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Preview of 14.7- 14.9

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14.7 Kinetic Energy

• Kinetic energy of a system of particles,

 

 

n

i

i i n

i

i i

i v v m v

m T

1

2 2

1 2 1

1  

i G

i v v

v    

• Expressing the velocity in terms of the centroidal reference frame,

   

 





n i

i i G

n i

i i n

i

i i G

G n

i i n i

i G i

G i

v m v

m

v m v

m v

v m

v v

v v

m T

1

2 2

2 1 2 1

1

2 2

1 1

2 2 1

1 2 1 1

• Kinetic energy is equal to kinetic energy of mass center plus kinetic energy relative to the centroidal frame.

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14.8 Work-Energy Principle. Conservation of Energy

• Principle of work and energy can be applied to each particle Pi ,

2 2

1

1 U T

T

where represents the work done by the internal forces and the resultant external force acting on Pi . ij

fFi

2 1

U

• Principle of work and energy can be applied to the entire system by adding the kinetic energies of all particles and considering the work done by all external and internal forces.

• Although are equal and opposite, the work of these forces will not, in general, cancel out.

ji

ij f

f  and

• If the forces acting on the particles are conservative, the work is equal to the change in potential energy and

2 2

1

1 V T V

T   

which expresses the principle of conservation of energy for the system of particles.

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14.9 Principle of Impulse and Momentum

2 1

1 2

2

1 2

1

L dt

F L

L L

dt F

L F

t t t

t

 

 

 

2 1

1 2

2

1 2

1

H dt

M H

H H

dt M

H M

t t

O t

t

O

O O

 

 

 

• The momenta of the particles at time t1 and the impulse of the forces from t1 to t2 form a system of vectors equipollent to the system of momenta of the particles at time t2 .

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Sample Problem 14.4

Ball B, of mass mB, is suspended from a cord, of length l, attached to cart A, of mass mA, which can roll freely on a frictionless horizontal tract. While the cart is at rest, the ball is given an initial velocity

Determine (a) the velocity of B as it reaches it maximum elevation, and (b) the maximum vertical distance h

through which B will rise.

.

0 2gl v

SOLUTION:

• With no external horizontal forces, it follows from the impulse-momentum principle that the horizontal compo- nent of momentum is conserved. This relation can be solved for the velocity of B at its maximum elevation.

• The conservation of energy principle can be applied to relate the initial

kinetic energy to the maximum potential energy. The maximum vertical distance is determined from this relation.

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Sample Problem 14.4

SOLUTION:

• With no external horizontal forces, it follows from the impulse-momentum principle that the horizontal compo- nent of momentum is conserved. This relation can be solved for the velocity of B at its maximum elevation.

2 1

2

1

L dt F L

t t

 

(velocity of B relative to A is zero at position 2)

2 , 2

, 2

, 2

,

0 1 , 1

, 0

A A

B A

B

B A

v v

v v

v v

v

Velocities at positions 1 and 2 are

 

,2

0 A B A

Bv m m v

m

0 2

, 2

, v

m m

v m v

B A

B B

A

x component equation:

2 , 2

, 1

, 1

, B B A A B B

A

Av m v m v m v

m

x y

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Sample Problem 14.5

Ball A has initial velocity v0 = 10 m /s parallel to the axis of the table. It hits ball B and then ball C which are both at rest. Balls A and C hit the sides of the table squarely at A’ and C’ and ball B hits obliquely at B’.

Assuming perfectly elastic collisions, determine velocities vA, vB, and vC with which the balls hit the sides of the table.

SOLUTION:

• There are four unknowns: vA, vB,x, vB,y, and vC.

• Write the conservation equations in terms of the unknown velocities and solve simultaneously.

• Solution requires four equations:

conservation principles for linear

momentum (two component equations), angular momentum, and energy.

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Sample Problem 14.5

x y

i v v

j v i v v

j v v

C C

y B x

B B

A A

, ,

SOLUTION:

• There are four unknowns: vA, vB,x, vB,y, and vC.

• The conservation of momentum and energy equations,

y B A

C x

B mv mv mv

mv mv

L dt F L

, ,

0

2 1

0



2 2,

21 2

2 , 2 1 2 2 1 2 0 1

2 2

1 1

C y

B x

B

A m v v mv

mv mv

V T

V T

 

O

 

O A

 

B y

 

C

O

mv mv

mv mv

H dt

M H

m 3 m

7 m

8 m

2 0 ,

2 , 1

,

   

Solving the first three equations in terms of vC,

C x

B C

y B

A v v v v

v , 3 20 , 10 Substituting into the energy equation,

   

0 800 260

20

100 10

20 3

2

2

2 2

2

C C

C C

C

v v

v v

v

2m/s4

m s8m s 4.47m s

4

B B

C A

v j

i v

v v  

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