PHYSICS QUESTIONS by Evi Dwi Pratiwi/11/XI IA 3

ROTATION DYNAMICS1.

A very light rod, the length is 120 cm. In the rod, there are 3 forces work on each F1=15N, F2=20N, and F3=30N with direction and position as in the picture. When the rod with

the axis located at the point of catching force F2 perpendicular penetrate the paper, then the total moment of the net force acting on the rod is : Data : F1=15N Ltot = 120 cm

F2=20N LF3-F2 (d2)= 100 cm F3=30N LF1-F2 (d1)= 20 cm

Problem : τtotal ?Answer : τtotal = τ1 + τ3 (Because F2 is the axis. So we can ignore F2)

= F1.d1 + F3.d3

= 15.20 + 30.100 = 300 + 3000τtotal = 3300 N. cm = 33 N.m

2. A solid sphere with 30 cm in diameter is rotating with the axis through the center of sphere. The equation of anguler speed sphere is ω =(10+20t) rad/s with t second. If mass of sphere is 5 kg, so the moment of force acting on the sphere is :Data : d = 30 cm → r = 15 cm = 15x10-2 m Solid sphere

ω =(10+20t) rad/s I = 25 mr2

m = 5 kgProblem : τ?Answer : ω =(10+20t) rad/s → α = 0 + 20 = 20 rad/s2

τ = I. α = 25 mr2 . α

= 25 5 (15x10-2)2 . 20 = 2 . 225x10-4 . 20 = 9000x10-4

τ = 0,9 N.m

F3=30N

100 cm F1=15N

3. A solid sphere is rolling on a flat surface rough due to be drawn with a force F=12N as in the picture

If the diameter is 20 cm and its mass is 4 kg. So the anguler acceleration of the solid ball is : Data : F =12N Solid sphere

d = 20 cm → r = 10 cm = 10x10-2 m I = 25mr2

m = 4kgProblem : α?Answer : Rotation Translation

∑τ = I. α ∑F = m . α fges. r = 25 mr2 ar F- fges = m. a

fges = 25mr2 a

rr

F - 25 m a = m . a

fges = 25 m a F = m.a + 25 m a

= m.a (1+25 ¿

a = F

m(1+ 25) α = ar

= 12

4 . 75

= 2,1410x 10−2

= 125,6 α = 21,4 rad/s2

= 2,14 4. A wheel which has moment inertia 5x10-3 kg m2 is rotating at a speed

20 rotation per second. The torque that required to stop the rotation of the wheel in 10 second is :Data : I = 5x10-3 kg m2

ω0 = 20 rot/s → 1 rot = 2π rad/s = 20 . 2π = 40π rad/s ωt = 0 t = 10 s

Problem : τ?Answer : ωt = ω0 - α.t τ = I. α

0 = 404π – α 10 = 5x10-3 . 4π α = 4π = 20 x 10-3π = 2π x10-2 N.m

fs F=12N

5. A solid sphere its diameter is 40 cm is rotating with axis through the center of sphere. The angular velocity equation of sphere is ω =(10+25t) rad/s with t second. If mass of sphere is 4 kg, determine the moment of force that works on the sphere :Data : d = 40 cm → r = 20 cm = 20x10-2 m Solid sphere m = 4 kg I = 25

mr2

ω =(10+25t) rad/s→ α = 0 + 25 = 25 rad/s2 Problem : τ?Answer : τ = I. α = 25 4 (20x10-2 )2 . 25

= 25 4 40080x10-4 . 25 = 2 . 4. 80x10-4 . 25 = 16000x10-4

τ = 1,6 Nm6. A solid cylinder is rolling on a flat floor with velocity 10 m/s. Mass of the

solid cylinder is 4 kg and the diameter is 80 cm. Determine the total kinetic energy of the solid cylinder : Problem : v = 10 m/s Solid cylinder m = 4 kg I = 12mr2

d = 80 cm → r = 40 cm = 40x10-2 mProblem : Ektot ? Answer : Ektot = Ektrans + Ekrot

= 12 m v2 + 12 I ω2

= 12 (m v2 + 12 mr2 v2

r2 )

= 12 (m v2 + 12 mv2 )

= 12 (1 + 12 ) mv2

= 34 mv2

= 34 4 102

= 3. 100 Ektot = 300 J

S = 3mA

B

m= 2 kg

7. A thin hollow sphere is rolling from rest, down on inclined plane which form 30◦ in angle. Determine the linier velocity of a thin hollow sphere after routed as far 3 m :Data : θ = 30◦ VA = 0

g = 10 m/s2 HB = 0 A thin hollow sphere I = 23mr2

(Because rolling)Problem : v after routed 3 m?Answer : EKA + EPA = EKB + EPB

12 m vA2 + mghA = ( 12 m vB2 + 12 I ω2 ¿ + mghB

0 + mghA = ( 12 m vB2 + 12 23mr2v 2

r 2¿+¿0

ghA = 12 vB2 + 13vB2

ghA = 56 vB2

vB2 = 6g h5 → vb = √ 6 gh5sin θ = hs → h = s. sinθ = 3. Sin 30◦ = 3. 12 = 32

vb = √ 6 gh5 = √ 6.10. 325 = √ 905 = √18 = 3√2 m/s

RIGID BODY BALANCE8. If g = 10 m/s, then the rope tension T1 and

T2 is :

Data : g = 10 m/s2

Θ = 53◦

m = 4 kg → W =24. 10 = 20 NProblem : T1 and T2 ?

Answer : ∑Fx = 0 ∑Fy = 0 T2x – T1x = 0 ….. (1) T1y + T2y – W = 0

T1y + T2y = W …… (2)

53◦

T2T1

30◦

T2x – T1x = 0 T2x = T1x T2 cos 37◦ = T1 cos 53◦ T2 0,8 = T1 0,6

T2 = T 10,60,8

T1y + T2y = W T2 = T 10,60,8

T1 sin 53◦ + T2 sin 37◦ = 20 = 120,60,8

T1 0,8 + T2 0,6 = 20 = 7,20,8T1 0,8 + T 10,60,8 0,6 = 20 x 0,8 T2 = 9 NT1 0,64 + T1 0,36 = 12

T1 = 12 N9. System is in balance as on the picture. C rope

Tsin30◦

Homogeneous AB rod 6 m in length with a mass of 12kg.Determine the force of rope tension when the load mass Is 15 kg : A BData : AB = 6 m 3cm x

mrod = 12 kg→Wrod = 120 N 150N m = 15 kg→W= 150 N

Problem : T? Answer : ∑τ = 0 Wrod . XA + Wload . AB – Tsin30◦ . AB = 0 (The axis is on A) 120 . 3 + 150 . 6 – T 12 6

3 = 0 360 + 900 – 3T = 1260 = 3T T = 12603 T = 420 N

10. Both front wheels axis and rear wheels axis of a truck which has mass 1200 kg and distance is 2 m. The center of the truck mass of truck is 1,5 m behind the front wheel. If the earth’s gravitational acceleration is 10 m/s2. the load that borned by the 2 rear wheels of the truck is : Data : mtruck = 1200 kg → Wtruck = 12000 N r = 2 m Center of mass = 1,5 m behind the front wheel g = 10 m/s2

6 m

6

3

6

Problem : Wthe rear wheels ? Answer : -F . r = Wtruck . Center of mass -F . 2 = 12000 . 1,5

-2F = 18000 F = 18000 : -2 F = -9000 N F = 9000 N (negative sign indicates the direction)

11. Pay attention to the picture. If the position of center of gravity wake that shaded is (24,14) then the wide of the shaded wake is : y Data :

The position of center gravity wake that shaded = (24,14)

40 X1 = 12x A1=10x

X2 = 13x A2 = 15 10 Problem : Wide of the shaded wake (A) ?10

x Answer : x0 =

x1 . A1+x2 . A 2A1+A2 LI = p. l = 60 . 10 = 600 cm2

24 = 12x .10x+ 1

3x .15 x

10x+25 xLII = a . t2 = 60 .30

15

2 = 900 cm2

24 = 5 x2+5 x2

10x+15 x Ltot = 600 cm2 + 900 cm2

24 = 10x2

25 x = 1500 cm2

24 = 2x5 2x = 120 x = 60

12. From the picture, the location of the center of gravity of the shaded wide to x-axis is : Y (cm) Data :

X1 = 3 cm Y1 = 1,5 cm A1 = 18 cm2

X2 = 4,5 cm Y2 = 4 cm A2 = 4,5 cm2

Problem : Y0? (the center of gravity of to x-axis)

X

I

II

I

II

3

m1 = 100→ (0,0) m2 = 300→ (40,0)

m3 = 100→ (0,30)

Answer : Y0 = y1 . A1+ y 2. A 2A 1+A2 = 1,518+4 4,518+4,5 = 27+1822,5 = 4522,5 = 2 cm 13. 3 pieces of point masses each 100 g, 300 g, and 100 g occupies the

coordinates (0,0), (40,0), and (0,30). Coordinat of the center point of the system is :Data :

Problem : Xpm and Ypm?Answer : Xpm = m1. x1+m 2.x 2+m3. x3m1+m 2+m3 Ypm = m1. y1+m2. y 2+m 3. y 3m1+m2+m3

= 100.0+300.40+100.0100+300+100 = 100.0+300.0+100.30100+300+100

= 0+12000+0500 = 0+0+3000500

= 12000500 = 3000500 Xpm = 24 Ypm = 6

14. Determine the distance of the center of gravity from a wire in the form of arc which calculated from the center of curvature, if the angle of the arc center is 60◦ and its radius is R : Data : θ = 60◦ arc r = RProblem : Y0 (the distance of the center of gravity)?Answer : Y0 =

The lenght of ABArc AB . R

=

R1632πR . R

= R2

13π R

= 3Rπ

FLUID STATICS15. A tube with high 50 cm filled full of water. The density is water is1

gr/cm3. If g = 10 m/s2, the hydrostatic pressure in the basic tube is :Data : ρ = 1 gr/cm3 = 1000 kg/m3

h = 50 cm = 0,5 mg = 10 m/s2

Problem : Ph ? Answer : Ph = ρ . g. h

= 1000 . 10. 0,5Ph = 5000 N/m2 = 5x103 N/m2

16. Mercury in related vessel has surface difference 2 cm and the height liquid in the left leg is 25 cm. If the density of mercury is 13.6, How much does density of the liquid?

Data : h1 = 25 cm h2 = 2cm ρ2 = 13,6 g/cm3 Problem : ρ1?Answer : ρ1h1 = ρ2h2

25 ρ1 = 13,6 . 225 ρ1 = 27,2 ρ1 = 27,2 : 25 ρ1 = 1,088 g/cm3 = 1088 kg/m3

17. F1 Cross-sectional diameter of the vacuum has a ratio 1 : 10. If F1=20N, How much does load that can be transported: A1 A2

Data : d1 : d2 = 1 : 10 → r1 : r2 = 0,5 : 5 F2 F 1 = 20 N Problem : F2?Answer : F1A1 = F2A 2

F1π r2

= F2π r2

20π 0,52

= F2π 52

200,25 = F 225

0,25F2 = 500 F2 = 5000,25 = 2000 N

18. A cylindrical hydraulic pump with each diameter 8 cm and 20 cm. If the small tube pressed with force 500 N, then the force that can be produced by the big tube is :Data : d1 = 8 cm → r = 4 cm = 0,04 m

2 cm

25 cm

d2 = 20 cm → r = 10 cm = 0,1 mF1 = 500 N

Problem : F2 ?Answer : F1A1 = F2A 2

F1π r2

= F2π r2

500π 0,042

= F2π 0,12

5000,0016 = F20,01

0,0016F2 = 5F2 = 5

0,0016 = 3125 N

19. Piece coin dipped in a fluid A with ρA = 0.8 g/cm3 is experience upward force as big as FA and if dipped in a fluid B with ρB = 0.7 g/cm3 is experience Archimedes force as big as FB. How much does FA/FB?Data : ρA = 0.8 g/cm3

ρB = 0.7 g/cm3

Problem : FAFB = ρV gρV g

FAFB = 0,8 .V .100,7 .V .10

FAFB = 8720. An object placed in water. Turns out 25% of the

objects floating above the water surface. How much does density of the object?Data : hb = 100% hbf = 100% - 25% = 75% ρf = 1000 kg/m3

Problem : ρb?Answer : ρb = hbfhb . ρf

= 75%100% . 1000

= 75100 . 100010 = 75. 10 ρb = 750 kg/m3

21. Mercury has a density of 13,6 g/cm3. At the mercury placed the small tube with diameter 5 mm. It turns out that the mercury in the tube 2

cm lower than the mercury in the outer tube. If the contact angle is 127◦, How much does surface tension of mercury?Data : ρm = 13,6 g/cm3 = 13600 kg/m3

d = 5 mm = 5x10-3 m → r = 2,5x10-3 m h = 2 cm = 2x10-2 m θ = 127◦

g = 10 m/s2

Problem: γ ?Answer : h = 2 γ cosθρgr

2x10-2 = 2 . γ cos 12713600.10 .2,5 x10−3

2x10-2 . 13600 . 10 . 2,5x10-3 = 2 . γ−0,6 6,8 = -1,2 γ γ = 6,8−1,2 γ = -5,67 N/m = 5,67 N/m

22. An object has weight in the air 4,9 N. But when put in kerosene the weight change into 4,74 N. The upward force that experienced by the object is :Data : W = 4,9 N

Wbk = 4,74 NProblem : FA?Answer : Wbk = W - FA

FA = W - Wbk

= 4,9 – 4,74FA = 0,16 N

23. A tube with diameter 0,4 cm inseeted vertically into the water. Its contact angle is 60◦. If the surface tension of water is 0,5 N/m and g = 10 m/s2. So, water in the tube will rise by :Data : d = 0,4 cm → r = 0,2 cm = 0,002 m

Θ = 60◦

γ = 0,5 N/mg = 10 m/s2

ρ = 1000 kg/m3

Problem : h?Answer : h = 2 γ cosθρgr

= 2 .0,5.cos601000.10 .0,002

= 2.0,5 .0,51000.10 .0,002

A2 v2A1 v1

= 0,520h = 0,025 m

FLUID DYNAMICS24. A pipe each end has a radius of 1 cm and 1,5 cm. If the velocity of

water in a small cross section is 9 m/s, how much does the velocity of water in a large cross-section?Data : r1 = 1 cm= 1x10-2 m r2 = 1,5 cm = 1,5x10-2 m v1 = 9 m/sProblem : v2?Answer : A1 . v1 = A2. v2

Πr12 . v1 = πr22 . v2

π(1x10-2)2 . 9 = π(1,5x10-2)2 . v2

9x10-4 = 2,25x10-4 v2

v2 = 9x 10−4

2,25 x 10−4

v2 = 4 m/s25. The velocity of water in a pipe with diameter 6

cm is 0,25 m/s. That the water out with speed 4 m/s, the pipe must be connected with another pipe, which one end is a smaller diameter. So, the diameter of that pipe is?Data : d1 = 6 cm = 6x10-2 m → r = 3x10-2 m v1 = 0,25 m/s v2 = 4 m/sProblem : d2?Answer : A1 . v1 = A2. v2

Πr12 . v1 = πr22 . v2

π(3x10-2)2 . 0,25 = πr2 . 4 2,25x10-4 = r2 . 4 r2 =

2,25x 10−4

4 r2 = 5,625 x10−5 r = √5,625 x 10−5 r = 7,5x10-3 m d = 2 x 7,5x10-3 = 0,015 m = 1,5 cm

26.80 cm

In the following picture, water flowing in the venturi meter. If g = 10 m/s2, cross-sectional areas A1 and A2, respectively 5 cm2 and 3 cm2. How much does the water velocity (v1) entering the venturi meter?Data : A1 = 5 cm2 = 5x10-4 m2

A2 = 3 cm2 = 3x10-4 m2

h = 80 cm = 80x10-2 m g = 10m/s2

Problem : v1?

Answer : v1 = A2√ 2ghA12−A22

= 3x10-4 √ 2 .10 .8 x 10−2

(5 x 10−4 )2−(3 x10−4 )2

= 3x10-4 √ 1625 x 10−8−9x 10−8

= 3x10-4 √ 1616 x 10−8

= 3x10-4 . 1x104

v1 = 3 m/s27. An aircraft moving at a certain speedso that air

passing through the upper and lower wing with large surface area is 60 m2 moving at a pace respectively 320 m/s and 290 m/s. If the air density of 1,3 kg/m3 how much does the lifting force of the plane's wing?Data : A = 60 m2

ρ = 1,3 kg/m3

v1 = 290 m/s (lower)v2 = 320 m/s (upper)

Problem : F1-F2?Answer : F1-F2 = 12 ρ A (v22- v12)

= 12 1,3 6030 (3202- 2902) = 1. 1,3 . 30 (102400-84100) = 39 . 18300 F1-F2 = 713.700 N

28. A water pipe, the cross-section is 0,5 cm2. If the speed of the water is 1 m/s, then the volume of water that comes out for 5 minutes is :Data : A = 0,5 cm2 = 5x10-5 m2

v = 1 m/s t = 5 minutes = 300 sProblem : V?

Answer : Q = A.v = 5x10-5 . 1 = 5x10-5

Q = V∆ t V = Q . ∆ t = 5x10-5 300 V = 0,015 m3

29. A water pipe each end of radius Ra = 2 cm and Rb = 4 cm. If speed of water in cross-section A is 1,6 m/s, then speed of water in cross-section B is?Data : Ra = 2 cm = 0,02 m

Rb = 4 cm = 0,04 mva = 1,6 m/s

Problem : vb? Answer : Aa . va = Ab. vb Πra2 . va = πrb2 . vb π(0,02)2 . 1,6 = π(0,04)2 . vb 6,4x10-4 = 16x10-4 vb Vb = 6,4 x10

−4

16 x10−4

vb = 0,4 m/s30. A waterfall with high 20 m is used for

hydropower (PLTA). Every second the water flowing at 10 m3. If the generator efficiency 55% and acceleration of gravity g = 10 m/s2, so the average power that can be produced is : (in kW)Data : h = 20 m

ρ = 1000 kg/m3

g = 10 m/s2

η = 55%V = 10 m3

Problem : Plistrik?Answer : Q = Vt = 101 = 10 m3/s

P = η ρ Q g h = 55% 1000 10 10 20

= 0,55 1000 10 10 20 = 1100000 W

Plistrik = 1100 kW