##### Document Text Contents

Page 1

V. Zubov and V. Shalnov

PROBLEMS

IN

PHYSICS

V. Zubov and V. Shalnov

PROBLEMS

IN

PHYSICS

Page 150

{46 ANSWERS AND SOLUTIONS

16. u = 8 m/s;E»,.s12°.

Solution. As in t e lprevious problem the velocities of the current

and of the launch shou d be reso ved into components along the line

AB and perpendicularly to it (Fig. 191). In order that the moving

launch is always on the straight line AB the components of the velo-

city of the current and of the launch in the direction perpendicular

to AB should be equal, i.e.,

usinB= vsinoc (1)

When the launch moves from A to B its velocity relative to the banks

g 6

" /3 ` {Ex

¢Z\x\ ZL U

1 A

Fig. 191 Fig. 192

will be u cos fl + v cos on and the time of motion will be determined

from the equation

S= (u cosB-|- vcos ot)t, (2)

The time of motion from B to A (Fig. 192) can be found from the

equation

S= (u cosB- vcos ot)t2 (3)

From the known conditions,

t, -|- tz = t (4)

Solving the system of the four equations we find

_ z '

B=arc cot -——————-—-————S + VS2 vzt COSZ OL at 12°

vt sin on

u == v m 8 m/s

17. u = 40 km/hr.

Solution. All the points on the wheel rim simultaneously perform

two motions: translational, with the entire bicycle, and rotational,

around the wheel axis. The total velocity of each point will be the

sum of the linear velocities of the translational and rotational motions.

If the wheel of the bicycle moves without sliding the linear velocity

of the rotational motion of the wheel rim will be equal in magnitude

to the velocity of the translational motion of the bicycle, These velo-

{46 ANSWERS AND SOLUTIONS

16. u = 8 m/s;E»,.s12°.

Solution. As in t e lprevious problem the velocities of the current

and of the launch shou d be reso ved into components along the line

AB and perpendicularly to it (Fig. 191). In order that the moving

launch is always on the straight line AB the components of the velo-

city of the current and of the launch in the direction perpendicular

to AB should be equal, i.e.,

usinB= vsinoc (1)

When the launch moves from A to B its velocity relative to the banks

g 6

" /3 ` {Ex

¢Z\x\ ZL U

1 A

Fig. 191 Fig. 192

will be u cos fl + v cos on and the time of motion will be determined

from the equation

S= (u cosB-|- vcos ot)t, (2)

The time of motion from B to A (Fig. 192) can be found from the

equation

S= (u cosB- vcos ot)t2 (3)

From the known conditions,

t, -|- tz = t (4)

Solving the system of the four equations we find

_ z '

B=arc cot -——————-—-————S + VS2 vzt COSZ OL at 12°

vt sin on

u == v m 8 m/s

17. u = 40 km/hr.

Solution. All the points on the wheel rim simultaneously perform

two motions: translational, with the entire bicycle, and rotational,

around the wheel axis. The total velocity of each point will be the

sum of the linear velocities of the translational and rotational motions.

If the wheel of the bicycle moves without sliding the linear velocity

of the rotational motion of the wheel rim will be equal in magnitude

to the velocity of the translational motion of the bicycle, These velo-

Page 151

CHAPTER I. MECHANICS 147

cities are opposite at the point A (Fig. 193). The total velocity of the

point A will therefore be equal to zero. At the point B the velocities

of the translational and rotational motion will be directed one way,

and the total velocity of the point B will be equal to 2v, i.e., to

40 km/hr.

18. The bobbin will roll in the 8

direction of motion of the end of the

thread with a velocity of u= Egg; v.

Solution. The displacement of the ·——

thread end depends on two things:

the movement of the bobbin axis and

the change in the length of the thread

during winding (or unwinding). It is |

clear that the motions of the end of

the thread due to these causes will Fig. 193

always be in opposite directions. If,

for example, the bobbin moves to the right the motion of its axis

will cause the thread end to move to the right and to the left due to

the thread winding.

Since r < R, the change in the length of the thread during one

full revolution will always be less than the motion of the bobbin axis

during the same time.

If one adds these motions, the end of the thread should always be

moving in the same direction as the axis of the bobbin. During one

full revolution T the axis of the bobbin will be dis laced to the right

by a distance 2nR and the length of the thread will be reduced by 2m-.

If the velocity of the axis is u = any-E the velocity with which the length

of the thread diminishes will be 2-;; = gf-gg- -%· = -1% u. The velocity

of motion of the thread end will be equal to the difference between

these velocities, i.e., v = u — gi u.

Hence,

uz--—i¥-— v"

R-—r

The axis of the bobbin moves faster than the end of the thread.

19. The bobbin moves in the direction of motion of the thread end

with a velocity u = % v.

Note. (See the solution to the previous problem.) The axis of the

bobbin moves at a slower speed than the end of the thread.

20. v, = 1.1 m/s; v2 = 0.5 m/s.

Solution. In the first case the velocity of one body relative to the

other will be

m=w+w m

and in the second case

uz = vi ··· U2 (2)

10*

CHAPTER I. MECHANICS 147

cities are opposite at the point A (Fig. 193). The total velocity of the

point A will therefore be equal to zero. At the point B the velocities

of the translational and rotational motion will be directed one way,

and the total velocity of the point B will be equal to 2v, i.e., to

40 km/hr.

18. The bobbin will roll in the 8

direction of motion of the end of the

thread with a velocity of u= Egg; v.

Solution. The displacement of the ·——

thread end depends on two things:

the movement of the bobbin axis and

the change in the length of the thread

during winding (or unwinding). It is |

clear that the motions of the end of

the thread due to these causes will Fig. 193

always be in opposite directions. If,

for example, the bobbin moves to the right the motion of its axis

will cause the thread end to move to the right and to the left due to

the thread winding.

Since r < R, the change in the length of the thread during one

full revolution will always be less than the motion of the bobbin axis

during the same time.

If one adds these motions, the end of the thread should always be

moving in the same direction as the axis of the bobbin. During one

full revolution T the axis of the bobbin will be dis laced to the right

by a distance 2nR and the length of the thread will be reduced by 2m-.

If the velocity of the axis is u = any-E the velocity with which the length

of the thread diminishes will be 2-;; = gf-gg- -%· = -1% u. The velocity

of motion of the thread end will be equal to the difference between

these velocities, i.e., v = u — gi u.

Hence,

uz--—i¥-— v"

R-—r

The axis of the bobbin moves faster than the end of the thread.

19. The bobbin moves in the direction of motion of the thread end

with a velocity u = % v.

Note. (See the solution to the previous problem.) The axis of the

bobbin moves at a slower speed than the end of the thread.

20. v, = 1.1 m/s; v2 = 0.5 m/s.

Solution. In the first case the velocity of one body relative to the

other will be

m=w+w m

and in the second case

uz = vi ··· U2 (2)

10*

Page 300

HANDBOOK OF PHYSICS

B. Yavorsky, D. Sc. and A. Detlaf, Cand. Sc.

A companion volume to Vygodsky’s Handbook of Higher Mathematics, designed

for use by engineers, technicians, research workers, students, and teachers. of

physics. Includes definitions of basic physical concepts, brief formulations of

physical laws, concise descriptions of phenomena, tables of physical quantities in

various systems of units, universal physical constants, etc.

This is a third English edition. _

Contents. Physical Basis of Classical Mechanics. Fundamentals of Ther-

modynamics and Molecular Physics. Fundamentals of Fluid Mechanics. Electrici-

ty and Magnetism. Wave Phenomena. Atomic and Nuclear Physics.

HANDBOOK OF PHYSICS

B. Yavorsky, D. Sc. and A. Detlaf, Cand. Sc.

A companion volume to Vygodsky’s Handbook of Higher Mathematics, designed

for use by engineers, technicians, research workers, students, and teachers. of

physics. Includes definitions of basic physical concepts, brief formulations of

physical laws, concise descriptions of phenomena, tables of physical quantities in

various systems of units, universal physical constants, etc.

This is a third English edition. _

Contents. Physical Basis of Classical Mechanics. Fundamentals of Ther-

modynamics and Molecular Physics. Fundamentals of Fluid Mechanics. Electrici-

ty and Magnetism. Wave Phenomena. Atomic and Nuclear Physics.

V. Zubov and V. Shalnov

PROBLEMS

IN

PHYSICS

V. Zubov and V. Shalnov

PROBLEMS

IN

PHYSICS

Page 150

{46 ANSWERS AND SOLUTIONS

16. u = 8 m/s;E»,.s12°.

Solution. As in t e lprevious problem the velocities of the current

and of the launch shou d be reso ved into components along the line

AB and perpendicularly to it (Fig. 191). In order that the moving

launch is always on the straight line AB the components of the velo-

city of the current and of the launch in the direction perpendicular

to AB should be equal, i.e.,

usinB= vsinoc (1)

When the launch moves from A to B its velocity relative to the banks

g 6

" /3 ` {Ex

¢Z\x\ ZL U

1 A

Fig. 191 Fig. 192

will be u cos fl + v cos on and the time of motion will be determined

from the equation

S= (u cosB-|- vcos ot)t, (2)

The time of motion from B to A (Fig. 192) can be found from the

equation

S= (u cosB- vcos ot)t2 (3)

From the known conditions,

t, -|- tz = t (4)

Solving the system of the four equations we find

_ z '

B=arc cot -——————-—-————S + VS2 vzt COSZ OL at 12°

vt sin on

u == v m 8 m/s

17. u = 40 km/hr.

Solution. All the points on the wheel rim simultaneously perform

two motions: translational, with the entire bicycle, and rotational,

around the wheel axis. The total velocity of each point will be the

sum of the linear velocities of the translational and rotational motions.

If the wheel of the bicycle moves without sliding the linear velocity

of the rotational motion of the wheel rim will be equal in magnitude

to the velocity of the translational motion of the bicycle, These velo-

{46 ANSWERS AND SOLUTIONS

16. u = 8 m/s;E»,.s12°.

Solution. As in t e lprevious problem the velocities of the current

and of the launch shou d be reso ved into components along the line

AB and perpendicularly to it (Fig. 191). In order that the moving

launch is always on the straight line AB the components of the velo-

city of the current and of the launch in the direction perpendicular

to AB should be equal, i.e.,

usinB= vsinoc (1)

When the launch moves from A to B its velocity relative to the banks

g 6

" /3 ` {Ex

¢Z\x\ ZL U

1 A

Fig. 191 Fig. 192

will be u cos fl + v cos on and the time of motion will be determined

from the equation

S= (u cosB-|- vcos ot)t, (2)

The time of motion from B to A (Fig. 192) can be found from the

equation

S= (u cosB- vcos ot)t2 (3)

From the known conditions,

t, -|- tz = t (4)

Solving the system of the four equations we find

_ z '

B=arc cot -——————-—-————S + VS2 vzt COSZ OL at 12°

vt sin on

u == v m 8 m/s

17. u = 40 km/hr.

Solution. All the points on the wheel rim simultaneously perform

two motions: translational, with the entire bicycle, and rotational,

around the wheel axis. The total velocity of each point will be the

sum of the linear velocities of the translational and rotational motions.

If the wheel of the bicycle moves without sliding the linear velocity

of the rotational motion of the wheel rim will be equal in magnitude

to the velocity of the translational motion of the bicycle, These velo-

Page 151

CHAPTER I. MECHANICS 147

cities are opposite at the point A (Fig. 193). The total velocity of the

point A will therefore be equal to zero. At the point B the velocities

of the translational and rotational motion will be directed one way,

and the total velocity of the point B will be equal to 2v, i.e., to

40 km/hr.

18. The bobbin will roll in the 8

direction of motion of the end of the

thread with a velocity of u= Egg; v.

Solution. The displacement of the ·——

thread end depends on two things:

the movement of the bobbin axis and

the change in the length of the thread

during winding (or unwinding). It is |

clear that the motions of the end of

the thread due to these causes will Fig. 193

always be in opposite directions. If,

for example, the bobbin moves to the right the motion of its axis

will cause the thread end to move to the right and to the left due to

the thread winding.

Since r < R, the change in the length of the thread during one

full revolution will always be less than the motion of the bobbin axis

during the same time.

If one adds these motions, the end of the thread should always be

moving in the same direction as the axis of the bobbin. During one

full revolution T the axis of the bobbin will be dis laced to the right

by a distance 2nR and the length of the thread will be reduced by 2m-.

If the velocity of the axis is u = any-E the velocity with which the length

of the thread diminishes will be 2-;; = gf-gg- -%· = -1% u. The velocity

of motion of the thread end will be equal to the difference between

these velocities, i.e., v = u — gi u.

Hence,

uz--—i¥-— v"

R-—r

The axis of the bobbin moves faster than the end of the thread.

19. The bobbin moves in the direction of motion of the thread end

with a velocity u = % v.

Note. (See the solution to the previous problem.) The axis of the

bobbin moves at a slower speed than the end of the thread.

20. v, = 1.1 m/s; v2 = 0.5 m/s.

Solution. In the first case the velocity of one body relative to the

other will be

m=w+w m

and in the second case

uz = vi ··· U2 (2)

10*

CHAPTER I. MECHANICS 147

cities are opposite at the point A (Fig. 193). The total velocity of the

point A will therefore be equal to zero. At the point B the velocities

of the translational and rotational motion will be directed one way,

and the total velocity of the point B will be equal to 2v, i.e., to

40 km/hr.

18. The bobbin will roll in the 8

direction of motion of the end of the

thread with a velocity of u= Egg; v.

Solution. The displacement of the ·——

thread end depends on two things:

the movement of the bobbin axis and

the change in the length of the thread

during winding (or unwinding). It is |

clear that the motions of the end of

the thread due to these causes will Fig. 193

always be in opposite directions. If,

for example, the bobbin moves to the right the motion of its axis

will cause the thread end to move to the right and to the left due to

the thread winding.

Since r < R, the change in the length of the thread during one

full revolution will always be less than the motion of the bobbin axis

during the same time.

If one adds these motions, the end of the thread should always be

moving in the same direction as the axis of the bobbin. During one

full revolution T the axis of the bobbin will be dis laced to the right

by a distance 2nR and the length of the thread will be reduced by 2m-.

If the velocity of the axis is u = any-E the velocity with which the length

of the thread diminishes will be 2-;; = gf-gg- -%· = -1% u. The velocity

of motion of the thread end will be equal to the difference between

these velocities, i.e., v = u — gi u.

Hence,

uz--—i¥-— v"

R-—r

The axis of the bobbin moves faster than the end of the thread.

19. The bobbin moves in the direction of motion of the thread end

with a velocity u = % v.

Note. (See the solution to the previous problem.) The axis of the

bobbin moves at a slower speed than the end of the thread.

20. v, = 1.1 m/s; v2 = 0.5 m/s.

Solution. In the first case the velocity of one body relative to the

other will be

m=w+w m

and in the second case

uz = vi ··· U2 (2)

10*

Page 300

HANDBOOK OF PHYSICS

B. Yavorsky, D. Sc. and A. Detlaf, Cand. Sc.

A companion volume to Vygodsky’s Handbook of Higher Mathematics, designed

for use by engineers, technicians, research workers, students, and teachers. of

physics. Includes definitions of basic physical concepts, brief formulations of

physical laws, concise descriptions of phenomena, tables of physical quantities in

various systems of units, universal physical constants, etc.

This is a third English edition. _

Contents. Physical Basis of Classical Mechanics. Fundamentals of Ther-

modynamics and Molecular Physics. Fundamentals of Fluid Mechanics. Electrici-

ty and Magnetism. Wave Phenomena. Atomic and Nuclear Physics.

HANDBOOK OF PHYSICS

B. Yavorsky, D. Sc. and A. Detlaf, Cand. Sc.

A companion volume to Vygodsky’s Handbook of Higher Mathematics, designed

for use by engineers, technicians, research workers, students, and teachers. of

physics. Includes definitions of basic physical concepts, brief formulations of

physical laws, concise descriptions of phenomena, tables of physical quantities in

various systems of units, universal physical constants, etc.

This is a third English edition. _

Contents. Physical Basis of Classical Mechanics. Fundamentals of Ther-

modynamics and Molecular Physics. Fundamentals of Fluid Mechanics. Electrici-

ty and Magnetism. Wave Phenomena. Atomic and Nuclear Physics.