PHYSICS Diyan Ricky Warisle.pdf

8/11/2019 PHYSICS Diyan Ricky Warisle.pdf

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PHYSI S Diyan Ricky Warisle / XI IPA 3 / 8

- ROTATIONAL DYNAMICS NO. 110

1.

A solid cylinder rolls up an inclined plane as shown. Initial speed of the cylinder when going up the

incline is 20 m/s. If the energy lost due to friction can be ignored, it is able to reach a height of

cylinder h before reversing direction. How tall h ...? (Solid cylinder, I = MR2)

Given : V0= 20 m/s I = MR2

= 30o

Asked : h?

Solution : EMawal = Emakhir

mv1

2+

I1

2+ mgh1=

mv2

2+

I2

2+ mgh2

mv1

2+

(

mR

2) (

) + mg(0) =

m(0) +

(

mR

2) (

) + mgh2

mv12

+ mv12= mv2

2+ mgh

2

mv1

2=

mv2

2+ mgh

2

m(20)

2=

m(0)

2+ mgh2

. 400 = gh2

= h2 = h2 h = 30 m

2. In the figure below, the pulley wheel C spinning detach winding rope. The mass of the wheel C is

300 g. If the acceleration of gravity is 10 m/s2, what is the rope tension T?

Given : m = 300 g = 0,3 kg g = 10 m/s2

Asked : T?

Solution : T =mg =

. 0,3 kg . 10 m/s

2 T = 1 N

3. A homogeneous solid roll on a flat surface with a velocity v = 10 m / s. The ball then rolled up the

incline to try a slice according apai turning point B as high as h. If the acceleration of gravity g = 10

m/s2, then h = ?

T

C
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Given : V0= 10 m/s I = mR2

g = 10 m/s2

Asked : h?

Solution : EMawal = Emakhir

mv1

2+

I1

2+ mgh1=

mv2

2+

I2

2+ mgh2

mv1

2+

(

mR

2) (

) + mg(0) =

m(0) +

I(0)

2+ mgh2

mv12

+ mv12

= mv22

+ mgh2

mv1

2= mgh

2

m(10)

2= mgh2

. 100 = gh2 = h2

= h2 h = 7 m

4. A yo-yo of mass 500 grams perform motion as shown. What is the acceleration experienced? (yo-yo is

considered solid cylinder).

Given : m = 500 g = 0,5 kg

Asked : a?

Solution : Translasi

F = m.a

wT = m.a Substitusi

T = wm.am.a = wm.a

Rotasi m.a = m.g = I . a =

g =

. 10

T . r = mr2.

a = 6,67 m/s

2

T = m.a

5. A solid cylinder initially at rest rolls down an inclined plane which height is 15 m. The liniear speed

when it arrived at the foot of the field if the acceleration of gravity is 9,8 m/s2

is...

Given : h1= 15 m I = MR2

g = 9,8 m/s2
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Asked : v2?

Solution : EMawal= EMakhir

mv1

2+ mgh1+

I1

2 =

mv2

2+

I2

2+ mgh2

m(0)

2+ mgh1+

(

mR

2(

)) =

mv2

2+

(

mR

2(

))

+ mg(0)

mgh1 +mv1

2=

mv2

2+

mv2

2

mgh1 +m(0)

2=

mv2

2

gh1 =v2

2

v22=

gh1

v2 = 1 = = v2 =14 m/s6.

People in the figure below has a mass of 90 kg. Both his arms stretched and his left hand holding a

load of mass M = 10 kg. Determine the horizontal and vertical position of the center of gravity of the

load plus M. (Select center point coordinate on the ground, right in the middle between the legs)

Given : mpeople= 90 kg Wpeople= 90.10 = 900 N

mmass= 10 kg Wmass= 10.10 = 100 N

Asked : X0 , Y0?

Solution :

X0= = = = 0,1 m

Y0=

=

=

= 1,14 m

7. In the figure below, C is a wheel pulley and the mass of the load B is greater than the mass of the load

A. If the acceleration of gravity = 10 m/s2and strap tension T1 = 24 N, the rope tension T2 is ...

Given : Mkatrol= 4 kg g = 10 m/s2

mB> mA

mA= 2 kg T1= 24 N I = MR

2

Asked : T2?

A

T1

B

T2

2 kg

M = 4 kg


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Solution :FA= mA. a

T1WA= 2a

2420 = 2a

4 = 2a a = 2 m/s2

katrol = I .

T2.RT1.R =. mkatrol. R

2.

T224 =. 4 . 2

T2= 4 + 24 = 28 N T2 = 28 N

8.

Four particles connected by a light rigid rod which mass can be ignored. Determine the moment of

inertia of the shaft system of particles: a) the axis AA b) the axis BB

Given :

Asked : a) I at the axis AA?

b) I at the axis BB?

Solution : a) Itotal= I1+ I2+ I3+ I4

= m1r12+ m2r2

2+ m3r3

2+ m4r4

2

= m(0)2+ 2m(b)

2+ m(2b)

2+ 3m(3b)

2

= 0 + 2mb2+ m4b

2+ 3m 9b

2

= 2mb2+ 4mb

2+ 27mb

2

Itotal= 33mb2

b) Itotal = I1+ I2+ I3+ I4

= m1r12+ m2r2

2+ m3r3

2+ m4r4

2

= m(2b)2+ 2m(b)

2+ m(0)

2+ 3m(b)

2

= 4mb2+2mb

2+ 3mb

2

Itotal= 9mb2

9. A homogeneous wooden board AB has a length of 120 cm and a weight of 1.20 N. The board is placed

at two sustainer C and D, each of them is placed 10 cm from the ends of the board. A load of 0.20 N

was hung by a thread as far as 30 cm from the end of A and a load of 0.90 N was hanged as far as 40

cm from the end of B. Determine the reaction forces of each sustainer which is done on the wooden

board.

A

A B

B

b b b

2m m 3mm

A

A B

B

b b b

2m m 3mm


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Given :

F = W

Asked : FAand FB?

Solution : A= Q+ R+ S

FA. l = FQ. l + FR. l + FS. l

FA. CD = 0,2 N . QD + 1,2 N . RD + 0,9 N . SDFA. 1 = 0,2 N . 0,8 + 1,2 N . 0,5 + 0,9 N . 0,3

FA= 0,16 + 0,6 + 0,27

FA= 1,03 N

B= Q+ R+ S

FB. l = FQ. l + FR. l + FS. l

FB. CD = 0,2 N . CQ + 1,2 N . CR + 0,9 N . CS

FB. 1 = 0,2 N . 0,2 + 1,2 N . 0,5 + 0,9 N . 0,7

FB= 0,04 + 0,6 + 0,63

FB= 1,27 N

10. A homogeneous field PQRS looks like in the picture. Determine the location of the center of gravity of

the object to the side PQ!

Given : PQ = 40 cm OZ = 30 cm

PS = 60 cm

Asked : Y0?

Solution : A1= Lpersegipanjang= p x l = 40 cm x 60 cm = 2400 cm2

Y1= = 30 cm

A2= Lsegitiga=

=

= 600 cm2

Y2= 60( ) = 50 cmY0=

=

= = = 23,33 cm

Y0= 23,33 cm

A BC D

L = 1,2 m

l = 0,3 m l = 0,4 m

W1= 0,2 NWbtg= 1,2 N

W2= 0,9 N

Q R S0,1 m 0,1 m

S R

QP40 cm

60cm

30c

O

Z


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- FLUIDS NO. 1120

11. Three types of liquids that cant be mixed poured into a cylindrical container with a cross-sectional

area 50 cm2. The volume and density of each type of fluid is: 0,50 litre, 2,6 g/cm

3; 0,25 litre, 1,0 g/cm

3;

dan 0,40 litre, 0,80 g/cm

3

. How much is the absolute pressure on the bottom of the container?Given : A = 50 cm

2= 5 x 10

-3m

2

v1= 5 x 10-4

m31= 2,6 x 10

3kg/m

3

v2= 2,5 x 10-4m

32= 1,0 x 10

3kg/m

3

v3= 4 x 10-4

m33= 0,8 x 10

3kg/m

3

Asked : Pmutlak?

Solution : h1= =

= 0,1 m

h2= = = 0,05 m

h3= =

= 0,08 m

Pmutlak= 1gh1+ 2gh2+ 3gh3 + Patm

= 2,6 x 103

(10)(0,1) + 1,0 x 103

(10)(0,05) + 0,8 x 103

(10)(0,08) + 1,01 x 105

= 2600 + 500 + 640 + 101000 = 104740 Pa

Pmutlak = 104740 Pa

12.

A bar with a size 0,2 m x 0,1 m x 0,3 m was hung up with a piece of wire vertically. Determine the

buoyant force of the balok if:

a) Entirely immersed in oil ( = 800 kg/m3)

b) Dippedparts into water ( = 1000 kg/m

3)

c) Dippedparts into mercury( = 13600 kg/m

3)

Given : vbalok= 0,2 m x 0,1 m x 0,3 m = 0,006 m3

g = 10 m/s2

Asked : FA?

Solution : a) FA= minyak. v . g = 800 . 0,006 . 10

= 48 N

b) FA= air.v . g = 1000 . 0,006) . 10= 45 N

c) FA= raksa.

v . g = 13600 .

0,006) . 10= 163,2 N

13.A hot air balloon containing hot air has a volume of 500m3

. The balloon was moving upward with

constant speed and the density remains in the air is 1,2 kg/m3

. When the density of the hot air inside

the balloon is 0,8 kg/m3

. How much is the total mass of the balloon and hot air in it? And how much is

the upward acceleration of the balloon when the temperature of the air inside the balloon is raised so

that the density is 0,7 kg/m3

?

Given : vudara= 500 m3 balon= 0,8 kg/m

3

udara= 1,2 kg/m3

Asked : a) mtotal? b) a?balon2= 0,7 kg/m3


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Solution : a) FA= udara. v . g

mtotal. g = 1,2 ( 500 ) . g

mtotal= 600 kg

b) FA= udara. v . g

= udara g= 1,2 10 = 6857,14 N

a=

=

=

= 1,42 m/s2

14. Water flows through a horizontal pipe that has 2 different parts. If the cross-sectional area of X is 8,0

cm2the velocity of water is 3,0 cm/s. Calculate : a) the velocity of Y cross-sectional area of 2,0 cm

2b)

the pressure difference between X and Y

Given :

Asked : a) v2?A2= 2,0 cm2 B) P?

Solution : a) A1v1= A2v2

8 ( 3 ) = 2 . v2

V2= 12 m/s

b) P1+v1

2= P2 +

v2

2

P1P2= ( v2

2v1

2)

=. 1000 ( 123 ) x 10

-2

= 45 Pa

15.

The water flows from the first floor of a two-floored house through 2 pipe which the diameter is 2,8cm. Water flowed into the bathroom on the second floor through a faucet with diameter 0,7 cm

located 3 m above the pipe on the first floor. If the speed of the water of the pipe in the first floor is

0,15 m/s and a pressure 1,8x105Pa. Determine : a) the speed of the water in the supply pipe b) the

pressure in the pipe

Given : d1= 2,8 cm = 0,028 m h2= 3 m

d2= 0,7 cm = 0,007 m P1= 1,8 x 105 Pa

h1= 0 m v1= 0,15 m/s

Asked : a) v2?

b) P2?

Solution : a) A1v1= A2v2 . d12. v1= . d22. v2 d12. v1= d22. v2

0,0282( 0,15) = ( 0,007 )

2(v2)

0,0001176 = 0,000049 . v2

v2 =

m/s

v2= 2,4 m/s

X

Y


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b) P1+ gh1+v1

2= P2 + gh2+

v2

2

P2= P1+ g ( h1h2) + ( v1

2v2

2)

P2= 1,8 x 105+ 1000 ( 9,8 ) ( 03 ) +

1000 ( 0,152,4 )

P2= 1,54 x 105Pa

16. Waterfall as high as h is used for hydroelectric power generation. Each second water flow 10 m3. If

the generator efficiency 55% and average power generated is 1100 kW. Define h?

Given : Q = 10 m3/s = 55%P = 1100 kW

Asked : h?

Solution : P = . . Q . g . hh = =

h = 20,4 m

17.A spherical object with a diameter of 2 cm dropped freely into a fluid that its density is 700 kg/m3.

From the experiments it was found that the greatest speed of the object is 4,9 m/s. If the density of

the object is 7900 kg/m3 and the gravitational acceleration 9,8 m/s

2. Determine the coefficient of

viscosity of the liquid?

Given : d = 2 cm = 0,02 m vT= 4,9 m/s

f= 700 kg/m3 b= 7900 kg/m3

a = 9,8 m/s2

Asked : ?Solution : vT=

(bf)

= (bf) =

( 7900700 )

= 0,3218.

A copper has cavity inside, the weight in air is 264 g dan 221 g in water. If the density of copper is 8,8

g/cm3. Determine the volume of the cavity?

Given : mudara= 264 g = 0,264 kg benda= 8,8 g/cm3= 8,8 x 10

3kg/m

3

mfluida= 221 g = 0,221 kg

Asked : vrongga?

Solution : FA= WudaraWfluida= ( 0,264 . 10 )( 0,221 . 10 ) = 2,642,21 = 0,43 N

FA= A. vB. g

= A. ( ) . g= A. vrongga. g + A(

)

0,43 = 1000 . vrongga. 10 + 1000 (

)

0,43 = 10000 . vrongga+ 0,3

0,13 = 10000 . vrongga

vrongga= 1,3 . 10-5

m3= 13 cm

3


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19. A balloon with a total volume of 50 m3filled with a hydrogen with density of 0,08 kg/m

3. the density

of the air around the balloon is 1,3 kg/m3

and the gravitational acceleration g = 9,8 m/s2. Determine

the force that is needed to lift the balloon?

Given : v = 50 m3 udara= 1,3 kg/m3

gas= 0,08 kg/m3

g = 9,8 m/s2

Asked : FA?

Solution : gaya apung = udara . vgas . g

= 1,3 . 50 . 9,8 = 637 N ... 1

gaya berat = m.g = gas . vgas . g

= 0,08 . 50 . 9,8 = 39,2 N ... 2

FA= gaya apunggaya berat

= 637 N39,2 N = 597,8 N

20. A tank filled with water that is above the mercury. A cube with a side of 60 mm stands perpendicular

in balance inside the fluids. Determine the height of each cube is submerged in the liquid. The density

of iron and mercury in a row 7,7 x 103kg/m

3dan 13,6 x 10

3kg/m

3

Given : besi= 7,7 x 103kg/m

3 rkubus= 60 mm

raksa= 13,6 x 103kg/m

3

air= 1000 kg/m3

Asked : hair dan hraksa?

Solution : FA= W

air. vbair. g + raksa. vbraksa. g = mg

air. hair. A . g + raksa. hraksa. A . g = mg

Ag ( air. hair+ raksa. hraksa) = mg

air. hair+ raksa. hraksa=

air. hair + raksa. hraksa=

= benda. rkubus ... 1

air( rkubushraksa) + raksa . hraksa= benda. rkubus

air. rkubusair. hraksa+ raksa. hraksa= benda. rkubus

air. rkubus+ hraksa( raksaair) = benda. rkubus

hraksa( raksaair) = benda. rkubusair. rkubus

hraksa ( raksaair) = rkubus( bendaair)

hraksa= rkubus( )

hraksa= 60 . hraksa= 31,9 mm

hair= rkubushraksa

= 60 mm31,9 mm

= 28,1 mm

air

raksa

hraksa

hair besi

PHYSICS Diyan Ricky Warisle.pdf

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