Further Mechanics

Water jets follow parabolic paths — the classic projectile motion under gravity.

A projectile 抛射体 moves freely under gravity, so it has constant acceleration 匀加速 $g$ downwards and no horizontal acceleration. Treat the horizontal and vertical motions separately. If it is launched at speed $u$ and angle $\alpha$:

Eliminating $t$ gives the Cartesian equation 直角坐标方程 of the trajectory 轨迹 (the path), which is a parabola. For a fixed launch speed, all the trajectories lie under a bounding parabola 包络抛物线 (the envelope of reachable points). The range on level ground is $\dfrac{u^2\sin 2\alpha}{g}$ and the greatest height is $\dfrac{u^2\sin^2\alpha}{2g}$.

The path is a parabola; the launch speed $u$ splits into a steady horizontal part and a vertical part slowed by gravity.

Worked example. A ball is thrown at $u = 20\ \text{m s}^{-1}$ at $30^\circ$ to the horizontal. Find the range and greatest height.

The moment 力矩 of a force about a point is force $\times$ perpendicular distance; it measures turning effect. The weight of a body acts at its centre of mass 质心, which you can find using symmetry 对称 for a uniform flat shape (a lamina 薄片), or by treating a composite body as a set of particles.

A rigid body under coplanar forces 共面力 is in equilibrium 平衡 when two conditions both hold: the vector sum of the forces is zero, and the sum of the moments about any point is zero. A body may also be on the edge of toppling 翻倒 (turning over) or sliding 滑动 (slipping).

Taking moments about the pivot $A$ balances the turning effects: $F\times4 = 100\times2$.

Worked example. A uniform beam $AB$ of length $4\ \text{m}$ and weight $100\ \text{N}$ rests on a pivot at $A$. A vertical force $F$ at $B$ keeps it horizontal. Find $F$.

Take moments about $A$ (the weight acts at the centre, $2\ \text{m}$ from $A$):

For a particle moving in a circle of radius $r$, the angular speed 角速度 $\omega$ links to the speed by $v = r\omega$. The acceleration points towards the centre — the centripetal acceleration 向心加速度 — with size

The velocity points along the tangent; the acceleration points inward to the centre.

The hammer thrower is a moving diagram of this page. The taut wire pulls the ball toward the centre — that inward pull is the centripetal force — while the ball's velocity points along the tangent. Let go, and with no inward force left the ball flies straight off that tangent

In a horizontal circle 水平圆 the speed is constant — as in a conical pendulum 圆锥摆 (a mass swung on a string). In a vertical circle 竖直圆 use energy conservation, because the speed changes with height; the normal contact force 法向接触力 or string tension provides the centripetal force.

Worked example. A particle moves in a horizontal circle of radius $2\ \text{m}$ with angular speed $3\ \text{rad s}^{-1}$. Find its speed and acceleration.

Hooke's law 胡克定律 says the tension in an elastic string or spring is proportional to its extension $x$:

where $L$ is the natural length and $\lambda$ is the modulus of elasticity 弹性模量. Stretching stores elastic potential energy 弹性势能:

Tension rises in proportion to extension; the shaded triangle is the stored elastic energy.

Worked example. An elastic string of natural length $2\ \text{m}$ and modulus $50\ \text{N}$ is stretched by $0.5\ \text{m}$. Find the tension and the stored energy.

When the force depends on position $x$, use acceleration in the form $a = v\dfrac{dv}{dx}$, which turns Newton's law into a differential equation 微分方程 relating $v$ and $x$.

Use dv/dt when the force depends on time, v dv/dx when it depends on position

Worked example. A particle of mass $8\ \text{kg}$ moves along a line under a variable force 变力 of size $(x^3 + 4x)\ \text{N}$ acting in the direction of motion. When $x = 0$, $v = 1$. Find $v$ in terms of $x$.

Newton's law gives $8v\dfrac{dv}{dx} = x^3 + 4x$. Separating and integrating:

At $x = 0$, $v = 1$ gives $C = 4$, so $4v^2 = \tfrac14 x^4 + 2x^2 + 4 = 4\left(\tfrac{x^2}{4} + 1\right)^2$. Hence $v = \tfrac14 x^2 + 1$.

A position-dependent force becomes a differential equation in v and x

A Newton's cradle demonstrates conservation of momentum in collisions.

In a collision, total linear momentum is conserved — the conservation of linear momentum 动量守恒. The bounciness is measured by Newton's experimental law 牛顿实验定律, which defines the coefficient of restitution 恢复系数 $e$:

Here $e = 1$ is perfectly elastic (no energy lost) and $e = 0$ is inelastic (the bodies stay together). In an oblique impact 斜碰撞, resolve the velocities along and perpendicular to the line of impact, applying restitution along it.

The coefficient of restitution $e$ compares how fast the bodies separate with how fast they approached.

Worked example. A sphere $A$ of mass $2\ \text{kg}$ moving at $5\ \text{m s}^{-1}$ hits a stationary sphere $B$ of mass $3\ \text{kg}$, with $e = 0.5$. Find the speeds afterwards.

Momentum: $2(5) = 2v_A + 3v_B$, so $2v_A + 3v_B = 10$. Restitution: $v_B - v_A = 0.5(5) = 2.5$. Solving together gives $v_A = 0.5\ \text{m s}^{-1}$ and $v_B = 3.0\ \text{m s}^{-1}$.

• For projectiles, resolve into horizontal (constant velocity) and vertical ($a = g$) motion, linked by the same time.
• For a rigid body in equilibrium, take moments about a point that removes an unknown force.
• For circular motion, use $F = mv^2/r = m\omega^2 r$ towards the centre; in a vertical circle check the minimum speed at the top.
• With a variable force, use $a = v\,\frac{dv}{dx}$ and integrate; elastic PE $= \frac{\lambda x^2}{2L}$.