Physic 231 Lecture 21 • • •

Main points of last lecture: Analogies between rotational and translational motion Solutions involving:

• • •

Main points of today’s lecture: Angular momentum L = Iω Tensile stress and strain

•

Shear stress and strain:

I

τ = Iα KErot

1 2 = Iω 2

L = Iω •

Rolling motion: 1 1 KEtotal = Mv 2 + Iω 2 2 2

∆F ∆L =Y A L0

∆F ∆x =S A h • Bulk stress and strain: ∆F ∆V = ∆P = B A V • Pressure in fluids:

P=

F ; Pbot = Ptop + ρgh A

Conceptual quiz •

A solid disk and a ring roll down an incline.The ring is slower than the disk if – a) mring = mdisk, where m is the inertial mass. – b) rring = rdisk, where r is the radius. – c) mring = mdisk and rring = rdisk. – d) The ring is always slower regardless of the relative values of m and r.

v=

Mgh 1 I  1 M +  2 2 R 2 

ring : I = MR 2 1 disk : I = MR 2 2

ring : v = gh 4 disk : v = gh 3

quiz •

A diver leaps from the 10 m platform and executes a triple forward somersault dive. While in the tuck, the diver has a moment of inertia of about 3.3 kg⋅m2 and during the final layout, the diver stretches her body and increases her moment of inertia to 10 kg⋅m2. If the diver achieves an angular velocity of 2 rev/s while in the tuck, what is the angular velocity during the layout? – a) 6 rev/s I0 – b) 66 rev/s ω f = ω0 If – c) 0.66 rev/s – d 0.17 rev/s 3 .3

ωf =

10

2rad / s = 0.66rad / s

Stress and strain in solids •

All materials deform when under stress, ie. when a force is applied to them. The atoms in a solid are held in place by forces that act like springs.

•

If you pull on a bar of cross-sectional area A with a force F, the bar will stretch a distance ∆L. In this case, the force is perpendicular to the surface, and the force F and distance ∆L are related by Young’s modulus Y:

F ∆L F ∆L =Y ; = Stress; = Strain A L0 A L0 •

The force will be linearly proportional to the strain up to the elastic limit. Beyond this point, the bar will be permanently deformed. Stress units Pa=N/m2.

Example •

Bone has a Young’s modulus of about 18 x 109 Pa. Under compression, it can withstand a stress of about 160 x 106 Pa before breaking. Assume that a femur (thigh bone) is 0.50 m long and calculate the amount of compression this bone can withstand before breaking. – a) 4.4 mm – b) 6.4 mm – c) 1.6 cm – d) 2.0 cm

∆L F = 160 x106 N / m 2 = Y L0 A F L0 0.5m 6 2 ⇒ ∆L = = 160 x10 N / m = 4.4mm 9 2 AY 18 x10 N / m

Shear •

You can also exert a force on an object with a force that is parallel to its surface. Friction and viscous drag are two example of such forces, which act on the surface of the object and can cause a shear stress.

•

Such shear forces cause the kind of deformation indicated in the figure. Over a distance h, from one surface, there is a shift ∆x in the direction of the applied force. This shift is governed by the shear modulus S.

F ∆x F ∆x =S ; = Shear stress; = Shear strain h A h A •

The shear modulus governs how easily an object can be bent.

Example •

A child slides across a floor in a pair of rubber-soled shoes. The friction force acting on each foot is 20 N, the footprint area of each foot is 14 cm2, and the thickness of the soles is 5.0 mm. Find the horizontal distance traveled by the sheared face of the sole. The shear modulus of the rubber is 3.0 x 106 Pa.

F ∆x =S h A F h 20 N 0.005m −5 ⇒ ∆x = = = 2 . 4 x 10 m 2 6 2 A S .0014m 3.0 x10 N / m

Pressure in fluids •

Consider a vessel filled with liquid and a smaller rectangular volume V of the same liquid contained within the vessel. If the liquid is at rest, there can be no shear force on V, because it would cause the liquid to move, but there is a pressure force perpendicular to v all 6 surfaces. In equilibrium : ∑ F = 0

x : Fleft − Fright = 0; Fleft = Pleft Ax ; Fright = Pright Ax ⇒ Pleft = Pright

y z

x

z : F front − Fback = 0; F front = Pfront Az ; Fback = Pback Az ⇒ Pfront = Pback W y : Fbot − Ftop − W = 0; Fbot = Pbot Ay ; Ftop = Ptop Ay ⇒ Pbot = Ptop + Ay • More detailed considerations show that pressure always exerts a force perpendicular to a surface and that pressure is the same everywhere at the same height, i.e. Pfront =Pback, and Pleft =Pright and Pfront= Pright, etc. • Defining density ρ of the liquid by ρ=M/V we can simplify the dependence of P on height as follows:

W = Mg = ρVg = ρAy hg Pbot = Ptop +

ρAy hg Ay

⇒ Pbot = Ptop + ρgh

Pascal’s principal • •

•

Because F=PA, one can make hydraulic machines that can generate large forces. Units: One atmosphere (1 atm) = – 76.0 cm of mercury – 1.013 x 105 Pa = 1.013 x 105 N/m2 – 14.7 lb/in

Quiz: A simple hydraulic lift is employed to lift a truck that weighs 30000N. If A1=1 cm2 and A2= 1000 cm2, what force F1 is needed to lift the truck? – a) 3N P1 = P2 – b) 300N F1 F2 F2 2 30000 N = ⇒ F = A = 1 cm = 30 N 1 1 – c) 30N 2 A1 A2 A2 1000cm – d) .0.33N

Pressure and Depth equation

Pbot = Ptop + ρgh •

•

If the top is open, Pbot is normal atmospheric pressure – 1.013 x 105 Pa = 14.7 lb/in2 The pressure does not depend upon the shape of the container

Bulk modulus •

If you drop a metal block in the ocean, it will sink to bottom where the pressure is much greater than that of the atmosphere. This will compress the block; the amount of compression is governed by the bulk modulus B.

∆P = B •

∆V ∆V ; ∆P = pressure change; = Volume strain V V

Example: A steel block with a volume of 8 cm3 is dropped in the ocean and comes to rest at the bottom, which is 3 km below the surface. 1) What is the pressure at the bottom of the ocean? 2) By what amount ∆V is the volume decrease. (The Bulk modulus for steel is 1.6x1011 N/m2. The density of water is 1000kg/m3. 1 atm. =1.01x105 N/m2) ∆V 5 2 b) P B ∆ = – 1a) 2x10 N/m V – 1b) 4x105 N/m2 a) Pbot = Ptop + ρgh6 V Pbot − Ptop ∆V = – 1c) 2x10 N/m2 5 3 2 B Pbot = 1–.0 x1d) Pa +7 1000 10 3x10 N/m2kg / m 9.8m / s 3000m 3

(

Pbot = 3.0 x107 N / m 2

)

(

)

8cm 7 2 3 x 10 N / m 1.6 x1011 N / m 2 = 1.5 x10−3 cm 3 =