Mechanics Exercise Class Ⅱ

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1 Mechanics Exercise Class Ⅱ

2 Review Ⅰ Angular Quantities Angular displacement Ⅱ Rotational Inertia
Angular velocity Angular acceleration Ⅱ Rotational Inertia Ⅲ The Parallel-Axis Theorem

3 Angular Momentum of a System of Particles
Review Ⅳ The Kinetic Energy of the rolling （克尼希定理，漆安慎p240） Ⅴ Torque Ⅵ Angular Momentum Angular Momentum of a Particle Angular Momentum of a System of Particles Angular Momentum of a Rigid Body

4 Review Ⅶ Work and Rotational Kinetic Energy
Ⅷ Newton’s Second Law in Angular Form Particle: System of Particles: A Rigid Body With fixed axis: Ⅸ Conservation of Angular Momentum a constant Ⅶ Work and Rotational Kinetic Energy

5 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? P Solution : The key idea here is that the angular acceleration is constant so we can use the rotation equation: ( Eq11-13) Substituting the given data and solving for we find,

6 (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,

7 The key idea here is each blade can be
[2.] Each of the three helicopter rotor blades shown in the Fig. is 5.20m long and has a mass of 240kg. The rotor is rotating at350rev/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? P242 38 Solution : 5.20m 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 (p227 Tab) (2) Do it by yourselves and remember it!

8 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

9 The law of conservation of mechanical energy
[3.]A uniform spherical shell of mass M and radius R rotates about a vertical axis on frictionless bearings. A massless cord passes around the shell, over a pulley of rotational inertia I and radius r, and is attached to a small object of mass m. There is no friction on the pulley’s axel; the cord does not slip on the pulley. What is the speed of the object after it falls a distance h from rest? Use energy consideration. P The law of conservation of mechanical energy Solution : The key idea is we consider the spherical shell, the pulley and the small object as a system, so the system is isolated ,its total energy can not change. We apply the law of conservation of energy to this system potential energy rotational kinetic energy (1) 1 2

10 The second key idea is the cord does not slip on the pulley , so the spherical shell and the pulley has the same liner speed and we can write (2) Combine the Eq (1) and (2) and substitute the ,we can get speed of the object

11 The law of Conservation of Angular Momentum
[4 .] A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axis through the center of the disk. The platform has a mass of 150kg, a radius of 2.0m, and a rotational inertia of 300kg.m2about the axis of rotation. A 60kg student walks slowly from the rim of the platform toward the center .If the angular speed of the system is 1.5rad/s when the student starts at the rim , what is the angular speed when she is 0.5m from the center? P272 54 Solution : The law of Conservation of Angular Momentum R The key idea is we consider the platform and the student as a system, the external Torque acting on the system is zero, so the angular momentum of the system remains constant. (1)

12 R Substituting the and into the Eq (1) and solving for we find The law of conservation of angular momentum can only be used in inertial system !

13 [例5]已知：轻杆，m 1 = m , m 2 = 4m , 油灰球 m， m 以水平速度v 0 撞击 m 2 ，发生完全非弹性碰撞
m v 0 = (m + m 2 ) v 解二： m 和 (m1 + m 2 )系统动量守 恒 m v 0 = (m + m 1 + m 2 ) v 解三： m v 0 = (m + m 2 ) v + m 1 • 2v 以上解法对不对？ m2 m1 m A

14 因为相撞时轴A作用力不能忽略不计，故系统动量不守恒。 因为重力、轴作用力过轴，对轴力矩为零，故系统角动量守恒。
m2 m1 m A Ny Nx 由此列出以下方程： 或： 得：

15 轴作用力不能忽略，动量不守恒，但对 o 轴合力矩为零，角动量守恒
注意：区分两类冲击摆 （1） o l m M 质点 质 点 柔绳无切向力 （2） Fx Fy 质点 定轴刚体（不能简化为质点） 水平方向： Fx =0 ， px 守恒 m v 0 = ( m + M ) v 对 o 点： ， 守恒 m v 0 l = ( m + M ) v l 轴作用力不能忽略，动量不守恒，但对 o 轴合力矩为零，角动量守恒 (漆安慎P256, 7.4.2)

16 = = [例6] 计算下列刚体的转动惯量I:P254 7.3.2 7.3.3 1.图示实验用的摆， ，
近似认为圆形部分为均质圆盘，长杆部分为均质细 杆.求：对过悬点且与摆面垂直的轴线的转动惯量. 解: ） 均匀细杆（ 均匀圆柱（ 将摆分为两部分： 则 = 用平行轴定理 = I= = 2.65

17 2.在质量为M半径为R的均质圆盘上挖出半径为r的两个圆孔，
与盘面垂直的轴线的转动惯量. 解： 设未挖两个圆孔时大圆盘转动惯量为I 如图半径为r的小圆盘转动惯量为 和 则有 （ ）

18 [例7]质量为m长为 的均质杆，其B端放在桌面上，A端用手支住，使杆成水平
[例7]质量为m长为 的均质杆，其B端放在桌面上，A端用手支住，使杆成水平.突然释放A端，在此瞬时，求：（1）杆质心的加速度，（2）杆B端所受的力. P [解 答] 取杆为隔离体，受力分析及建立 坐标如图。规定逆时针为转动正方向。依据质心运动定理有： B 依据转动定理： ？ 依据角量与线量关系：

19 此外， B 联立上述四个方程求得：