
ω + ∆ω ω
2 π (2.41031 × 10−9 rad/s)
=
(6.71305 × 10−5 rad/s)
1
×
(6.7128 × 10−5 rad/s)
before
I after = Iplanet
.
Warning: At intermediate stages of this calculation, do not round off intermediate results
and keep at least 7 significant digits.
Correct answer: 3.3607 s (tolerance ± 1 %).
= 3.3607 s ,
where
ω + ∆ω = (6.7128 × 10−5 rad/s)
+(2.41031 × 10−9 rad/s)
= 6.71305 × 10−5 rad/s .
Explanation:
The net angular momentum is conserved in
this collision, so after the impact the planet
(with the asteroid stuck in its crust) has
~′ = L
~p +L
~a .
L
p
~
Focusing on the Northward component of L
(the other two components vanish), we have
Question 17, chap 13, sect 3.
part 1 of 2
10 points
L′p = Lp + La
= Ip ω + Rp Ma va cos θ .
At the same time,
L′p = IP′ ω ′ ,
and since we assume unchanged moment of inertia Ip′ = Ip , it follows that after the impact,
the planet rotates at new angular velocity
ω′ =
L′p
Ip
=ω+
Rp Ma va cos θ
.
Ip
A uniform rod, supported and pivoted at
its midpoint, but initially at rest, has a mass
2 m and a length ℓ. A piece of clay with mass
m and velocity v hits one end of the rod, gets
stuck and causes the clay-rod system to spin
about the pivot point O at the center of the
rod in a horizontal plane.
Viewed from above the scheme is
v
m
In other words,
∆ω = ω ′ − ω
Rp Ma va cos θ
=
Ip
5 Ma va cos θ
=
2 Mp Rp
= 2.41031 × 10
0
ℓ
−9
ω
ω
(a) 2 m
rad/s .
(b)
(c)
Figure: The piece of clay and rod:
(a) before they collide,
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 3 2007, 3:00 pm
The final moment of inertia If of the clay-rod
system is the moment of inertia of the rod
plus the moment of inertia of the clay
(b) at the time of the collision, and
(c) after they collide.
After the collisions the clay-rod
system has an angular velocity ω
about the pivot.
If = Irod + Iclay
1
=
2 m ℓ2 + m
12
With respect to the pivot point O, what
is the magnitude of the initial angular momentum Li of the piece of clay and the final
moment of inertia If of the clay-rod system?
8
m ℓ2
12
4
If =
m ℓ2
12
4
If =
m ℓ2
12
5
If =
m ℓ2 correct
12
3
If =
m ℓ2
12
7
If =
m ℓ2
12
5
If =
m ℓ2
12
8
If =
m ℓ2
12
1. Li = m v ℓ ,
2. Li = m v
If =
ℓ
,
2
3. Li = m v ℓ ,
4. Li = m v
ℓ
,
2
5. Li = m v ℓ ,
6. Li = m v
ℓ
,
2
7. Li = m v ℓ ,
ℓ
,
2
Explanation:
Basic Concepts: Conservation of angular
~
momentum L
8. Li = m v
~ = ~r × ~p = m~r × ~v
L
~
dL
dt
Lz = I ω .
X
~τext =
Therefore, if the net external torque acting on
a system is zero, the total angular momentum
of that system is constant.
Since the total external torque acting on
the clay-rod system is zero, the total angular
momentum is a constant of motion. The total
initial angular momentum Li is simply the
angular momentum of the clay, since the rod
is at rest initially
Li = k~r × ~pk = m r v
=
mvℓ
.
2
12
(1)
=
5
m ℓ2 .
12
2
ℓ
2
(2)
Question 18, chap 13, sect 3.
part 2 of 2
10 points
The final angular speed ωf of the rod-clay
system is
12
v.
7
12 v
2. ωf =
.
5 ℓ
6
3. ωf = v .
5
12 v
4. ωf =
.
7 ℓ
5 v
5. ωf =
.
6 ℓ
6 v
.
6. ωf =
2 ℓ
3 v
7. ωf =
.
5 ℓ
4
8. ωf = v .
6
5 v
9. ωf =
.
12 ℓ
6 v
. correct
10. ωf =
5 ℓ
Explanation:
The total final angular momentum is the
same as the total initial angular momentum.
According to L = I ω, we have
1. ωf =
Lf = Li
mvℓ
=
2
= I f ωf
5
m ℓ 2 ωf ,
=
12
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 3 2007, 3:00 pm
so
X
6v
ωf =
.
5 ℓ
13
~ = const
L
From conservation of angular momentum
Question 19, chap 13, sect 3.
part 1 of 1
10 points
A cylinder with moment of inertia I1 rotates with angular speed ω0 about a frictionless vertical axle. A second cylinder, with
1
moment of inertia I2 = I1 , initially not ro3
tating, drops onto the first cylinder. Since the
surfaces are rough, the two eventually reach
the same angular speed ω.
I1 ω0 = (I1 + I2 ) ω , so
I1
ω0
ω=
I1 + I2
I1
ω0
=
1
I1 + I1
3
3
= ω0 ,
4
(1)
since K is less than K0 , kinetic energy is lost.
Question 20, chap 14, sect 3.
part 1 of 1
10 points
1
I1
3
I2
I1
ω0
ω
After
Before
Calculate the final angular speed ω with
respect to the initial angular speed ω0 .
3
ω0 correct
4
2
2. ω = ω0
3
A solid bar of length L has a mass m1 . The
bar is fastened by a pivot at one end to a
wall which is at an angle θ with respect to the
horizontal. The bar is held horizontally by a
vertical cord that is fastened to the bar at a
distance xcord from the wall. A mass m2 is
suspended from the free end of the bar.
1. ω =
3. None of these
4. ω =
5. ω =
6. ω =
7. ω =
8. ω =
9. ω =
1
ω0
3
4
ω0
5
3
ω0
5
2
ω0
5
3
ω0
7
4
ω0
7
Explanation:
T
θ
m1
xcord
m2
L
Find the tension T in the cord.
1
L
1. T = m1 + m2
g
2
xcord
2. T = (m1 + m2 ) g cos θ
1
L
3. T = m1 + m2
g sin θ
2
xcord
g
L
4. T = (m1 + m2 )
x
2
cord
1
L
5. T = m1 + m2
g cos θ
2
xcord
oldmidterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 3 2007, 3:00 pm
1
L
6. T =
m1 + m2
g correct
2
xcord
L
1
m1 + m2
g cos θ
7. T =
2
xcord
L
1
m1 + m2
g sin θ
8. T =
2
xcord
9. T = (m1 + m2 ) g sin θ
10. T = 0
Explanation:
X
Under static equilibrium
F = 0 and
X
τ = 0.
Since we know little about the reaction
force between the bar and the wall, it is easiest to begin by examining the torques around
the connection between the bar and the wall,
because the reaction forces will produce no
torques around that point.
X
L
m1 g = 0 .
2
L
1
m1 + m2
g .
T =
2
xcord
τ = −L m2 g + xcord T −
14