Mechanics Exercise Class Ⅲ

Review Ⅰ The Requirements of Equilibrium Balance of forces Balance of torques In xy plane Ⅱ Elastic Moduli Tension and Compression Shearing Hydraulic Stress Young’s modulus Shear modulus Bulk modulus

Review Ⅲ The Law of Gravitation Ⅳ Gravitational Potential Energy Ⅴ Kepler’s Law 1. The law of orbits 2. The law of areas 3. The law of periods All planets move in elliptical orbits with the sun at one focus.

Ⅸ Bernoulli’s Equation Review Ⅵ Pascal’s Principle Ⅶ Archimedes’ Principle (buoyant force) Ⅷ Flow of Ideal Fluids =a constant =a constant Ⅸ Bernoulli’s Equation =a constant

[例1] As shown in Fig ,a thin rod with a small mass and a length l can rotate in the vertical plane about the axis going through the mid-point O and perpendicular to the plane. When the thin rod is resting at horizontal position , a small insect vertically drops with a speed vo on the rod is at a distance l/4 from point O, and then starts to crawl away from O toward point A at the end of the rod . Assume that the mass of the insect and the mass of the rod both are m . If we want to rotate the thin rod with a constant angular velocity, what is the speed at which the insect should be crawling toward the end of the rod ? Solution Key ideas: The process can be viewed as a complete inelastic collision . 2. The impulsive torque of the gravity can be ignored for the short time. 3. The angular momentum of the system of the insect and the thin rod is conserved.

So we can get the follow equation (1) From the Eq(1) we obtain the angular velocity of the thin rod The external acting on the system of the thin rod and the insect is only the gravitational torque (2) Since the angular velocity is required to be constant ,from the theorem of angular momentum we have (3)

 = t , the above equation becomes When the insect is at point P the moment of inertia of the system About the rotational axis is (4) Combine Eq2-4, we can get  = t , the above equation becomes Since

[例2]已知：两平行圆柱在水平面内转动， 求：接触且无相对滑动时 .o1 m1 R1 .o2 R2 m2 o1. o2.

o1. . o2 解一：因摩擦力为内力，外力过轴 ，外力矩为零， 则：I1 + I2 系统角动量守恒 ，以顺时针方向旋转 为正方向： 接触点无相对滑动： o1. . o2 又： 联立1、2、3、4式求解，对不对？

问题：（1）式中各角量是否对同轴而言？ （2）I1 +I2 系统角动量是否守恒？ 分别以m1 , m2 为研究对象，受力如图： o2 o1. F2 o1. F1 f1 f2 系统角动量不守恒！ 此解法不对。 解二：分别对m1 , m2 用角动量定理列方程

分别以m1 , m2 为研究对象，受力如图： 设：f1 = f2 = f ， 以顺时针方向为正 o2 o1. m1对o1 轴： 接触点：

[例3] 杂技演员M从h高处下落, 弹起N. 设跷板长l,质量为m. C为转动支点. M、N质量都为m. M与板为完全非弹性碰撞 [例3] 杂技演员M从h高处下落, 弹起N. 设跷板长l,质量为m. C为转动支点. M、N质量都为m. M与板为完全非弹性碰撞.问N可弹起多高? 解:M落到A处,速率为 l A B C h N M 碰撞后M、N有共同的线速度 以M、N板为系统, 合外力矩为零,碰撞过程角动量守恒. 而

于是 演员N以速率u 跳起, 达到高度h´ l A B C h N M

[例4] A 150.0kg rocket moving radially outward from Earth has a speed of 3.70km/sWhen its engine shuts off 200km above Earth’s surface. (a) Assuming negligibe air drag , find the rocket’s kinetic energy of the rocket is 1000km above the Earth’s surface. (b) What maximum height above the surface is reached by the rocket? P319 36 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)

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

[例5] Determine the mass of Earth from the period T (27.3days) and the radius r of the Moon’s orbit about Earth .Assume the Moon orbits the center of Earth rather than the center of mass of the Earth-Moon system solution Applying the law of period to the solve this problem, we can have Rewriting the above equation , we obtain Substituting the known data, we finally get

[例6] A tank is filled with water to a height H [例6] A tank is filled with water to a height H . A hole is punched in one of the walls at a depth h below the water surface .(a) Show that distance x from the resulting stream strikes the floor is given by (b) Could a hole be punched at another depth to produce a second stream that would have the same range? (c) At what depth should the hole be placed to make the emerging stream strike the ground at the maximum distance from the base of the tank?p344.54 (a) Prove From the sample problem15-9 p338, we know The motion of the stream, flowing from the hole , is the projectile motion. So, when the stream Strikes the floor we can get the time interval Combining the above two equations the distance can be obtained as

(b) Assuming a hole be punched at another depth to produce a second stream that would have the same range, we can write the follow equation (1) Solving the above unary quadratic equation (1) (c) From the answer (a) ,we know that distance from the resulting stream strikes the floor is

If the emerging stream strikes the ground at the maximum distance from the base of the tank, we can get So the hole be punched at the distance is maximal. Substituting the into the , we can obtain

[例7]关于流动流体的吸力的研究,若在管中细颈处开一小孔, 用细管接入容器A中液内,流动液体不但不漏出,而且A中液体 可以被吸上去.为研究此原理,做如下计算:设左上方容器很大, 流体流动时,液面无显著下降,液面与出液孔高度差为 表示管横截面,用 表示液体密度,液体为理想流体,试证: (漆11.4.3 P418) 即 处有一真空度,因此可 将A内液体吸入. [解 答] 以出液孔处为势能零点,选取过C中液面和截面 各一点的流 由题设条件可知,对此流线可用伯努力方程.流体在管中流动又满 连续性方程. 因此 (C,A,B流线)

(A,B水平流线) 由此得: 又 因此可将A内液体吸入.

[例8]容器A和B中装有同种液体.可视为理想流体.水平管截面 .求E管中的液柱高度( ). (漆11.4.4 P418) 以管子D处中心为势能零点,如图 选取过点1,2,3的一条流线,由已 知条件可知,对此流线可用伯努力 方程: [解 答] 3 h 2 由连续性方程: E管中为静止流体,由已知

故可近似认为E管上部气体压强处处相等.因此由静液压强得: 由以上方程解得: