11.53. Global warming is a cause for concern because even small changes in the Earth’s temperature can have significant consequences. For example, if the Earth’s polar ice caps were to melt entirely, the resulting additional water in the oceans would flood many coastal cities. Would it appreciably change the length of a day? Calculate the resulting change in the duration of one day. Model the polar ice as having mass 2.30 x1019 kg and forming two flat disks of radius 6.00 x105m. Assume the water spreads into an unbroken thin spherical shell after it melts.

Penyelesaian

The moment of inertia of the rest of the earth is

I=25

M R2=25

5.98 X 1024 kg (6,37 x106 )2=9.71 x1037kg . m2

For the original ice disks

I=12

M r 2=12

2.30 x 1019 kg (6 x105m )2=4.14 x1030kg .m2

For the final thin shell of water

I=23

M r2=23

2.30 x1019 kg (6.37 x 106 m )2=6.22 x1032 kg . m2

Conservation of angular momentum for the spinning planet is expressed by I iωi=I f ωf

( 4.14 x1030+9.71 x1037) 2 π86400 s

=(6.22 x1032+9.71 x1037 ) 2π(86400 s+δ )

(1+ δ86400 s )(1+ 414 x 1030

9.71 x1037 )=(1+ 6.22 x1032

9,71 x1037 )δ

86400 s=6.22 x 1032

9.71 x 1037−4.14 x1030

9.71 x 1037

δ=0,550 s

11.54. A solid cube of wood of side 2a and mass M is resting on a horizontal surface. The cube is constrained to rotate about an axis AB (Fig. P11.54). A bullet of mass m and speed v is shot at the face opposite ABCD at a height of 4a/3. The bullet becomes embedded in the cube. Find the minimum value of v required to tip the cube so that it falls on face ABCD. Assume m ≪M.

Penyelesaian

For the cube to tip over, the center of mass (CM) must rise so that it is over the axis of rotation AB .To to this , the CM must be raised a distance of a (√2−1 ).

∴Mga (√2−1 )=12

I cube ω2

From conservation of angular momentum

4 a3

mv=( 8 M a2

3 )ω

ω= mv2 Ma

12 ( 8 M a

2

3 ) m2 v2

4 M 2 a2 =Mga (√2−1 )

v=Mm

√3 ga (√2−1 )

14.62. In about 1657 Otto von Guericke, inventor of the air pump, evacuated a sphere made of two brass hemispheres. Two teams of eight horses each could pull the hemispheres apart only on some trials, and then “with greatest difficulty,” with the resulting sound likened to a cannon firing (Fig. P14.62). (a) Show that the force F required to pull the evacuated hemispheres apart isπ R2 ( P0−P ), where R is the radius of the hemispheres and P is the pressure inside the

hemispheres, which is much less than P0. (b) Determine the force if P=0.100 P0 and R=0.300 m.

Penyelesaian :(a) The pressure on the surface of the two hemispheres is constant at all points, and the force on

each element of surface area is directed along the radius of the hemispheres.The applied force along the axis must balance the force on the “effective” area, which is the projection of the actual surface onto a plane perpendicular to the x axis.

A=π R2

Therefore F=( P0−P ) π R2

(b) For the values given F=( P0−0.100 P0 ) [ π (0.300 )2 ]=0.254 P0=2.58 x 104 N

15.60. A particle with a mass of 0.500 kg is attached to a spring with a force constant of 50.0 N/m. At time t=0 the particle has its maximum speed of 20.0 m/s and is moving to the left. (a) Determine the particle’s equation of motion, specifying its position as a function of time. (b) Where in the motion is the potential energy three times the kinetic energy? (c) Find the length of a simple pendulum with the same period. (d) Find the minimum time interval required for the particle to move from x=0¿ x=1.00 m.

Penyelesaian :(a)I n x=A cos ωt+ϕ v=−ωAsin (ωt+ϕ )we have at t=0 v=−ωAsin ϕ=−vmax

this requires ϕ=900 , so x=A cos ( ωt+900 )¿ this is equivalent ¿ x=−A sin ωt

Nummerically we have ω=√ km

=√ 50 N /m0,5 kg

=10 s−1

And vmax=ω Α 20ms=(10 s1 ) Α A=2 m

So x=(−2 m )sin [ (10 s1 ) t ](b)

¿ 12

m v2+ 12

k x2=12

k A2

12

k x2=3( 12

m v2)Implies

13

12

k x2+ 12

k x2=12

k A2

43

x2=A2

x=±√ 34

A=± 0.866 A=±1.73m