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PHYSICS (PHY101)

TOTAL MARKS: 0

Due Date: 29/05/2017

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Question. No: 1

A pendulum consists of a uniform disk with radius 10.0 cm and mass 500 g attached to a uniform rod with length 500 mm and mass 270 g (see figure below). (a) Calculate the rotational inertia of the pendulum about the pivot. (b) What is the distance between the pivot and the center of mass of the pendulum? (c) Calculate the period of oscillation.                                                    Marks =10

 

                                                                                                                                                                                                                 

 

Question. No: 2

 

Explain the following terms.                                                                Marks = 2*5 + 5 =15

a)      Steady flow, Unsteady flow, Compressible fluid, Incompressible fluid and Viscous fluid.

b)      Why do airplane pilots prefer to take off into the wind?

 

                                                 

 

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have any solution plz

ASSIGNMENT # 2 (PHY101)

STUDENT ID: bc160404071

Solution 1.

 Itot = Iro + Ido = (1/3)*Mr*Lr^2 + [ Idcm + Md*(Rd + Lr)^2 ] 
 PART (a)

 Itot = (1/3)*Mr*Lr^2 + [ (1/2)*Md*Rd^2 + Md*(Rd + Lr)^2 ] 
where the parallel axis theorem has been used to get the moment of inertia 
Ido of the disk about the pivot point O. Substituting the values 
Mr = 270 g = 0.270 kg

 Lr = 500 mm = 0.5 m

Md = 500 g = 0.500 kg , 
Rd = 10 cm = 0.10 m

we get

Itot = 0.205kg*m^2 
ANSWER: 0.205 kg*m^2 

PART (B) 
The position of the center of mass relative to the pivot point O is given by

Ycm = [ Mr*(Lr/2) + Md*(Lr +Rd) ] / [ Mr + Md ] 
Substituting the values, we get

Ycm = 47.7 cm 
ANSWER:Ycm= 47.7 cm

PART (C) 
The period of oscillation of a physical pendulum is given by 
T = 2*π*sqrt [ I /(M*g* D ) ] 
where D is the distance of the mass M from the axis of rotation and I is 
the moment of inertia of the disk about the axis of rotation. Substituting 
the values

 I = 0.205 kg*m^2

M = 0.770 kg

D = Ycm = 47.7 cm 
we get

 T = 1.53 s 
ANSWER: 1.53 s

………….

SOLUTION  2.

a)1.  A flow in which the velocity of the fluid at a particular fixed point does not change with time also called stationary flow.

a)2. If at any point in the fluid, the conditions change with time, the flow is described as Unsteady Flow.

a)3. Compressible fluid (gas dynamics) is that deals with flows having significant changes in fluid density. Gases mostly display such behavior.

a)4. incompressible fluid (isochoric flow) refers to a fluid in which the material density is constant within a fluid.

a)5. The viscosity of a fluid is a measure of its resistance to gradual deformation by shear stress or tensile stress. For liquids, it corresponds to the informal concept of "thickness"; for example, honey has a much higher viscosity than water.

 

 

b) Airplanes fly thanks to lifting force that is generated by the wings. The wings do this by moving forward through the air. All else being equal, the faster the wings move through the air, the greater the lift. 
Flying into the wind causes the wings to move through the air more quickly without increasing the speed of the aircraft in relation to the ground. This means that an aircraft flying into the wing can get into the air more quickly.

 

 

 

 

 

Hey brother what is meant by the non-graded assignment?

means physically u'll not get the marks but the image. :)

assignment

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