Exercise of mechanical vibration (1-dof vibration ...ย ยท (2) Find the natural angular frequency of...
Transcript of Exercise of mechanical vibration (1-dof vibration ...ย ยท (2) Find the natural angular frequency of...
Q1: Consider a balancer depicted below. Answer its equation of motion and natural angular frequency. The mass of the stick can be ignored and a small angle assumption is available: sin ๐ โ ๐.
๐ ๐
Exercise of mechanical vibration (1-dof vibration (rotation)
Basic
Q2: A stick with a mass m is supported by a frictionless pin, around which the stick rotates. An end of the stick is connected to the wall by a spring. When ๐ 0, its length is natural. Gravity force acts on the mass. Answer the equation of motion of the system about ฮธ and the period of free vibration.
m
๐๐
๐k
Basic
Q3: A pulley, which rotates around the center of mass, with a string hangs mass m. The mass moment of inertia of the pulley is denoted by I. There is no slippage between the pulley and string. The string is supported by a spring of coefficient k. Answer the period of free vibration of the system.
m
R
k
g
Basic
๐oZY
X
ab
Q4: Consider a homogeneous stick of which volume can be negligible. Its lengths and mass are l and m, respectively. Answer its mass moment of inertia about the center O.
Q5: Consider a rectangle plate made of a homogeneous material. Its thickness is small enough to be neglected. The mass and dimensions are m and 2a by 2b, respectively. X and Y axes are on the surface and along with its edges. Z axis is normal to the plate. Answer the mass moments of inertia about each of the three axes.
Basic
Q6: A slender rod of mass m and length l, moves around a circular trackunder the influence of gravity. The small rollers at the ends of therod are confined to the circular track and roll without friction.Neglect the mass of the rollers. The radius of the track is r and ๐3๐. Find an equation of motion of the rod as a function of thecoordinate, ฮธ, as shown in the figure.
๐
๐ 3๐๐๐2Basic
Q7: A cuckoo clock pendulum consists of two pieces glued together:ใป a slender rod of mass m1 and length l, andใป a circular disk of mass m2 and radius r, centered at the slenderrodโs midpoint.
The pendulum is attached at one end to a fixed pivot (center ofrotation), O. , as shown below. The gravity acts.a) Find the mass moment of inertia about O.b) Find the pendulumโs angular momentum about O.c) Find the equation of motion of the pendulum about ฮธ.
Basic
ฮธ
O
rl
m1
m2
Basic
Q8: Consider a pendulum made by a homogeneous plate. Its dimensionsare, a, b, and h as shown in the figure. The plate rotates around Ounder the gravity force, and smallโangleโassumption applies. Thematerial density is ฯ = m / (abh) as the mass and thickness of theplate are m and h, respectively. ฮธ is 0 at the equilibrium position ofthe system.
(1) Obtain the plateโs mass moment of inertia about O.(2) Obtain the equation of motion in terms of ฮธ.(3) Find the natural angular frequency of the pendulum.(4) Solve the rotational behavior of the object ฮธ(t) with the initial
conditions ๐ 0 ๐ and ๐ 0 0.(5) Solve the rotational behavior of the object with the initial
conditions ๐ 0 0 and ๐ 0 ๐ .
Basic
a ฮธ
O
b/2b/2
Basic
Q9: Consider a homogenous plate of skewed Tโshape as is shown in thefigure. It is composed of two parts M1 and M2, each of which hasthe same dimensions and mass m.
(1) Obtain the mass moment of inertia about an axis passingthrough O.
(2) Obtain the centroid (center of gravity) of the plate using x and y.
Oa
bb/2
M1
M2
X
Y
Basic
Q10: Consider a cylinder that rolls without slipping. Let x = 0 denote therest position of the cylinder. The mass of the cylinder is m. Neglectthe mass of the spring. The mass moment of inertia of the cylinderis I.
Rk
ฯ
x
ฮธ
Basic
(1) Obtain the equation of motion in terms of x.
(2) Determine the natural angular frequency of the cylinder.
(3) Let ฮด is the static deflection of the spring. Answer the kinetic and potential energy of the system.
(4) Suppose ๐ฅ ๐ดcos ๐๐ก , answer the maximum kinetic and potential energy of the cylinder as well as the minimum potential energy.
(5) Solve ๐ ๐ ๐ for p to obtain the natural angular frequency of the cylinder.
Basic
Q11: As shown in the figure, one end of the spring of constant k is fixed to the wall. The other end hangs the mass of m through the a pulley. The radius and mass of the pulley are a and M, respectively. The pulley rotates smoothly around its fixed central axis and the mass vibrates along the vertical direction. The string does not slip on the pulley. x and ฮธ are the displacement of the mass and the rotation angle of the pulley from their statistically equivalent position.
1) What is the moment of inertia of the pulley about its rotation axis?
2) When the displacement of the mass is x, answer tension Tusing x and k. Furthermore, answer the relation between x and ฮธ.
Basic
x
MT
Tโ
TโDisplacement from the statically equivalent position
ฮธ
3) Answer the equation of motion about x. Use m, M, k, and a, and not T and Tโ.
4) Answer kinetic energy of K in this system.5) Answer potential energy of U in this system.6) By Lagrangeโs method, answer the motion equation.
* Lagrangeโs method is not the scope of mini quiz.
Basic
Q12: Consider a disk with a handle. The disk rotates around its center, which is fixed to the ground. The length of the handle is l, and its width is negligibly small. Two ends of the disk are linked to the wall by springs of coefficient k. Answer the following problems.1) Mass of the disk and handle is m, for each. Answer the mass
moment of inertia of the disk with a handle around the rotation axis.
2) Establish the equation of motion about ฮธ.3) Answer the natural frequency of the disk.
ฮธ
Radius r
Length l
Basic
Q13: Consider a thin circular disk of radius a and mass m. Answer the mass moment of inertia around axis Y.
X
Y d๐dr
rdrd๐a
Q14: Consider a cylinder of radius a and height l. Answer the mass moment of inertia around axis Y.
Y
Z
a
Basic
Q15: A simplified model of a rackโandโpinion steering system is shown inthe figure. A gear of radius r and mass moment of inertia I isattached to a shaft of torsional stiffness kt, for which the restoringtorque is given by ๐ ๐ ฮ๐. The gear rolls without slip on therack of mass m. The rack is attached to a spring of coefficient k. x isthe horizontal displacement of the rack from the systemโsequilibrium position.
k
x
Rack (m)
Pinion (I)
ฮธ
Basic
(1) Derive the differential equation governing the motion of thesystem using x.
(2) Looking at the rackโandโpinion system as a 1โDOF springโmasssystem, the above equation is rewritten as ๐ ๐ฅ ๐ ๐ฅ 0.Answer the equivalent mass me and spring constant ke.
(3) Answer the kinetic energy of the system using x.(4) Answer the potential energy of the system using x.(5) Suppose ๐ฅ ๐ดcos ๐๐ก , solve p based on the energy
conservation law.
Basic
mm
l
a a
R1: Consider a balancer that has two weights at each side and rotates around O. The weight of the stick is ignorable. Find the equation of motion of a balancer (ใใใในใ) and its natural angular frequency p.
Given that ฮธโช 1, sinฮธ โ ฮธ and cosฮธ โ 1 โ ฮธ2/2
O
Intermediate
R2: When the arm of the balancer is circular, answer l such that the natural frequency becomes 1 Hz.
mm l
l
l
Intermediate
R3: Consider the following system which consists of a disk of mass m andspring. The spring is fixed to the disk at its center of gravity, around whichthe disk rotates. The motion of the disk is constrained along xโaxis. Theslippage between the disk and ground does not occur. Mass moment ofthe disk is I.
rk
x
(1) Answer the equation of motion about x.
(2) Answer the kinetic energy of translation (Tt) and rotation (Tr) using ๐ฅ and ๐, respectively.
(3) Answer the potential energy of the spring (U) using x.
(4) When ๐ฅ =๐ด cos ๐ ๐ก, find the natural frequency of the system (๐ ) based on the relationship of T ๏ผ๐ .
m
Intermediate
R4: ๅๅพRใฎๅ็ญๅ ใง่ปขใใใชใใๅพฎๅฐๆฏๅใใใ่ณช้m๏ผๅๅพr ใฎๅๆฑใ่ใใ๏ผๅ็ญใฏ๏ผๆปใใชใใใฎใจใใ๏ผ(1)โ(4)ใฎๅใใซ็ญใใ๏ผๅ็ญใซใฏ๏ผฮฆใฏ็จใใฆใฏใชใใชใ๏ผ
(1) ไธฆ้ฒใฎ้ๅใจใใซใฎใผ๐ ใๆฑใใ๏ผ
(2) ๅ็ญใฎๆ ฃๆงใขใผใกใณใใ I ใจใใใจใ๏ผๅ็ญใฎๅ่ปขใฎ้ๅใจใใซใฎใผ๐ ใๆฑใใ๏ผ
(3) ไฝ็ฝฎใจใใซใฎใผU ใๆฑใใ๏ผฮธใๅๅ
ใซๅฐใใใจใ๏ผcos ๐ 1 ใฎ่ฟไผผใ
็จใใใใจ๏ผ
(4) ๐=๐ด cos ๐๐ก ใจใใฆ๏ผๅ็ญใฎๆๅคง่งๅบฆ๐ ๐ด ใงใใ๏ผๆๅคง่ง้ๅบฆ ๐๐ด๐ ใจใชใ๏ผใใฎใจใ๏ผ๐ ๐ ใ ๐ใซใคใใฆ่งฃใ๏ผใชใ๏ผ๐ ใฏใใฎ็ณปใฎๅบๆ่งๅจๆณขๆฐใงใใ๏ผ
๐R ๐ r
ๅบ้ก: ใใญในใใฎๆผ็ฟๅ้ก 2.8aIntermediate
R5๏ผ Figure (1) shows a simplified seesaw. The main plate is of length 2land mass M. Its width and thickness are small enough to be neglected. The supporting pivot O is at d from the plate. The gravity acceleration is g. Using the small inclination of the seesaw ๐, establish the equation of motion. Also, find the natural frequency.Furthermore, as is shown in Fig. (2), when a mass point is at
each end, answer the equation of motion and the natural frequency.
๐d
ll
(1) ๐d
ll
(2)
m
mM M
Intermediate
R6: The inverted pendulum is connected by two springs as shown in the figure. Assume small angles of vibration and neglect the rod mass.
a) Derive the equation of motion of the system about ๐.
b) What is the relation among ๐ , ๐ , ๐, and ๐ for the characteristic roots to include at least one positive real value?
c) Describe the behavior of the system when the characteristic roots include positive real values.
๐๐ ๐๐๐ ๐
๐
๐๐
Intermediate
R7: A Tโshaped plate is made from two solid plates M1 and M2 ofuniform density. Their dimensions are shown in the figure. M1 andM2 have mass 2m and 2/3m, respectively. G1 and G2 are thecenters of mass of each plate.a) Find the location of the center of mass of the combined plate,
G12, which lies somewhere on the vertical line. Express it as adistance s measured from the top of the object.
b) Find the mass moment of inertia of the plate about its gravitycenter, G12.
M1
M2
a
b G1
G2
sG12
b
bIntermediate
R8: A grandpa accidentally dropped a rice ball (weight m, radius r), andit rolled on a curved surface of which radius was R. The rice ball did not slip on the surface. The angle of the rice ball is denoted by ฮธ.
(1) Assume that the rice ball is a homogeneous sphere. Answer the moment of inertia of the rice ball around its centroid.
(2) Answer the equation of motion of the rice ball about ฮธ.
(3) Using a small angle assumption (sinฮธ ~ ฮธ), determine the natural frequency of the ball.
R
mg
ฮธ
hr
Riceball
Intermediate
R9: The mass moment of inertia of a tire can be investigated by rolling it on a slope, i.e., release it from rest at a certain height. Answer what quantities and which formula should be used for calculating the mass moment of inertia. Note that the density of the tire is not disclosed.
ฮธg
Intermediate
R10: A pendulum of length l and mass m is fixed to a pin on a rodrotating at a constant angular speed ๐. The pendulum rotatesaround the pin and its rotation angle is denoted by ๐. ๐ is smallenough ๐ 0 when the pendulum aligns with the gravitationalacceleration.(1) Find the equation of
motion about ๐.(2) Find the natural frequency
of the pendulum when ๐ is small enough and the smallโangle approximation of trigonometric functions is available.
๐๐
๐Frictionlesspin
m๐
Intermediate
R11: We calculate the natural frequency of a rocking chair. The base ofthis rocking chair is a circular arc, and its center of mass (m) moveson the xโy plane as the chair rocks back and forth. The radius of thearc is R. The position of the center of mass is L from the bottom.The origin of the coordinate system lies at the contact pointbetween the arc and floor when ๐ 0.(1) Express the coordinate (x and y) of the center of mass as a
function of ๐.(2) Find the kinetic energy of the center of mass.(3) From the principle of energy conservation, find
the natural frequency of the chair when ฮธ issmall. Use g as the gravitational acceleration.
ใญใใญใณใฐใใงใขใฎใใใช่ปขใใๆฏๅญใ๏ผ่ปขๅๆฏๅญ๏ผ่ปขใใๆฏๅญ๏ผใจๅผใถ๏ผIntermediate
Rm
L
m
x
y
(x, y)๐
O O
Intermediate
R12: Consider a physical pendulum made of a homogeneous rod of which length and mass are l and m, respectively. The rod rotates about the frictionless pivot. The gravitational acceleration is g. The rotational angle of the rod is ๐, for which ๐ 0 at its equilibrium position.
(1) Find the mass moment of inertia of the rod about the pivot.(2) Find the natural angular frequency of the pendulum, when ๐ is
small enough that small angle assumptions are available.(3) Find the position of pivot d such that the natural angular
frequency becomes maximum. d is the distance between the pivot and the centroid of the rod.
Intermediate
๐3
๐๐2๐3 ๐2
๐ Distance between the pivot and the mass center of the rod.
Intermediate
R13: Inertial mass damper. The mechanism in the figure realizes an effective mass significantly greater than the actual mass. When the displacement arises at the tip of the ball screw, it revolves the flywheel, of which radius is r, in the chamber. When the mass and mass moment of inertia of the flywheel are m and 1/2 ๐๐ . Find the effective mass.
[Sugimoto et al., J. Struct. Constr. Eng., 2018]
rTorque: ฯ
Force: fDisplacement: x
Flywheel: I
Intermediate
(1) Ld is the displacement of the screw when it revolves 2๐ rad. Find the angular velocity ๐ of the flywheel when the velocity of the ball screw is ๐ฅ.
(2) Ld also determines the ratio of force f and torque ๐: ๐ ๐. Find the relationship between f and ๐.
(3) Find the effective (apparent) mass by using following values: ๐ฟ 0.02 m, ๐ 0.3 m, and ๐ 1.0 kg.
Intermediate
S1: A rod of mass M and length 2a revolves around its frictionless center O. Thickness of the rod is small enough to be neglected. A weight m is attached at b from the center. The gravity force acts on the mass and rod, and they behave as a pendulum. Answer the following problems.(1) Answer the mass moment of
inertia of the pendulum around O.
(2) Answer the centroid of the pendulum from O.
(3) Using a small angle assumption (sinฮธ ~ ฮธ), determine the natural frequency of the pendulum.
(4) Determine b, such that the pendulum rotates most quickly.
O
a
am
b
M
g
ๅคงๅญฆ้ขๅ ฅ่ฉฆใ่งฃใใฆใฟใใ! ๅๅคๅฑๅคงๅญฆ 2018ๅนดๅคๅฝไบบ็ๅญฆ็่ฉฆ้จ
S2: Consider a cube made by solid bars of which lengths are l and masses are negligible. The mass points of m or 2m are distributed on the corners as in the figure. The origin O is set on one of the corners. Answer the following problems.(1) Find the center of mass of the cube.(2) Find the mass moment of inertia of the cube with respect to
the Xโaxis.(3) Find the mass moment of inertia about an axis passing through
the center of mass and parallel to the Xโaxis.(4) When the gravitational acceleration g acts on the masses in the
negative Z direction, answer the moment vector about O.
ๅคงๅญฆ้ขๅ ฅ่ฉฆใ่งฃใใฆใฟใใ! ๅๅคๅฑๅคงๅญฆ 2018ๅนดๅคๅฝไบบ็ๅญฆ็่ฉฆ้จ
X
Y
Z
2m
2m
2mm
m
l
l
l
O
ๅคงๅญฆ้ขๅ ฅ่ฉฆใ่งฃใใฆใฟใใ! ๅๅคๅฑๅคงๅญฆ 2018ๅนดๅคๅฝไบบ็ๅญฆ็่ฉฆ้จ
S3: Consider a mass particle and a stepped cylinder wound by masslesscables. Their motions are linear along the Zโaxis. A gravitationalforce is acting in the downward direction, and the acceleration ofgravity is g. The cylinder has mass m1 and the moment of inertia Iwith respect to the rotational axis located at the mass center of thecylinder. The upper cable is fixed on the ceiling, and its positivetension is denoted by T1. r1 and r2 denote the radii of the cylinder.z1 and z2 are the positions from the ceiling, and ๐ denotes therotational angle of the cylinder. The mass particle of the mass m2 isattached to the lower cable, and its positive tension is denoted byT2. Answer the following questions.(1) Find the equation of translational motion of the mass particle.
You may use T2 in the answer.(2) Find the equations of translational and rotational motions of
the cylinder. You may use T1 and T2 in the answer.
ๅคงๅญฆ้ขๅ ฅ่ฉฆใ่งฃใใฆใฟใใ! ๅๅคๅฑๅคงๅญฆ 2018ๅนดๅคๅฝไบบ็ๅญฆ็่ฉฆ้จ
๐
๐๐๐
๐๐
๐
๐ ๐๐ง
๐ง
(3) Represent the velocity ๐ง using the angular velocity ๐.(4) The angular acceleration ๐ can be given by ๐ ๐ผ๐, where ๐ผ is
a constant. Find ๐ผ, not using T1 and T2.
ๅคงๅญฆ้ขๅ ฅ่ฉฆใ่งฃใใฆใฟใใ! ๅๅคๅฑๅคงๅญฆ 2018ๅนดๅคๅฝไบบ็ๅญฆ็่ฉฆ้จ
S4: Two masses are supported by four rigid links. This system is symmetric about zโaxis and revolves around zโaxis at constant angular velocity ๐. Points O and B are connected by a spring of coefficient k. The springโs natural length is 2๐ฟ. The position of B is limited to 0 ๐ง 2๐ฟ. The gravitational acceleration is g. The angle between link OA is denoted by ๐.(1) Find the equation of force at B using ๐ which is the tension of
link AB.(2) Find the equation of motion of the mass about x and z using ๐
and ๐ which is the tension of link OA.(3) Find the equation of motion of the mass about ๐ by removing ๐ and ๐ from the equations found in (1) and (2).(4) The generalized coordinate of the system is ๐. Find the kinetic
energy of the system.(5) The generalized coordinate of the system is ๐. Find the
potential energy of the system.
ๅคงๅญฆ้ขๅ ฅ่ฉฆๅ้กใ่งฃใใฆใฟใใ! H31ๅนดๅบฆๅคง้ชๅคงๅญฆโไธ้จ
S4: Cont.
ๅคงๅญฆ้ขๅ ฅ่ฉฆๅ้กใ่งฃใใฆใฟใใ! H31ๅนดๅบฆๅคง้ชๅคงๅญฆโไธ้จ
A: Translate the following paragraph on pulley dynamics into Japanese.
The figure provides an example of pulley dynamics. The pulleyโscenter is fixed, and forces f1 and f2 are the tensions in the cord oneither side of the pulley. From law of rotational motion,
The tension forces are equal if ๐ผ๐ is negligible.This condition is satisfied if either the pulleyrotates at a constant speed or the pulleyinertia is negligible compared to the otherinertias in the system. The pulley inertia willbe negligible if either its radius or its mass issmall. The force on the support at the pulleycenter is f3 = f1 + f2. If ๐ผ๐ is negligible, then thesupport force is 2f1.
๐ผ๐ ๐ ๐ ๐ ๐ .R
ฮธ
f1 f2
f3
COLUMN: Monkey climb puzzle from Lewis Carrol
ไธๆ่ญฐใฎๅฝใฎใขใชในใง็ฅใใใ Lewis Carrol ใฎๆๅใชใใบใซใใใ๏ผ
ๆฉๆฆใฎ็กใๆป่ปใซ่ณช้ใ็ก่ฆใงใใใญใผใใๆใใใใฆใใ๏ผใใฎไธก็ซฏใซๅใ่ณช้ใฎใใใใจใตใซใใถใไธใใฃใฆใใ๏ผใตใซใฏ๏ผใญใผใใใคใใฃใฆไธใซ็ปใใใจใใ๏ผใใไบบใฏ๏ผใตใซใฏใญใผใใ็ปใใใจใใงใใใซๅใๅ ดๆใซ็ใพใ๏ผใใใใฏไธใซ่ฝใกใฆใใพใใจ่จใ๏ผๅฅใฎไบบ
ใฏ๏ผใตใซใฏใญใผใใ็ปใใใจใใงใใใจ่จใ๏ผใใฆ๏ผใตใซใฏใญใผใใ็ปใใใจใใงใใใ ใใใ๏ผใใฎ็็ฑใฏ๏ผ
mmๆผ็ฟๅ้กใใใณ่งฃ็ญใซใฏใในใใใใพใ๏ผใในใๅใใซ่ฆใคใใฆๆใใฆใใใไบบใซใฏ๏ผๆ็ตๆ็ธพใซๅ ็นใใพใ๏ผ1ๅ็ฎใชใ2็น๏ผ2ๅ็ฎไปฅ้ใฏ1็นใใคใๅ ็นใใพใ๏ผ
Solution (Basic)Q1: ๐๐ ๐ ๐๐ ๐ 0
๐ ๐๐ ๐๐๐ ๐๐๐ ๐ ๐๐๐ ๐ 0๐ ๐๐๐ ๐ ๐๐๐
Q2:
Q3: ๐ผ ๐๐ ๐ ๐ ๐๐ 0๐ ๐ ๐๐ผ ๐๐
๐ 2๐๐๐ 2๐๐
๐ผ ๐๐12Q4: Q5: ๐ผ 13 ๐๐๐ผ 13 ๐๐๐ผ 13 ๐ ๐ ๐ ๐ผ ๐ผ
๐๐ ๐sin๐ 0Q6:
๐ผ 13 ๐ ๐ ๐ ๐2 ๐4Q7:
a)
๐ผ ๐b)
13 ๐ ๐ ๐ ๐2 ๐4 ๐ ๐ ๐ ๐ ๐2 sin๐ 0c)
Q8:(1)
22412
bamIO
(2)
(3)
(4)
(5)
Q9:(1)
a
b
O X
Y
COG2
COG1r1
r2
a/4
(3a+b)/4
S
(2)ๅ่ปขไธญๅฟOใใไธ่จใฎไฝ็ฝฎS
Q10:R3ใจๅใ้ๅๆน็จๅผใซใชใใพใ๏ผ
๐ ๐ก ๐ cos ๐๐ก๐ ๐ก ๐๐ sin ๐๐ก
๐ผ ๐12 20๐ 12๐๐ 5๐๐ ๐๐๐2๐ผ ๐ 0๐ ๐๐๐2๐ผ 6๐๐4๐ ๐
Q11:
I 12
Ma2 T kxx a
1) 2)
021
kxxmM 3)
U 12
kx2
4)
5)
6)
Q12:1 ๐ผ ๐ผ ๐ผ๐ผ 12 ๐๐ ๐ผ 13 ๐๐3 ๐ 2๐ ๐ ๐ผ
2 ๐ 2๐ ๐ ๐ผ ๐ 0
Q13: 14 ๐๐ 14 ๐๐ 112 ๐๐Q14:
Q15:
๐ผ๐ ๐ ๐ฅ ๐ ๐๐ ๐ฅ 0(1)
(3) ๐ 12 ๐ ๐ผ๐ ๐ฅ๐ 12 ๐ ๐๐ ๐ฅ(4)
(2) ๐ ๐ผ๐ ๐ ๐ ๐ ๐๐
๐ ๐ ๐๐๐ผ ๐๐(5)
R1: ๐ ๐๐๐ ๐ ๐ 0๐ ๐๐๐ ๐
R2: ๐ ๐8๐
(1)๐ผ 12 ๐๐๐ ๐ผ๐ ๐ฅ ๐๐ฅ 0(3)๐ 12 ๐๐ฅ
R3:
(2)๐ 12 ๐๐ฅ ,๐ 12 ๐ผ๐ 12 ๐ผ ๐ฅ๐(4)๐ ๐๐ ๐ผ๐
R4:
(1) ๐ 12 ๐ ๐ ๐ ๐ ่ณช้ไธญๅฟใฎไฝต้ฒ้ๅบฆใฏ๏ผ ๐ ๐ ๐ใงใใ
(2) ๐ 12 ๐ผ ๐ ๐๐ ๐ ๐ใฏๆฅ่งฆ็นใใใฎ่งๅบฆใงใใใฎใง๏ผๅ็ค่ชไฝใฎๅ่ปข่งๅบฆใฏ๏ผ๐ ๐ใงใใ๏ผ๐ ๐ ๐๐ ใง
ใใใใ๏ผๅ็คใฎ่ง้ๅบฆ ๐ ๐ 1 ๐ใจใชใ๏ผ
(3) ๐ ๐๐ ๐ ๐ 1 cos ๐โผ ๐๐ ๐ ๐ ๐2(4) ๐ ๐ ๐ ๐ 12 ๐ ๐ ๐ ๐ ๐ผ๐ ๐ ๐ด๐
๐ ๐ ๐ด๐2 ๐ ๐ ๐ ๐ผ๐
๐ ๐ โผ ๐๐ ๐ ๐ ๐ด2 ๐ ๐ด๐ ๐ ๐ ๐ ใใ๏ผ
๐ ๐๐๐๐ ๐ ๐๐ ๐ผ
(1)ๆฃใฎไธญๅฟๅจใใฎๆ ฃๆงใขใผใกใณใใฏ
ใงใใ๏ผpivotๅจใใงใฏ ๐๐๐ ๐๐๐3 ๐ ๏ผ
(2)ๆฃใฎไธญๅฟๅจใใฎๆ ฃๆงใขใผใกใณใใฏ2๐๐ ใงใใ๏ผpivotๅจใใงใฏ2๐๐ ๐ 2๐ ๐
R5: R6: ๐๐ ๐ 2๐๐ ๐๐๐ ๐ 0a) 2๐๐ ๐๐๐ 0b)
c) The pendulum falls down. ใชใใชใใฐ๏ผ๐ ๐ดexp ๐๐กใงใใใจใใฆ๏ผ๐ใๆญฃใฎๅฎ้จใๆใใใชใ๏ผๆ้ใจใจใใซ๏ผฮธใๅคงใใใชใ็ถใใใใ๏ผ
๐ ๐ 2๐ ๐๐๐ ๐3 ๐ 2๐(d2+l2)
๐ 34 ๐R7:
a) ๐ผ ๐ผ ๐ ๐ ๐ผ ๐ ๐b) ๐ 14 ๐ ๐ 34 ๐๐ผ 2๐12 ๐ ๐๐ผ ๐๐9d1 and d2 are the distances from G12 to G1and G2, respectively.
๐ผ ๐ ๐6 79 ๐
(1)
(2)
2
52 mrI
(3)
R8:
ๅๅพrใฎ็ใฎๆ ฃๆงใขใผใกใณใใจใใฆ
๐๐ ๐ ๐๐sin๐ ๐น๐ผ ๐ ๐ ๐๐น๐ ๐ ๐ ๐๐๐ 5๐7๐ ๐ 0
๐ 5๐7๐ ฮธ
h
h๐๐ใ็ใฎๅ่ปข่งๅบฆ๏ผๅฐ้ขใจใฎๆฅ็นใใใฎ๏ผใจใใใจ๏ผ๐ ๐ ๐ ๐๐ใจใชใใใจใซๆณจๆ๏ผQ7ใๅ่ใซใใฆไธใใ๏ผ
๐
R9: We should measure the following parameters:m: mass of the tirer: radius of the tirex: position of the tire on the slope
and use the formula:
ฮธ
x
๐ ๐๐sin๐ ๐ ๐๐๐ฅ ๐ผ ๐ฅ๐
๐ผ ๐ ๐๐sin๐๐ฅ ๐fx: Force along xโaxisft: Force for translationfr: Force for rotation
R9: From the preservation of energy, we may use the following relationships. In this case, we need to measure the velocity of the tire after the slope and the height of the slope.
ฮธ
x
๐ ๐๐๐Initial energy
๐ 0
l
After the slope๐ 0๐ 12 ๐๐ฅ 12 ๐ผ๐12 ๐ ๐ผ๐ ๐ฅ๐ ๐
R10:
๐๐
m
๐๐๐๐ ๐sin๐Centrifugalforce
(1)
๐๐ ๐ ๐๐๐sin๐ ๐๐ ๐sin๐๐cos๐(2)
๐๐ ๐ ๐๐๐๐ ๐๐ ๐ ๐
้ ๅฟๅใฏ๏ผๅธธใซฮธใฎๆญฃใฎๆนๅใซๅใใใจใซๆณจๆใใฆ
ๅ่ปข่งๅบฆใฎ่ฟไผผใ็จใใใจ๏ผ(1)ใฎๅผใฏไธ่จใฎ้ใใซใชใ๏ผ
๐๐ ๐ ๐ ๐๐ 0๐ ๐ ๐๐๐
R11:(1)
m๐
O
๐ฅ ๐ ๐ ๐ ๐ฟ sin๐๐ ๐ ๐ ๐ฟ sin๐๐ฆ ๐ ๐ ๐ฟ cos๐
๐ ๐ฟ cos๐(2) ๐ฅ ๐ ๐ ๐ ๐ฟ ๐cos๐๐ฆ ๐ ๐ฟ ๐sin๐๐ 12 ๐๐ฃ 12 ๐ ๐ฅ ๐ฆ12 ๐๐ ๐ 2๐ ๐ ๐ฟ cos๐ ๐ ๐ฟ
~ 12 ๐๐ฟ ๐ cos๐ 1 ใฎใจใ
(3)ๅทฆ่พบ: ้ๅใจใใซใฎใผใๆๅคงใซใชใ๏ผไฝ็ฝฎใจใใซใฎใผใๆๅฐใซใช
ใๅงฟๅข ๐ 0ๅณ่พบ: ้ๅใจใใซใฎใผใๆๅฐ๏ผ0๏ผใซใชใ๏ผไฝ็ฝฎใจใใซใฎใผใๆๅคง
ใซใชใๅงฟๅข ๐ ๐12 ๐๐ฟ ๐ ๐ฟ๐๐ 0 ๐ ๐ ๐ฟ cos๐ ๐๐๐ ใๅๅใซๅฐใใใฎใง๏ผ cos๐ ~1 ๐ ใจใใฆๆด็ใใใจ12 ๐๐ฟ ๐ ๐ ๐ฟ ๐๐๐2Rayleighโs method (๐ ๐sin๐๐ก) ใ็จใใฆๅบๆๅจๆณขๆฐใๅพใ
๐ ๐ ๐ฟ ๐๐ฟ
R12:(1) ๐ผ 19 ๐๐(2)
(3)
๐ 3๐2๐ ๐ผ ๐ 16 ๐๐๐๐ 0๐ ๐2 3
R13:(1)
(2)
(3)
Effective mass: kg
1 kg ใฎ่ณช้ใฎใใฉใคใใคใผใซใ๏ผ4 t ใใฎ่ณช้ใฎๅฝนๅฒใๆใใ๏ผ
S1:
(1) ๐ผ 13 ๐๐ ๐๐ (2) ๐๐๐ ๐(3) ๐๐๐13 ๐๐ ๐๐ (4) ๐ ๐ ๐3๐
S2:(1) ้ๅฟใฎไฝ็ฝฎใใฏใใซ rG, ่ณช็นiใฎไฝ็ฝฎใใฏใใซri
๐ ๐ ๐ ๐8๐๐ ๐๐ 100 ๐๐ 010 2๐๐ 101 +2๐๐ 111 ๐๐ 534 ๐ ๐8 534
(2) ๐ผ ๐๐ 2๐๐ 2๐ 2๐ 7๐๐(3) ๐ผ ๐ผ 8๐ 0, ๐ , ๐ ๐ผ 8๐ ๐64 3 4 ๐ผ 258 ๐๐
๐ผ ๐ผ 258 ๐๐ 56 258 ๐๐ 318 ๐๐
(4) ๐ 1 21 40 ๐๐๐ 350 ๐๐๐S3:
(1) ๐ ๐ง ๐ ๐ ๐(2) ๐ ๐ง ๐ ๐ ๐ ๐๐ผ๐ ๐ ๐ ๐ ๐(3) ๐ง ๐ง ๐ ๐๐ ๐ ๐ ๐๐ ๐ ๐
(4) (1)โ(3)ใงๅพใใใๅผใ้ฃ็ซใใใฆ๏ผ็ญๅ ้ๅบฆ้ๅใฎๅ ้ๅบฆใๅพใ๐ผ๐ ๐ ๐ ๐ ๐๐ ๐ ๐ ๐ ๐ ๐ ๐๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐๐ ๐ ๐๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐๐ผ ๐ ๐ ๐ ๐ ๐ ๐
S4:(1) 2๐๐ฟ 1 sin๐ 2๐ sin๐(2) ๐๐ฅ ๐๐ฟcos๐๐ ๐ cos๐ ๐ cos๐๐๐ง ๐๐ ๐ sin๐ ๐ sin๐(3) ๐ฅ ๐ฟcos๐ ๐ง ๐ฟsin๐๐๐ฟ ๐ ๐๐๐ฟcos๐ ๐๐ฟ ๐ cos๐sin๐ 2๐๐ฟ cos๐ 1 sin๐(4) ๐ ๐๐ฟ ๐ ๐ cos ๐(5) ๐ 12 ๐ 2๐ฟ 1 sin๐ 2๐๐๐ฟsin๐2๐๐ฟ 1 sin๐ 2๐๐๐ฟsin๐
1st comp: Inertia around point O2nd comp: Centrifugal force about zโaxis
A: Translate the following paragraph on pulley dynamics into Japanese.
ๅณใฏ๏ผใใผใชใฎๅๅๅญฆใฎไพใงใใ๏ผใใผใชใฎไธญๅฟใฏๅบๅฎใใใฆใใ๏ผf1 ใใใณ f2 ใฏใใใใ๏ผใใผใชใฎไธก็ซฏใฎ็ดใฎๅผตๅใงใใ๏ผๅ่ปข้ๅใฎๆณๅใใ๏ผ
๐ผ๐ ใ็ก่ฆใงใใใจใ๏ผ2ใคใฎๅผตๅใฏ็ญใใ๏ผใ
ใฎๆกไปถใฏ๏ผใใผใชใ็ญ้ใงๅ่ปขใใฆใใใจใใ๏ผใใผใชใฎๆ ฃๆงใ็ณปใฎไธญใฎไปใฎๆ ฃๆงใซๆฏในใฆ็ก่ฆใงใใใจใใซๆบใใใใ๏ผใใผใชใฎๅๅพใใใใฏ่ณช้ใๅฐใใใจใใฏ๏ผใใผใชใฎๆ ฃๆงใฏ็ก่ฆใงใใ๏ผใใผใชไธญๅฟใฎๆฏๆ้จใซๅ ใใๅใฏf3 = f1 + f2 ใจใชใ๏ผ๐ผ๐ ใ็ก่ฆใงใใใจใใฏ๏ผๆฏๆๅใฏ 2f1 ใจใชใ๏ผ
๐ผ๐ ๐ ๐ ๐ ๐ .R
ฮธ
f1 f2
f3
Q1:
ไธฆ้ฒใฎๆน็จๅผใใ๏ผ๐ ๐๐ sin ๐๐ ๐ ๐ ๐ sin ๐๐๐ก ๐๐ ๐๐ ๐ ๐ ๐๐๐ sin ๐ ๐๐ ๐๐๐ ๐ ๐๐ ๐ 0๐ ๐๐ ๐๐
ๅ่ปขใฎๆน็จๅผใใ๏ผ๐ผ ๐๐๐ ๐ผ๐ ๐๐ ๐๐๐ sin ๐ ๐๐ ๐๐๐ ๐ ๐๐ ๐ 0
Q2:๐ 0 ใฎใจใ๏ผใฐใญใฏ่ช็ถ้ทใซใชใ๏ผๆฃใฎ่ณช้ใฏ็ก่ฆใงใใ๏ผๅ่ปข้ๅฮธใซ้ขใใ้ๅๆน็จๅผใ็ซใฆ๏ผๅจๆTใๆฑใใ๏ผ
๐๐๐ ๐ ๐๐๐ ๐ ๐๐ ๐๐๐ ๐๐ ๐๐ ๐๐๐ ๐ ๐๐๐ ๐ 0๐ ๐๐๐ ๐ ๐๐๐
๐ 2๐๐
Q3:
m
R
k
g
่ณช้mใฎๅคไฝใxใใฒใใฎๅผตๅใTใจใใใจใใใคใใใใฎๅผใใใ๐๐ฅ ๐ๆป่ปๅจใใฎ้ๅๆน็จๅผใฏใ๐ผ๐ ๐น ๐๐ผ๐ ๐๐ ๐ ยท ๐ ๐๐ฅ๐ผ๐ ๐๐ ๐ ๐๐ ๐๐ผ ๐๐ ๐ ๐๐ ๐ 0ใใใใฃใฆๅบๆ่งๆฏๅๆฐใฏ๐ ๐๐ ๐ผ ๐๐
๐ 2๐๐ 2๐ ๐ผ ๐๐ ๐๐
I
๐ผ 2 ๐ฅ ๐๐ , ๐๐ ๐๐ ๐๐ฅ 2 ๐๐ ๐ฅ ๐๐ฅ ๐๐12
Q4: Q5:ZY
X
ab
ๅฏๅบฆฯใฏ ๐ ๐4๐๐ใใฎใจใใๅพฎๅฐ้ข็ฉใซใใใ่ณช้dmใฏd๐ ๐d๐ ๐d๐ฅd๐ฆ
ๅพฎๅฐ้ข็ฉd๐ d๐ฅd๐ฆ
x่ปธๅจใใฎๆ ฃๆงใขใผใกใณใ๐ผ ๐ฆ ๐d๐ฅd๐ฆ๐4๐๐ ๐๐ฅ ๐ฆ ๐๐ฆ13 ๐๐
y่ปธๅจใใฎๆ ฃๆงใขใผใกใณใ๐ผ ๐ฅ ๐d๐ฅd๐ฆ๐4๐๐ ๐ฅ ๐๐ฅ ๐๐ฆ13 ๐๐
z่ปธๅจใใฎๆ ฃๆงใขใผใกใณใ๐ผ ๐ฅ ๐ฆ ๐d๐ฅd๐ฆ๐4๐๐ ๐ฅ ๐๐ฅ ๐๐ฆ๐๐ฅ ๐ฆ ๐๐ฆ13 ๐ ๐ ๐ ๐ผ ๐ผ
Q8:
a ฮธ
O
b/2b/2๐ผ ๐ ๐3 ๐121)
๐ผ๐ ๐2 ๐๐sin๐2)
๐ ๐3 ๐12 ๐ ๐๐๐2 sin๐ 0
Q6: ๐ผ 112 ๐ 3๐ ๐ ๐4๐ ๐2๐ผ๐ ๐๐ ๐2 ๐๐๐ ๐๐ 0
3)
4)
๐ ๐๐๐2๐ ๐3 ๐12๐ 6๐๐4๐ ๐
๐ ๐3 ๐12 ๐ ๐๐๐2 sin๐ 0๐ 0 ๐ด๐ ๐๐ ๐๐ก ๐ ๐๐ 0 ๐ด๐๐๐๐ ๐๐ก ๐ 0
๐ 6๐๐4๐ ๐๐ 90ยฐ๐ด ๐๐ ๐ก ๐ cos 6๐๐4๐ ๐ ๐ก
๐ 0 ๐ด๐ ๐๐ ๐ ยท 0 ๐ 0๐ 0 ๐ด๐๐๐๐ ๐ ยท 0 ๐ ๐๐ 6๐๐4๐ ๐๐ 0ยฐ๐ด ๐๐๐ ๐ก ๐๐ sin 6๐๐4๐ ๐ ๐ก
5)
Q9:
a
b
O X
Y M1
COG2
da+b/2
a/2
a
(1)
IO IO1 IO2
From the last problem,๐ผ ๐12 4๐ ๐๐ผ ๐ผ ๐๐๐12 ๐ ๐ ๐ ๐2 ๐ ๐2๐ผ ๐12 20๐ 12๐๐ 5๐a
b
O X
Y
COG2
COG1r1
r2
a/4
(3a+b)/4
S
mSx mirx
mSy miry
m1 m2 S m1r1 m2r2
S Sx, Sy ,r1 0, a2
,r2 a
2, a b
2
X
Y
0 m1 a2
m2 Sx m1 m2
Sx a4
a2
m1 a b2
m2 Sy m1 m2
Sy 3a b
4
m1 m2 m
M2
XY
mi
COG
riS
(2) Definition of COG้ๅฟ
โ centroidโ center of gravity
mS miri
S, r1, r2 ใใใใใๅ็นOใใ๏ผๅๆฟใฎ้ๅฟ๏ผๆฟ1ใฎ้ๅฟ๏ผๆฟ2ใฎ้ๅฟใธใฎไฝ็ฝฎใใฏใใซใจใใ
1๐ผ 12 ๐๐ ๐ ๐ผ๐ ๐ฅ ๐๐ฅ 0(2) ๐(3)๐ 12 ๐๐ฟ ,๐ 12 ๐ผ๐ 12 ๐ผ ๐ฟ๐ ๐ 12 ๐๐ฟ
Q10:
Rk
ฯ
x
ฮธ
(4) ๐ฅ ๐ก ๐ด๐๐๐ ๐๐ก๐ฅ ๐ก ๐ด๐๐ ๐๐ ๐๐ก๐ 12 ๐๐ด ๐ ,๐ 12 ๐ผ ๐ด ๐๐ ๐ 12 ๐๐ด
๐ด ๐
๐ 0(5) 12 ๐๐ ๐ 12 ๐ผ๐ 12 ๐๐ 0
๐ ๐๐ ๐ผ๐
Q13: ๅใฟใฎ็ก่ฆใงใใๅ็คใฎYๅใใฎๆ ฃๆงใขใผใกใณใใๆฑใใใๅ็คใฎ่ณช้ใฏ m
X
Yd๐
drrdrd๐
a
ๆฅตๅบงๆจ ๐, ๐ ใฎๅพฎๅฐ่ณช้d๐d๐ ๐๐๐ ๐d๐d๐Y่ปธใใd๐ใพใงใฎ่ท้ข๐ cos ๐ๆฑใใY่ปธๅจใใฎๆ ฃๆงใขใผใกใณใ๐ผ ๐ cos ๐ d๐
๐ cos ๐ ๐๐๐ ๐d๐d๐ 14 ๐๐
Y
Z
a
Q14: ่งฃ่ชฌ
ๅบงๆจ ๐, ๐, ๐ง ใฎๅพฎๅฐ่ณช้d๐d๐ ๐๐๐ ๐ ๐d๐d๐d๐งY่ปธใใd๐ใพใงใฎ่ท้ข๐ cos ๐ ๐งๆฑใใY่ปธๅจใใฎๆ ฃๆงใขใผใกใณใ๐ผ ๐ cos ๐ ๐ง d๐
๐ cos ๐ ๐๐๐ ๐ ๐d๐d๐d๐ง 14 ๐๐ 112 ๐๐Z
Y๐ cos ๐ ๐ง
Q21ใฎๅๆฟใซใคใใฆz่ปธใๅ ใ๏ผๅๆงใซ่ใใ๏ผ
R1: Solution
๐sin๐
ๅนณ่ก่ปธใฎๅฎ็ใใ๐ผ 2๐๐ 2๐๐ 2๐ ๐ ๐๐ผ๐๐๐ ๐cos๐ ๐sin๐ ๐๐ ๐cos๐๐sin๐๐ โช 1, โด cos๐ 1, sin๐ ๐2๐ ๐ ๐๐๐ ๐ ๐๐ ๐๐ ๐ ๐๐โด ๐ ๐๐๐ ๐ ๐ 0๐ ๐๐๐ ๐
mm
l
a a
O
mg mg
R2:
Q4ใจๅใ่งๆฏๅๆฐใซใชใใใใ๐ ๐ใจใใฆใ๐ ๐2๐ๆฏๅๆฐ๐ ใงใใใฎใงใ
๐ ๐2๐2๐1 ๐2๐2๐๐ ๐8๐
R3: 1๐ผ 12 ๐๐๐ ๐ผ๐ ๐ฅ ๐๐ฅ 0(2)๐ 12 ๐๐ฅ ,๐ 12 ๐ผ๐ 12 ๐ผ ๐ฅ๐(3)๐ 12 ๐๐ฅ(4)๐ 12 ๐ ๐ผ๐ ๐ฅ 12 ๐ ๐ผ๐ ๐ด ๐ , ๐ 12 ๐๐ด๐ ๐ ใใใ ๐
๐ ๐๐ฅ๐ ๐ ๐๐ ๐๐ฅ๐๐ ๐ผ๐ใฐใญใฎๅพฉๅ ๅ f ใ๏ผไฝต้ฒใฎใใใฎๅ f1 ใจๅ่ปขใฎใใใฎๅ f2 ใซๅใใใ๏ผ f2 ใฏ๏ผๅฐ้ขใจใใฃในใฏใฎๆฉๆฆใใฎใใฎใงใใ๏ผf2ใซใใฃ
ในใฏใฎๅ่ปขไธญๅฟใพใงใฎใขใผใกใณใใขใผใ rใใใใ rf2 ใใขใผใกใณใใซใชใ๏ผ
f1
f2
๐ฅ ๐๐
R5:(1)ๅนณ่ก็ทใฎๅฎ็ใใ,I = M (2l)2/12+Md2 = M (l2 /3+d2)๏ผๅพฉๅ ใขใผใกใณใใฏMgdsin ๐๏ผ๐M (l2 /3+d2)๐๏ผMgd๐ =0๐ ๐๐๐3 ๐ ๏ผ
(2)I = M (2l)2/12+Md2 +2ml2+ 2md2 =M (l2 /3+d2) +2m(d2+l2)ๅพฉๅ ใขใผใกใณใใฏ (M+2m)gdsin ๐๐ ๐ 2๐(d2+l2) ๐๏ผ(M+2m)gd๐ =0
๐ ๐ 2๐ ๐๐๐ ๐3 ๐ 2๐(d2+l2)
๐๐ ๐๐๐ ๐
๐
๐๐
R6:(1) ๐ผ๐ ๐๐ sin ๐ ๐๐ sin ๐ ๐๐๐ sin ๐
๐๐ ๐ 2๐๐ ๐๐๐ ๐ 0(2) ฮป๐๐๐ 2๐๐ 0
(3) The pendulum is unstable, falling down.
R8:
1)
๐๐ ๐ ๐๐๐ ๐๐๐ ๐น2) ๐ผ ๐ ๐ ๐๐น
๐๐ ๐ ๐๐๐ ๐ผ๐ ๐ ๐ 0๐ ๐ ๐ ๐๐
๐ ๐ ๐๐ ๐๐๐ 25 ๐๐ ๐ ๐๐๐ 0
3) ๐ 5๐7๐
mg
ฮธ
Riceball ฯ
ฯ
F