1. Mechanics Exercise Class Ⅲ

2. Brief Review Ⅰ Rotation and Angular Momentum 1 Angular Quantities
Angular displacement
Angular velocity
Angular acceleration
2 Rotational Inertia
3 The Parallel-Axis Theorem

3. Angular Momentum of a System of Particles
4 The Kinetic Energy of the rolling
5 Torque
6 Angular Momentum
Angular Momentum of a Particle
Angular Momentum of a System of Particles
Angular Momentum of a Rigid Body

4. 7 Work and Rotational Kinetic Energy
8 Newton’s Second Law in Angular Form
Particle:
System of Particles:
A Rigid Body With fixed axis:
9 Conservation of Angular Momentum
a constant

5. Ⅱ Gravitation
1 The Law of Gravitation
2 Gravitational Potential Energy
3 Escape Speed

6. The key idea here is that the angular acceleration is
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 :
The key idea here is that the angular acceleration is
constant so we can use the rotation equation:
Substituting the given data and solving for we find, w

7. (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:
Inserting the given data and solving for ,we can obtain,

8. (b) What is the total kinetic energy of rotation?
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?
5.20m
Solution :
The key idea here is each blade can be
considered to be a thin rod rotated about one
end, so the total rotational inertia of the rotor
blades can be written as :
(1)
The second key idea is the rotational inertia of a thin rod rotated
about its end is
(2)

9. Combine the Eq (1) and (2) and substitute the given data ,we can
get the total rotational inertia
(b) Substituting and into Eq 11-27, we find

10. 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
Where v1 is the speed of the block before it collides with the rod

11. (2) The small block collides with the rod and sticks to it
(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! )
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
(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

12. 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 So
We have
Thus the angle θ we look for is

13. The key idea is that the mechanical energy of the rocket is constant
4. A 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?
solution
The key idea is that the mechanical energy of
the rocket is constant
(1)
When the rocket is 200km above the Earth’s
surface, the energy is
(2)
When the rocket is 1000km above the Earth’s surface, the energy is
(3)

14. Combining the Eq(1)-(3), we can get
(4)
(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
Rewriting the above equation , we obtain
Substituting the known data, we finally get

15. And the speed of the satellite must be
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.
solution
(a) The only force on the satellite is gravity, assuming the satellite moves in a circle, then
And the speed of the satellite must be
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):

16. We solve for r:
and, taking the cube root, r=4.23107 m, or km from the Earth’s center. We substrate the Earth’s radius of 6380 km to find that the satellite must orbit about km (about 6 rE ) above the Earth’s surface.
(b)
We get the same result if we use