midterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 4 2007, 11:00 pm
1
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midterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 4 2007, 11:00 pm
Question 1, chap 11, sect 2.
part 1 of 1
10 points
Consider the collision of two identical particles, where the initial velocity of particle 1
is v1 and particle 2 is initially at rest.
2
A horizontal boom of mass m and length
ℓ with a mass M suspended from one end is
held by a cord with tension T , fastened at a
distance D from the end of the boom that is
attached to a wall with a frictionless pin.
Boom and Weight
T : tension in the cord
v1
θ
1
2
T
h
m
After an elastic head-on collision, the final
velocity v2′ of particle 2 is given by
1. v2′ =
2. v2′ =
3. v2′ =
4. v2′ =
3 v1
4
5 v1
3
v1
3
2 v1
3
4 v1
3
ℓ
What is magnitude of the torque τ about
the pin exerted on the boom by the cord?
1. τ =
DT
cos θ
2. τ = D T tan θ
5.
=
6.
v2′
= v1 correct
3. τ = D T cos θ correct
v1
7. =
4
v
1
8. v2′ =
2
v2′
9.
D
M
v2′
v2′
Pin
4. τ = D T sin θ
5. τ = D T (1 − cos θ)
6. τ = D T (1 − sin θ)
=0
DT
1 − cos θ
DT
8. τ =
sin θ
DT
9. τ =
1 − sin θ
7. τ =
10. v2′ = 2 v1
Explanation:
For the final velocity v2′ of particle 2 after
an elastic collision, we have
v2′ = 2 vcm − v2 .
For the present case,
v1
m1 v1 + m2 v2
= .
vcm =
m1 + m2
2
So
v 1
′
− 0 = v1 .
v2 = 2
2
Question 2, chap 13, sect 1.
part 1 of 1
10 points
10. τ = D T (1 − tan θ)
Explanation:
The distance from the pin to the cord is
D cos θ, thus the magnitude of the torque is
~
~
T × D cos θ = D T cos θ .
midterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 4 2007, 11:00 pm
Question 3, chap 12, sect 2.
part 1 of 3
10 points
A beetle takes a joy ride on a pendulum.
The string supporting the mass of the pendulum is 162 cm long.
If the beetle rides through a swing of 32 ◦ ,
how far has he traveled along the path of the
pendulum?
Correct answer: 90.4779.
Explanation:
Arc length is defined as
s = rθ
= (162 cm)(0.558506 rad)
= 90.4779 cm ,
where theta is in radians.
Dimensional analysis for θ
cm ·◦ ·
π radians
= cm
180 ◦
Question 4, chap 12, sect 2.
part 2 of 3
10 points
If the pendulum swings through the same
angle 32 ◦ , what is the linear displacement
experienced by the beetle?
Correct answer: 89.3065.
Explanation:
Using the law of cosines, we have
p
~ = r 2 + r 2 − 2 r r cos θ
kRk
p
= r 2 (1 − cos θ)
p
= (162 cm) 2 [(1 − cos(32◦ )]
If the pendulum at some instant is swinging
at 2.3 rad/s, how fast is the beetle traveling?
Correct answer: 372.6.
Explanation:
Linear and angular velocity are related by
v = rω
= (162 cm)(2.3 rad/s)
= 372.6 cm/s ,
where ω is in radians per unit time.
Question 6, chap 13, sect 4.
part 1 of 1
10 points
A solid steel sphere of density 7.77 g/cm3
and mass 0.3 kg spins on an axis through its
center with a period of 2.3 s.
4
Given Vsphere = π R3 , what is its angular
3
momentum?
Correct answer: 0.000144115.
Explanation:
The definition of density is
ρ≡
= 89.3065 cm .
Question 5, chap 12, sect 2.
part 3 of 3
10 points
M
M
=
,
4
V
π R3
3
Therefore
1
3M 3
R=
4πρ
1
3
3 (0.3 kg)
=
4 π (7770 kg/m3 )
= 0.020967 m .
= 89.3065 cm .
Alternative Solution: Divide the isosceles triangle in half. Then
~ = 2.0 r sin θ
kRk
2
= 2.0 (162 cm) sin (0.279253 rad)
3
Using
2π
T
2π
=
(2.3 s)
= 2.73182 s−1
ω=
midterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 4 2007, 11:00 pm
and
2
I = M R2
5
2
= (0.3 kg)(0.020967 m)2
5
= 5.2754 × 10−5 kg m2 ,
we have
L≡Iω
4 π M R2
=
5T
4 π (0.3 kg)(0.020967 m)2
=
5 (2.3 s)
= 0.000144115 kg m2 /s .
Question 7, chap 12, sect 5.
part 1 of 1
10 points
A massless rod of length L has a mass m
fastened at its center and another mass m
fastened at one end. On the opposite end, the
rod is pivoted with a frictionless hinge.
The rod is released from rest from an initial
horizontal position; then it swings down.
L/
m
2
L
θ
radius
R≪L
m
What is the angular velocity as the rod
swings through its lowest (vertical) position?
r
3g
1.
2L
r
2g
2.
L
r
g
3.
L
4
r
6g
5L
r
g
5.
4L
r
12 g
correct
6.
5L
r
2g
7.
3L
r
5g
8.
6L
r
4g
9.
5L
r
g
10.
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
4.
Ui = (2 m) g L .
The potential energy at the bottom of the
swing,
L
Uf = m g
, so
2
1
3
∆U = 2 −
mgL = mgL.
2
2
The moment of inertia of the system is
2
5
L
+ m L2 = m L2 .
I=m
2
4
So the change in the kinetic energy is
∆K =
1
5
I ω 2 = m L2 ω 2
2
8
Therefore,
∆U = ∆K
5
3
m g L = m L2 ω 2
2
8
12 g = 5 L ω 2 , so
r
12 g
.
ω=
5L
midterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 4 2007, 11:00 pm
I1
ω0
I1 + I2
I1
ω0
=
1
I1 + I1
4
4
= ω0 .
5
ω=
Question 8, chap 13, sect 3.
part 1 of 2
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 ro4
tating, drops onto the first cylinder. Since the
surfaces are rough, the two eventually reach
the same angular speed ω.
1
I1
4
I2
Calculate the final kinetic energy K with
respect to the initial kinetic energy K0 .
2. K =
ω0
ω
After
Before
Calculate the final angular speed ω with
respect to the initial angular speed ω0 .
4
ω0
7
3
2. ω = ω0
5
3
3. ω = ω0
4
1. ω =
4. K =
5. K =
7. K =
8. K =
4
K0
7
3
K0
5
2
K0
3
3
K0
7
2
K0
5
4
K0 correct
5
3
K0
4
1
K0
3
9. None of these
1
ω0
3
4
6. ω = ω0 correct
5
2
7. ω = ω0
3
3
8. ω = ω0
7
2
9. ω = ω0
5
Explanation:
X
~ = const
L
5. ω =
From conservation of angular momentum
I1 ω0 = (I1 + I2 ) ω ,
3. K =
6. K =
4. None of these
(1)
Question 9, chap 13, sect 3.
part 2 of 2
10 points
1. K =
I1
5
so
Explanation:
Substituting ω from Eq. 1, the final rotational kinetic energy is
1
K = (I1 + I2 ) ω 2
2
I12
1
ω2
= (I1 + I2 )
2
(I1 + I2 )2 0
1 I12
=
ω2 ,
(2)
2 I1 + I2 0
and the initial rotational kinetic energy is
1
K0 = I1 ω02
2
1
K0
(3)
ω02 =
1
I1
2
midterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 4 2007, 11:00 pm
Substituting ω 2 from Eq. 3 into Eq. 2, we have
I12
1
2 I1 + I2
K=
K0
1
I1
2
I1
K0
=
I1 + I2
I1
=
K0
1
I1 + I1
4
4
K0 ,
=
5
since K is less than K0 , kinetic energy is lost.
Question 10, chap 13, sect 3.
part 1 of 1
10 points
Assume: A bullet of mass m and cube of
mass M undergo an inelastic collision, where
m ≪ M.
Note: The moment of inertia of this cube
(with edges of length 2 a and mass M ) about
8 M a2
.
an axis along one of its edges is
3
A solid cube is resting on a horizontal surface. The cube is constrained to rotate about
an axis at its bottom left edge (due to a small
obstacle on the table). A bullet with speed
vmin is shot at the left-hand face at a height
4
of a. The bullet gets embedded in the cube.
3
ω
2a
m~vmin
4
a
3
M
Mg
Find the minimum value of vmin required
to tip the cube so that it falls its right-hand
face.
r
√
m
2−1
1. vmin ≈
5ga
M
r
√
M
2. vmin ≈
3ga
3−1
m
r
√
M
2 − 1 correct
3ga
3. vmin ≈
m
6
r
√
M
4. vmin ≈
2ga
2−1
m
r
√
M
5ga
2−1
5. vmin ≈
m
r
√
m
2ga
6. vmin ≈
2−1
M
r
√
m
3ga
5−1
7. vmin ≈
M
r
√
M
3ga
8. vmin ≈
5−1
m
r
√
m
3ga
2−1
9. vmin ≈
M
r
√
m
3ga
3−1
10. vmin ≈
M
Explanation:
Basic Concepts:
X
~ = const
L
∆U + ∆K = 0
The moment of interia for a rectangular
plate, where the plate is a square and sides
are 2 a is
1
cm
Isquare
=
M (2 a)2 + (2 a)2
12
2
= M a2 , so
3
edge
Icube = Icm + M R2
√
2
= M a2 + M ( 2 a)2
3
2
+ 2 M a2
=
3
8
= M a2 .
3
For the cube to tip over the center of mass
(CM) must rise so that it is over the axis of
rotation AB. To do√
this the CM must be
raised a distance of a
2 −1 .
From conservation of energy, initial kinetic energy equals potential enegy at tipping
point, gives
√
1
2
Icube ω = M g a
2 −1 .
(1)
2
midterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 4 2007, 11:00 pm
We are given m ≪ M , so the mass m of
the bullet can be neglected when considering
the angular momentum of the bullet embeded
in the block. From conservation of angular
momentum
8 M a2
4a
ω
mv ≈
3
3
mv .
(2)
ω≈
2M a
Let : M = 7 kg ,
R = 13 cm ,
m1 = 26 kg ,
m2 = 12 kg ,
h = 3.9 m ,
v = ωR,
1
I = M R2 , and
2
1
1
Kdisk = I ω 2 = M v 2 .
2
4
From conservation of energy
Thus, substituting Eq. 2 into 1, we have
1
2
2 2 √
8 M a2
m vmin
≈
M
g
a
2
−
1
3
4 M 2 a2
√
2
1 m2 vmin
≈ ga
2 −a
3 M2
r
√
M
3ga
2 −1 .
vmin ≈
m
Question 11, chap 13, sect 1.
part 1 of 1
10 points
A 26 kg mass and a 12 kg mass are suspended by a pulley that has a radius of 13 cm
and a mass of 7 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 3.9 m apart. Treat
the pulley as a uniform disk.
The acceleration of gravity is 9.8 m/s2 .
13 cm
ω
7 kg
26 kg
3.9 m
12 kg
Determine the speeds of the two masses as
they pass each other.
Correct answer: 3.59075.
Explanation:
7
or
K1 + K2 + Kdisk = −∆U
m1 v 2 m2 v 2 M v 2
h
+
+
= (m1 − m2 ) g
2
2
4
2
M
v 2 = (m1 − m2 ) g h ,
m1 + m2 +
2
h
where is the height. Taking no slipping into
2
account, we can solve for v
s
2 (m1 − m2 ) g h
(1)
v=
2 (m1 + m2 ) + M
s
535.08 J
=
41.5 kg
= 3.59075 m/s .
Alternative Solution: The forces in the
vertical directions for m1 , m2 , and the torque
on the pulley give us
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
1
a h
= m1 (g − a)] R
M R2
2
R
i
− [m2 (g + a) R
1
M a = m1 g − m1 a − m2 a − m2 g
2
2 (m1 − m2 ) g
.
a=
M + 2 (m1 + m2 )
midterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 4 2007, 11:00 pm
Because the raindrops fall vertically, they do
not carry momentum horizontally. Assume
∆m of rain water accumulates on the cart:
The velocity is
vf2 = vi2 + 2 a ∆y
r
h
vf = 2 a
2
s
=
pi = pf
m v = (m + ∆m) v ′ .
2 (m1 − m2 ) g h
,
2 (m1 + m2 ) + M
Therefore
m
v
m + ∆m
v′ < v .
v′ =
which is the same as Eq. 1.
Question 12, chap 11, sect 4.
part 1 of 1
10 points
The speed of the cart will decrease because
of conservation of momentum.
An open cart on a level surface rolls without
frictional loss through a downpour of rain.
The rain falls vertically downward as shown
below.
As the cart rolls, an appreciable amount of
rain water accumulates in the cart.
rain
rain water
8
v
cart
The speed of the cart will
1. decrease because of conservation of momentum. correct
2. increase because of conservation of mechanic energy.
3. decrease because of conservation of mechanic energy.
4. remain the same because the raindrops
are falling perpendicular to the direction of
cart’s motion.
5. increase because of conservation of momentum.
Explanation:
This is an inelastic collision in the direction
along which the cart is rolling. Only momentum ~p along that direction is conserved.
Question 13, chap 11, sect 2.
part 1 of 1
10 points
A 9.35 g bullet is stopped by an initially
stationary 7.68 kg block of wood.
The speed of the bullet-plus-wood combination immediately after the collision is
0.577 m/s .
What was the original speed of the bullet?
Correct answer: 474.519.
Explanation:
mb = 9.35 g = 0.00935 kg ,
m = 7.68 kg , and
v = 0.577 m/s .
Let v0 be the initial velocity of the bullet.
Using momentum conservation,
mb v0 = (mb + m) v
mb + m
v0 =
v
mb
(0.00935 kg) + (7.68 kg)
(0.577 m/s)
=
(0.00935 kg)
= 474.519 m/s .
Question 14, chap 12, sect 3.
part 1 of 1
10 points
At t1 = 0, a wheel rotating about a
fixed axis at a constant angular acceleration α = −0.31 rad/s2 has an angular velocity ω = 1.8 rad/s and an angular position
φ1 = 9.5 rad.
midterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 4 2007, 11:00 pm
What is the angular position φ2 of the wheel
at time t2 = 2 s?
Correct answer: 12.48.
2.6 m
Explanation:
In the usual notation
s
φ≡
r
dφ
v
ω=
=
dt
r
a
dω
= ,
α=
dt
r
the only equation we need is the one for an1
gular position “ φ − φ0 = ω0 t + α t2 ”.
2
For our problem it reads
1
φ2 = φ1 + ω1 t2 + α t22
2
= (9.5 rad) + (1.8 rad/s) (2 s)
1
+ (−0.31 rad/s2 ) (2 s)2
2
= (9.5 rad) + (3.6 rad) + (−0.62 rad)
9
θ
1.3 m
b
µ=0
b
20.1 kg
µ = 0.5
Note: Figure is not to scale.
What is the minimum angle θmin (between
the horizontal and the ladder) so that the
person can reach a distance of 1.3 m without
having the ladder slip?
Correct answer: 45.
Explanation:
Let : d = 1.3 m ,
L = 2.6 m ,
W = m g = 196.98 N ,
µ = 0.5 .
= 12.48 rad .
and
Nw
b
Question 15, chap 14, sect 2.
part 1 of 1
10 points
A 20.1 kg person climbs up a uniform ladder
with negligible mass. The upper end of the
ladder rests on a frictionless wall. The bottom
of the ladder rests on a floor with a rough
surface where the coefficient of static friction
is 0.5 . The angle between the horizontal and
the ladder is θ . The person wants to climb up
the ladder a distance of 1.3 m along the ladder
from the ladder’s foot.
The acceleration of gravity is 9.8 m/s2 .
θ
f
mg
b
P ivot
Nf
Using the above equilibrium conditions, we
have (taking the sum of the torques at the
bottom of the ladder)
midterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 4 2007, 11:00 pm
X
Fx :
X
Fy :
X
τ◦ :
f − Nw = 0 ,
(1)
Nf − W = 0 , and
(2)
W d cos θ − Nw L sin θ = 0 , (3)
where d is the distance of the person from
the bottom of the ladder. Using Eq. 2, Eq. 3
becomes
f L sin θ = W d cos θ ,
Wd
so
f=
.
(4)
L tan θ
The ladder may slip when f = fmax = Nw ,
from Eq. 4
Wd
.
L tan θ
Thus, solving for θ , we have
d
θ ≥ arctan
µL
(1.3 m)
≥ arctan
(0.5) (2.6 m)
fmax ≡ µ W ≥
(5)
Find the weight of W3 .
Correct answer: 45.2833.
Explanation:
Let : W = 19 N ,
ℓr1 = 9 m ,
ℓl1 = 4 m ,
ℓr2 = 6 m ,
ℓl2 = 5 m ,
ℓr3 = 8 m ,
ℓl3
ℓl1 W
ℓr1
(4 m) (19 N)
=
(9 m)
= 8.44444 N .
W1 =
A mobile consisting of four weights hanging
on three rods of negligible mass.
W2
19 N
X
Apply
~τ = 0 about an axis through the
point of suspension of the middle part of the
mobile, we obtain
ℓlw W2 − ℓr2 [W + W1 ] = 0
8m
6m
4m
= 6 m.
ℓl1 W − ℓr1 W1 = 0
Question 16, chap 14, sect 3.
part 1 of 1
10 points
5m
and
X
We can apply the balance condition
~τ =
0 successively, starting with the lowest part of
the mobile, to find the value of each unknown
weight. X
Apply
~τ = 0 about an axis through the
point of suspension of the lowest part of the
mobile, we obtain
≥ 45 ◦ .
6m
10
W3
9m
W1
ℓr2 [W + W1 ]
ℓlw
(6 m) [(19 N) + (8.44444 N)]
=
(5 m)
= 32.9333 N .
X
Apply
~τ = 0 about an axis through the
point of suspension of the top part of the
mobile, we obtain
W2 =
ℓle [W + W1 + W2 ] − ℓr3 W3 = 0
midterm 03 – JYOTHINDRAN, VISHNU – Due: Apr 4 2007, 11:00 pm
ℓl3 [W + W1 + W2 ]
ℓr3
(6 m) h
(19 N)
=
(8 m)
Correct answer: 4.72.
W3 =
+ (8.44444 N) + (32.9333 N)
= 45.2833 N .
11
Explanation:
Basic Concept:
i
Question 17, chap 11, sect 1.
part 1 of 1
10 points
If one knows only the constant resultant
force acting on an object and the time during
which this force acts, one can determine the
1. acceleration of the object.
2. change in kinetic energy of the object.
3. change in momentum of the object. correct
4. change in velocity of the object.
5. mass of the object.
Explanation:
Impulse is the change in momentum, defined by
~I = F
~ ∆t = ∆(m~v) .
Therefore the change in momentum of the
~ and the
object can be determined from F
time.
To determine the changes in velocity or kinetic energy, we need additional information.
Question 18, chap 13, sect 4.
part 1 of 1
10 points
A particle is located at the vector position
~r = (1.6 m)ı̂ + (3.4 m) ̂
and the force acting on it is
~ = (2.8 N)ı̂ + (3 N) ̂ .
F
What is the magnitude of the torque about
the origin?
~
~τ = ~r × F
Solution: Since neither the position of the
particle, nor the force acting on the particle
have a z-component, the torque acting on the
particle has only a z-component
~τ = [x Fy − y Fx ] k̂
= [(1.6 m) (3 N) − (3.4 m) (2.8 N)] k̂
= [−4.72 N m] k̂ .