Download Homework 22 - University of Utah Physics

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HW#22
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Fall
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HW #22
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Problem 3
At t = 0 a block with mass M = 5 kg moves with a velocity v = 2 m/s at position xo = -.33 m from the
equilibrium position of the spring. The block is attached to a massless spring of spring constant k =
61.2 N/m and slides on a frictionless surface. At what time will the block next pass x = 0, the place
where the spring is unstretched? t1 = ?
Solution
We are given 𝑚 = 5.0 kg, and 𝑘 = 61.2 N/m, and so the angular frequency of oscillation is:
𝜔=�
Unit check:
�
𝑘
61.2 N⁄m
=�
= 3.50 s−1
𝑚
5.0 kg
N⁄m
1 1
kg ∙ m 1 1
1 1
= �N ∙ ∙
=� 2 ∙ ∙
=� 2=
kg
m kg
s
m kg
s
s
The general solution for the position as a function of time in general form is:
𝑥 = 𝐴 cos(𝜔𝜔 + 𝜙)
We are given that at 𝑡 = 0, 𝑥 = 𝑥0 = −0.33 m
→ 𝐴 cos 𝜙 = 𝑥0 … (1)
and differentiating with respect to time we get the velocity as a function of time in general form:
𝑣 = −𝜔𝜔 sin(𝜔𝜔 + 𝜙)
We are given that 𝑡 = 0, 𝑣 = 𝑣0 = +2.0 m/s
→ −𝜔𝜔 sin 𝜙 = 𝑣0 … (2)
In order to find 𝑡1 such that 𝑥(𝑡1 ) = 0, we need to first solve for 𝐴 and 𝜙 from equations (1), (2):
Taking the sum (1)2 + (2)2 ⁄𝜔2 gives
𝑣0 2
𝐴2 cos2 𝜙 + 𝐴2 sin2 𝜙 = 𝐴2 = 𝑥0 2 + � �
𝜔
𝐴 = �𝑥0 2 + �
2
𝑣0 2
2.0 m⁄s
� = �(−0.33m)2 + �
� = 0.66 m
𝜔
3.50 s−1
And substituting this value for A into (1) and (2), we have
𝑥0 −0.33m
1
(1) → cos 𝜙 =
=
= −0.50 = −
𝐴
0.66 m
2
𝑣0
2.0m/s
√3
(2) → sin 𝜙 =
=
= −0.866 = −
−1
−𝜔𝜔 −(3.50s )(0.66 m)
2
…continued
1
√3
cos 𝜙 = − ,
sin 𝜙 = −
→
2
2
Where 𝜙 is calculated in radians, and so we write
And so
tan 𝜙 = √3 and 𝜙 = −
𝑥 = 𝐴 cos(𝜔𝜔 + 𝜙) = (0.66m) cos �3.50 s−1 ∙ 𝑡 −
𝑥(𝑡1 ) = (0.66m) cos �3.50 s −1 ∙ 𝑡1 −
The first time this happens
at
cos �3.50 s−1 ∙ 𝑡1 −
2𝜋
𝜋
=−
3
2
𝜋
2𝜋
−3𝜋
4𝜋 𝜋
3.50 s −1 ∙ 𝑡1 = − +
=
+
=
2
3
6
6
6
𝜋
𝑡1 =
= 0.15 s
6 ∙ 3.50 s−1
3.50 s −1 ∙ 𝑡1 −
Answer: 𝑡1 = 0.15 s
2𝜋
�=0
3
2𝜋
�=0
3
2𝜋
�
3
2𝜋
3
Fall
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Problem 4
HW #22
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Problem 5
A simple pendulum with mass m = 1.5 kg and length L = 2.57 m
hangs from the ceiling. It is pulled back to a small angle of θ = 9.8°
from the vertical and released at t = 0.
1) What is the period of oscillation? _______ s
The moment of inertia about the pivot for a small particle is
𝐼 = 𝑚𝑟 2 = 𝑚𝐿2
The torque exerted by the force of gravity is
𝜏 = 𝑟𝑟 sin 𝜙𝑟𝑟 = 𝐿 ∙ 𝑚𝑚 ∙ sin(−𝜃) = −𝑚𝑚𝑚 sin 𝜃
Where we have observed that SINE is an odd function: sin(−𝜃) = − sin 𝜃, so the equation of
motion is:
Which has the form
𝑑2 𝜃 𝜏 −𝑚𝑚𝑚 sin 𝜃
𝑔
𝑔
= =
= − sin 𝜃 ≈ − 𝜃
2
2
𝑑𝑡
𝐼
𝑚𝐿
𝐿
𝐿
𝑑2𝜃
= −𝜔2 𝜃,
2
𝑑𝑡
𝑔
𝜔 = ,
𝐿
2
𝑔
9.81 m⁄s 2
�
�
→ 𝜔=
=
= 1.95 s −1
𝐿
2.57m
And the period is related to the angular velocity/frequency by (period is the time it takes to do
one cycle: i.e. 𝜔𝜔 = 2𝜋
𝑇=
2𝜋
2𝜋
=
= 3.22 s
𝜔
1.95 s −1
2) What is the magnitude of the force on the pendulum bob perpendicular to the string at
t=0? _______ N
The component of the force perpendicular to the string is also the “tangential component”
𝐹⊥ = 𝐹𝑇 = 𝑚𝑚 sin 𝜃 = 1.5 kg ∙ 9.81 m⁄s2 ∙ sin 9.8° = 2.50 N
3) What is the maximum speed of the pendulum?
_______m/s
𝑣𝑀𝑀𝑀 = 𝑟|𝜔𝜔| = 𝐿|𝜔𝜃𝑀𝑀𝑀 |
In this case we started at 𝑡 = 0 with
And so
𝜃 = 𝜃𝑀𝑀𝑀 = 9.8° = 9.8° ∙
𝜋 rad
= 0.171 rad
180°
𝑣𝑀𝑀𝑀 = 𝐿|𝜔𝜃𝑀𝑀𝑀 | = L = 2.57 m ∙ 1.95 s −1 ∙ 0.171 rad = 0.859 m⁄s
4) What is the angular displacement at t = 3.73 s? (give the answer as a negative angle if the
angle is to the left of the vertical) _______°
Since we started at t=0 from rest, we have
𝜃 = 𝜃𝑀𝑀𝑀 cos 𝜔𝜔
𝜃(3.73s) = 9.8° ∙ cos(1.95 s−1 ∙ 3.73 s) = 9.8° ∙ cos(7.29 rad) = 9.8° ∙ 0.537 = 5.26°
5) What is the magnitude of the tangential acceleration as the pendulum passes through the
equilibrium position? _______ m/s2
Equilibrium is the position at which the net force on the pendulum bob is zero. In this case this
means 𝐹𝑇 = 0
Hence, by Newton’s 2nd Law: the acceleration at equilibrium is
𝑎𝑇 =
𝐹𝑇
=0
𝑚
6) What is the magnitude of the radial acceleration as the pendulum passes through the
equilibrium position? _______ m/s2
𝑣2
𝑟
Here 𝑟 = 𝐿, and, at the equilibrium position: 𝑣 = 𝑣𝑚𝑚𝑚
𝑎𝑟 =
𝑎𝑟 =
𝑣𝑚𝑚𝑚 2 (0.859 m⁄s)2
=
= 0.287 m⁄s2
𝐿
2.57m
7) Which of the following would change the frequency of oscillation of this simple pendulum?
(1) increasing the mass
(2) decreasing the initial angular displacement
(3) increasing the length
(4) hanging the pendulum in an elevator accelerating downward
𝜔
1 𝑔
�
𝑓=
=
2𝜋 2𝜋 𝐿
This does NOT depend on mass, or the initial displacement
But it does depend on the length and the gravitational acceleration—which means it will change
in an accelerating elevator
So correct choices are (3) and (4)
Problem 6
A rigid rod of length L= 1 m and mass M = 2.5 kg is attached to a pivot
mounted d = 0.17 m from one end. The rod can rotate in the vertical
plane, and is influenced by gravity. What is the period for small
oscillations of the pendulum shown? T = ? seconds
Solution: We use the rotational equation of motion, where the angular
acceleration of the rod is given by
𝑑2𝜃
𝜏
≡𝛼=
2
𝑑𝑡
𝐼
Where 𝐼 is the moment of inertia of the rod about the pivot point P, given
by (using Parallel Axes Theorem – P.A.T.)
𝐼 = 𝐼𝑃 = 𝐼𝐶𝐶 + 𝑀𝐷2
Where 𝐷 is the distance between the pivot and the center-of-mass, given by
𝐷=
𝐿
− 𝑑 = 0.50m − 0.17m = 0.33m
2
And the moment of inertia of the rod about its own center-of-mass is given by
𝐼𝐶𝐶 =
1
𝑀𝐿2
12
The torque exerted by gravity about P is given by (𝑟 = 𝐷 is the distance from the pivot to where
the force of gravity acts: at the center-of-mass)
𝜏 = 𝑟𝑟 sin 𝜙𝑟𝑟 = 𝐷 ∙ 𝑀𝑀 ∙ sin(−𝜃) ≈ −𝑀𝑀𝑀𝑀
So we have
𝑑2𝜃 𝜏
𝑀𝑀𝑀
= ≈−
𝜃 = −𝜔2 𝜃
2
1
𝑑𝑡
𝐼
2
2
12 𝑀𝐿 + 𝑀𝐷
This then gives the angular frequency of the oscillations:
𝜔=�
𝑀𝑔𝑔
1
2
2
12 𝑀𝐿 + 𝑀𝐷
=�
𝑔𝑔
1 2
2
12 𝐿 + 𝐷
=�
9.81 m⁄s 2 ∙ 0.33m
= 4.10s −1
(1.0m)2 ⁄12 + (0.33m)2
And the period is related to the angular velocity/frequency by (period is the time it takes to do
one cycle: i.e. 𝜔𝜔 = 2𝜋
𝑇=
2𝜋
2𝜋
=
= 1.53 s
𝜔
4.10s −1
Problem 7
A circular hoop of radius 40 cm is hung on a narrow horizontal
hoop and allowed to swing in the plane of the hoop. What is the
period of its oscillation, assuming that the amplitude is small?
_______ s
Solution: We use the rotational equation of motion, where the
angular acceleration of the hoop is given by
𝜏
𝑑2𝜃
≡𝛼=
2
𝐼
𝑑𝑡
Where 𝐼 is the moment of inertia of the hoop about the pivot point
P, given by (using Parallel Axes Theorem – P.A.T.)
𝐼 = 𝐼𝑃 = 𝐼𝐶𝐶 + 𝑀𝐷2
Where 𝐷 is the distance between the pivot and the center-of-mass,
given here by
𝐷 = 𝑅 = 0.40m
And the moment of inertia of the hoop about its own center-of-mass is given by
𝐼𝐶𝐶 = 𝑀𝑅 2
The torque exerted by gravity about P is given by (𝑟 = 𝑅 is the distance from the pivot to where
the force of gravity acts: at the center-of-mass)
So we have
𝜏 = 𝑟𝑟 sin 𝜙𝑟𝑟 = 𝑟 ∙ 𝑀𝑀 ∙ sin(−𝜃) ≈ −𝑀𝑀𝑀𝑀
𝑑2𝜃 𝜏
𝑀𝑀𝑀
= ≈−
𝜃 = −𝜔2 𝜃
2
𝐼
𝑀𝑅 2 + 𝑀𝑅 2
𝑑𝑡
This then gives the angular frequency of the oscillations:
𝜔=�
𝑀𝑀𝑅
𝑔
9.81 m⁄s 2
�
�
=
=
= 3.50s−1
2𝑀𝑅 2
2𝑅
2 ∙ 0.40m
And the period is related to the angular velocity/frequency by (period is the time it takes to do
one cycle: i.e. 𝜔𝜔 = 2𝜋
𝑇=
2𝜋
2𝜋
=
= 1.79 s
𝜔
3.50s −1