mechanics exercise class Ⅲ. brief review 2 rotational inertia 3 the parallel-axis theorem angular...

Mechanics Exercise Class Ⅲ

Brief Review

2 Rotational Inertia 2 2di ii m

I m r r m

3 The Parallel-Axis Theorem

2 1

avg

ins

tddt

avg

ins

tddt

2comI I Mh

Angular displacement

Angular velocity Angular acceleration

1 Angular Quantities

Ⅰ Rotation and Angular Momentum

4 The Kinetic Energy of the rolling

2 21 1

2 2com comK I Mv

Angular Momentum of a System of Particles

5 Torque

r F ; rF sin rF r F

6 Angular Momentum

Angular Momentum of a Particler mv rmv sin rmv r mv

L I Angular Momentum of a Rigid Body

1 2

1

n

n i

i

L ....

����������������������������������������������������������������������

8 Newton’s Second Law in Angular Form

Particle: d

dnet

t

d

dnet

L

t

��������������

System of Particles:

A Rigid Body With fixed axis:

9 Conservation of Angular Momentum

0

L ��������������

a constant

7 Work and Rotational Kinetic Energy

2 21

2f

if iW d K I( )

Inet

Ⅱ Gravitation

1 The Law of Gravitation

221

r

mmGF

2 Gravitational Potential Energy

r

GMmU

3 Escape Speed R

GMv

2

1. A wheel rotating about a fixed axis through its center has a constant angular acceleration of 4.0rad/s2. In a certain 4.0s interval the wheel turns through an angle of 80rad. (a) What is the angular velocity of the wheel at the start of the 4.0s interval? (b) Assuming that the wheel starts from rest, how long is it in motion at the start of the 4.0 interval?

Solution :(a)The key idea here is that the angular acceleration is constant so we can use the rotation equation:

t 20 0

1

2

Substituting the given data and solving for we find, 0

srad

t

t/12

4

4421

8021 22

0

0

(b) If the wheel starts from rest, the key idea here is that the

initial angular velocity is 0, so we can use the rotation

equation:

t 21

2

Inserting the given data and solving for ,we can obtain,

rad 214 4 32

2

5.20m

2. Each of the three helicopter rotor blades shown in the figure is 5.20 m long and has a mass of 240 kg. The rotor is rotating at 350 rev/min. (a) What is the rotational inertia of the rotor assembly about the axis of rotation ? (Each blade can be considered to be a thin rod rotated about one end).

(b) What is the total kinetic energy of rotation?

Solution :

(a)The key idea here is each blade can be considered to be a thin rod rotated about oneend, so the total rotational inertia of the rotorblades can be written as :

( )I I I 1 2 3I I I I I 1 2 3 13 (1)

I ml 21

1

3

The second key idea is the rotational inertia of a thin rod rotated about its end is

(2)

. . .I I ml kg m 2 2 213 240 5 2 6489 6

Combine the Eq (1) and (2) and substitute the given data ,we can get the total rotational inertia

I(b) Substituting and into Eq 11-27, we find

.

.

K I

J

21

21 350 2

6489 62 601188678 4

3. (P206.43) In Fig.6-79, a small 50g block slides down a frictionless surface through height h=20 cm and then sticks to a uniform rod of mass 100 g and length 40 cm. The rod pivots about point O through angle θ before momentarily stopping. Find θ .

Solution : The whole process can be divided into three parts :(1) The small block slides down the frictionless surface through height h, in this part only the gravitational force being a conservative force, does work, so the law of conservation of mechanic energy holds

2111 2

1vmghm

Where v1 is the speed of the block before it collides with the rod

(2) The small block collides with the rod and sticks to it. During this interaction there is no net torque acting on the block–rod system relative to the point O, the angular momentum of the system is conserved (Note: since there is a net force acting on the rod at point by the pivot, the law of conservation of linear momentum does not hold! )

ILvm 11

Where ωis the angular speed of the system about point O just after the collision. I is the rotational inertia of the block-rod system about point O, which is

2 22 1

1

3I m L m L

(3) The block-rod system swings up until it momentarily stops . During this process the mechanic energy of the system is conserved , we thus write

comhgmmI )(2

121

2

Where Δh is the height change of the center of mass of the block-rod system in the swing up process. In the vertical pisition the center of mass of the system is below point O at

1 2

1 2

(0.5 ) (0.05 )(0.4 ) (0.1 )(0.2 )0.2667

0.05 0.10com

m L m L kg m kg mL m

m m kg kg

So )cos1( comcom Lh

cos 0.85 We have

Thus the angle θ we look for is

31.8

4. A 150.0 kg rocket moving radially outward from Earth has a speed of 3.70 km/s. When its engine shuts off 200 km above Earth’s surface. (a) Assuming negligible air drag , find the rocket’s kinetic energy of the rocket 1000 km above the Earth’s surface. (b) What maximum height above the surface is reached by the rocket?

solutionsolution

21

1

1

2 ( )i

GMmE mv

R h

When the rocket is 200km above the Earth’s surface, the energy is

When the rocket is 1000km above the Earth’s surface, the energy is

2( )f f

GMmE K

R h

(a)The key idea is that the mechanical energy of the rocket is constant

i i f fK U K U (1)(1)

(2)(2)

(3)(3)

(b) (b) The key idea is that when the rocket reaches the maximum height , the kinetic energy of it is zero , the Eq (4) is changed as

max 1034.9h km

21

1 max

1 1 1( ) 0

2fK mv GMmR h R h

Rewriting the above equation , we obtain

21

1 max

1 1

2

GMmv GMm

R h R h

Substituting the known data, we finally get Substituting the known data, we finally get

Combining the Eq(1)-(3), we can get

2 71

1 2

1 1 1( ) 3.8 10

2fK mv GMm JR h R h

(4)

5. A geosynchronous satellite is one that stays above the same point on the Earth, which is possible only if it is above a point on the equator. Such satellites are used for such purpose as cable TV transmission, for weather forecasting, and as communication relays. Determine (a) the height above the Earth’s surface such a satellite must orbit and (b) such a satellite’s speed.

solutionsolution

(a) The only force on the satellite is gravity, assuming the satellite moves in a circle, then

r

vm

r

mmG Sat

ESat2

2

And the speed of the satellite must be

T

rv

2

where T=1 day=(24h)(3600s/h)=86400 s. We put this into the first equation above and obtain (after canceling mSat on both sides):

2

2

2

2

rT

r

r

mG E

We solve for r:

2

242211

23

4

8640001098.5/1067.6

4 skgkgmNTGm

r E

3221054.7 mand, taking the cube root, r=4.23107 m, or 42300 km from the Earth’s center. We substrate the Earth’s radius of 6380 km to find that the satellite must orbit about 36000 km (about 6 rE ) above the Earth’s surface.

(b) sm

m

kgkgmN

r

Gmv E /3070

1023.4

1098.5/1067.67

242211

We get the same result if we use Trv /2