理论力学作业答案-试点班

10-16解1：刚体作平面运动C为基点v_

v_

v_B

C

BC

0r

9m/svBCvB

v0

vBC

7m/sFdvBC

v0vBFNmgaCmaCx

FxCyma

Fy(1)maC

Fd0

mg

FN

(2)(3)CCJ

M2dm

F

rFd

FN

f

(4)aC

fg

0.49m/s

2

fgr

/

2

5.1rad/s210-16aC

fg

const

fgr

/

2

const圆心匀加速直线运动轮匀

转动vC

v0

aCt

0

t纯滚动满足vC

rt

5.575svC

4.732m/s纯滚动后FSmgFNaCmaCx

FxmaC

FS

(1)2Sa

rJC

MCFS

0CaC

0m

F

r

(2)(3)

0

const纯滚下去．vC

const10-16FdmgO(固定点)FNLO

MO

0轮对Ｏ的动量矩守恒LO

const解２Ｏ为地面任意固定点．0mvJC002O1

0L

mv

r

m

状态１（又滚又滑）状态２（纯滚动）2O

2

CL

mv r

m

vC

rO

2O1L

L0

v

r02

r

2

2

2

r

2vC

r0

0

4.732m/s2

2

v

r10-16mv

mv0

Fdt

fmgtt

5.575sconst

r

2

2

2

v

r

0

0

r

2

2

2

v

rCv

r

0

0

const纯滚下去．10-19OABC解：轮A（刚体定轴转动）FTmgFOxOFOyAO

OJ

Mmr

2A

FT

r

(1)0

FOxmaCx

FxmaCy

FyT

F

(2)Oy0

mg

F轮Ｂ（刚体平面运动）JC

MC1FTBD

CaCmg2mr

2B

FT

r

(3)maCy

FymaC

mg

FT

(4)OABC补充运动学方程动点Ｃ，动系绳子（无常接触点）a_

a_

a_a

e

rae

r

Aaear

rBarFOx

0C4a

3

gaC

r

A

rB

5OyF

5

mg410-19解FTmgFOxFOyOAT

F

r

M

(1)Amr

22T

F

r

(3)B1

mr

2maC

mg

FT

(4)aC

r

A

rB5M

3mgrCa

0FTD

CBaCmg10-22解：A物体无质量受平衡力系.(二力杆)mgFAaAFNAB:刚体平面运动

a_

a_A

CACa_A为基点

0CAanaAFAaCAmaCx

FxmaCy

FyJC

MCmaA

aCA

cos

mg

sin

(1)

maCA

sin

mg

cos

FN

(2)(3)l21212Nml

F

sin10-222CAa

l

(4)1

3sin

2

mg

cosFN

4g

sin

1

3sin

2

aA11-32mgFNy解：刚体作平面运动aCyaCxxCxma

mg

cos

(1)maCx

FxmaCy

FyJC

MCNma

mg

sin

F

(2)Cy1(3)122Nml

F

b11-32aCxxaCyyaDD为基点aDaCDCDDCyCxa-

-a

a

a-

-aCyaCx

aD

(4)

aCD

b

(5)aCx

aD

g

cos

12b

2FDl

2

mgl

2

sin

l

212b2

12gb2

sin

aCyOCBA12-10解：以杆为研究对象OArBlFAxFAymACatCanmaCx

FxAxC

man

F

yCyma

FAyC

mat

FOOJ

MJO

mA

FAx

cosr

FAy

sin

r2273mrO12J

12m2r

m

2r

22ranC

2ratCAxF

2m

2r2mrFAy

3Am

4

mr

2

mr

2211-2OA(a)

圆盘与杆固结(杆和轮一起定轴）212OJ

T

21JO

13mR2

ml

2

ml

2

2241

3T

2

ml

2

mR

(b)

圆盘绕轴A相对杆OA以角速度

逆时针方向转动；A

0O定轴；Ａ平动OA(b)r

T

T1

T2

1

ml2

2

ml

222

3221

1ml

T

2

3OA(c)

r

(c)

圆盘绕轴A相对杆OA以角速度

顺时针方向转动。A

2O定轴；Ａ平面运动T

T1

T22211

1ml

2

3T

122

22222ml

m22mR

22T

1R

l21

122l

R2

31

2T

T

T

m211-3vvCA212CJ

CT

1

mv22A基点解:圆盘平面运动vC

v

vCA

10

51.3

16.51

1222

2

5

0.3

5

683.4JT

1

516.52

211-4FNFSrM

r

N2M

F

mg

1

F

F2W

Fr

F

cos

30x

1 3

FxrMr2

2r

W

M

mg

1

F

x

mg

FN

FS

不作功11-7解：系统为研究对象1T

0222A

21

AO

1J

12J

1

m

v

22T

Av12223121JO

m

R

r

212m1rJ

A

v

A

1

R

r

r222

112W

M

m g

R

r

sin

m

g

R

r

sin

2m

g1m

gT2

T1

W12代入求得21

29m

2m

R

r

12

M

12m1

6m2

R

r

g

sin

21求导得.1

1.

121

29m1

2m2

R

r6M

6m

3m

R

r

g

cos

1

11-9mgT2

T1

W122

b

1

b

122OJ

0

mg

2

2

212

6CJ

1

m

b2

b

2

1

mb

2mb

22322b

JO

JC

m2b3

2

1g

11-9mg2

b

1

b

122PJ

0

mg

2

22

2

1

5J

P

JC

m

2

b

12

mb5b12

2

1g

vCP水平方向不受力,质心铅直向下运动vOT2

T1

W1211-10mAg解:AB

水平,B瞬心B

0T2

T1

W121T

02212B

ABT

J

2W

m

g

l

sin3012

A3B

AJ

1

m l

2

4.95

rad/s

AB机械能守恒22

1