Circular motion

A turn needs a change in velocity

• A ball on a string changes direction even when its speed stays the same. The string pulls the ball inwards. Your hand feels the ball's pull outwards: those two forces act on different objects.
• In a frame fixed to the ground, draw the real forces on the ball. Its inertia 惯性 does not supply an extra outward force. The name centrifugal 离心 force belongs to a rotating-frame description, which we do not need here.

Link angular speed to distance along the circle

• An angle of $\theta$ radians 弧度 sweeps arc length $s=r\theta$. For fixed radius, angular speed 角速度 is the rate of turning; linear speed 线速度 is distance travelled per second.
• Therefore $v=r\omega$. A point farther from the axis has greater speed at the same $\omega$. Use radians per second in this formula; degrees per second must first be multiplied by $\pi/180$.

Convert a period or a revolution rate

• The period 周期 $\tau$ is the time for one full turn. The frequency 频率 $f$ is turns per second. Hence $\omega=2\pi/\tau=2\pi f$.
• At $30$ revolutions per minute, $f=1/2$ turn per second and $\omega=\pi\ \dfrac{\text{rad}}{\text{s}}$. At radius $0.4\ \text{m}$, $v=0.4\pi\ \dfrac{\text{m}}{\text{s}}$. Do not put $30$ straight into $v=r\omega$.

Separate inward acceleration from changing speed

• Velocity is along the tangent 切线. The radial 径向 component of acceleration points towards the centre and has magnitude $a_r=v^2/r=r\omega^2$. This inward component is the centripetal acceleration 向心加速度. A proof of the acceleration formulae is not required here.
• At constant speed this is the whole acceleration. If speed changes, there is also a tangential 切向 component. In a vertical circle under gravity, the full acceleration need not point exactly towards the centre.

Use both forms of the acceleration formula

• For $r=2\ \text{m}$ and constant $\omega=3\ \dfrac{\text{rad}}{\text{s}}$, the speed is $v=2(3)=6\ \dfrac{\text{m}}{\text{s}}$ and $a_r=2(3^2)=18\ \dfrac{\text{m}}{\text{s}^2}$.
• Checking with $v^2/r=6^2/2=18$ gives the same result. At fixed speed, doubling the radius halves $a_r$; at fixed angular speed, doubling the radius doubles it. State which quantity is fixed.

Resolve real forces towards the centre

• Draw a free-body diagram 受力图 and choose inwards as positive. Newton's law gives $\sum F_{\rm inward}=mv^2/r$. A centripetal force 向心力 is the inward resultant needed for the motion, not another arrow to add beside weight, tension or contact forces.
• A horizontal circle 水平圆 need not have constant speed. In this syllabus's steady horizontal-circle problems, vertical acceleration and tangential acceleration are zero; the inward acceleration is not zero. Resolve vertically and radially as separate equations.

Friction can provide the inward resultant

• A particle on a rough horizontal rotating disc has normal reaction 法向反力 $N=mg$. If friction is the only horizontal force, the required friction is $F_f=mr\omega^2$. It must satisfy $F_f\leq\mu N$.
• Therefore $\omega^2\leq\mu g/r$. For $r=1.2\ \text{m}$, $\mu=0.3$ and $g=10\ \dfrac{\text{m}}{\text{s}^2}$, the greatest angular speed is $\sqrt{2.5}\ \dfrac{\text{rad}}{\text{s}}$, about $15.1$ revolutions per minute. Below the limit, friction is not automatically $\mu N$.

An inclined string changes the disc reaction

• A $2\ \text{kg}$ particle moves at radius $1.5\ \text{m}$ on a rough disc. A string pulls towards a point above the axis, with tension 张力 $T=10\ \text{N}$, $\sin\alpha=3/5$ and $\cos\alpha=4/5$, where $\alpha$ is measured from vertical. Take $g=10\ \dfrac{\text{m}}{\text{s}^2}$.
• Vertical balance gives $N=20-10(4/5)=12\ \text{N}$. With $\mu=1/4$ and impending outward slipping, limiting friction 最大静摩擦 is $3\ \text{N}$ inwards. Thus $10(3/5)+3=2(1.5)\omega^2$, giving $\omega=\sqrt3\ \dfrac{\text{rad}}{\text{s}}$.

Friction may reverse when the speed changes

• Keep that particle, tension and geometry fixed. The string already supplies $6\ \text{N}$ inwards. Required signed inward friction is $F_f=3\omega^2-6$, while $|F_f|\leq3$.
• Hence $1\leq\omega^2\leq3$. At the lower limit friction acts outwards to reduce the inward resultant; at the upper limit it acts inwards. An outward real friction force is possible even though the net radial force is inwards.

A conical pendulum has two useful directions

• A conical pendulum 圆锥摆 has a light inextensible string of length $L$ making angle $\theta$ with the vertical. Its orbit radius is $r=L\sin\theta$, not $L$. The bob stays at one height.
• Resolve tension: $T\cos\theta=mg$ and $T\sin\theta=mr\omega^2$. For a non-zero orbit, substitution gives $\omega^2=g/(L\cos\theta)$. The tension has both vertical and inward components.

Find the orbit before finding its speed

• Let $L=2.5\ \text{m}$, $m=0.4\ \text{kg}$, $\cos\theta=4/5$, $\sin\theta=3/5$ and $g=10\ \dfrac{\text{m}}{\text{s}^2}$. Then $r=1.5\ \text{m}$ and $T=mg/\cos\theta=5\ \text{N}$.
• Now $\omega^2=10/[2.5(4/5)]=5$, so $v=1.5\sqrt5\ \dfrac{\text{m}}{\text{s}}$ and the period is $2\pi/\sqrt5\ \text{s}$. Using the string length as the orbit radius would give the wrong speed.

A second particle can fix the tension

• A string of total length $9a$ passes through a fixed smooth ring $R$. Particle $A$ of mass $2m$ hangs below it. Particle $B$ of mass $m$ moves in a horizontal circle whose centre is $A$. Let $RB=L$ and let $RB$ make angle $\theta$ with the vertical.
• For $A$, $T=2mg$. For $B$, $T\cos\theta=mg$, so $\cos\theta=1/2$. Geometry gives $RA=L\cos\theta=L/2$ and $RA+RB=9a$, so $L=6a$. Then $\omega^2=g/(L\cos\theta)=g/(3a)$.

A bowl radius is not the orbit radius

• On the smooth inner surface of a hemispherical bowl of radius $a$, let the normal make angle $\theta$ with upward vertical. The orbit radius is $a\sin\theta$. Vertical and inward equations are $N\cos\theta=mg$ and $N\sin\theta=m(a\sin\theta)\omega^2$.
• Therefore $\omega^2=g/(a\cos\theta)$. If $\cos\theta$ changes from $4/5$ to $5/13$, the angular-speed ratio is $\omega_1/\omega_2=\sqrt{(5/13)/(4/5)}=5/(2\sqrt{13})$. Compare squared speeds first, then take the square root.

A cone needs its own geometry

• A smooth hollow cone has its vertex downwards and semi-vertical angle $\alpha$. At height $h$ above the vertex the orbit radius is $r=h\tan\alpha$. The normal points upwards and inwards, making angle $\alpha$ above horizontal.
• With only gravity and contact, $N\sin\alpha=mg$ and $N\cos\alpha=mr\omega^2$. These give $\omega^2=g\cot^2\alpha/h$. A cone's radius is set by height; do not borrow the bowl's radius formula.

Add a string through the cone vertex

• Now attach the particle of mass $m$ to another mass $m$ hanging below the vertex through a smooth hole. The string pulls down the sloping surface towards the vertex. Take $\tan\alpha=4/3$, so $\sin\alpha=4/5$, $\cos\alpha=3/5$ and $r=4h/3$.
• The hanging mass gives $T=mg$. Then $N(4/5)=mg+T(3/5)$ gives $N=2mg$. Radially, $N(3/5)+T(4/5)=mr\omega^2$, hence $\omega^2=3g/(2h)$. The string's downward component increases the reaction.

Measure height in a vertical circle

• For a vertical circle 竖直圆 of radius $r$, take angle $\theta$ from the downward vertical and measure height from the bottom. At the right-hand point of the circle, $x=r\sin\theta$ and height $h=r(1-\cos\theta)$.
• At the bottom $\theta=0$ and $h=0$; at the side $\theta=\pi/2$ and $h=r$; at the top $\theta=\pi$ and $h=2r$. Mark the angle's reference direction before writing an energy equation.

Energy gives the speed at each height

• For a light inextensible string or a fixed smooth circular surface, the constraint force is perpendicular to velocity and does no work. With no resistance, energy conservation 能量守恒 gives $\tfrac12mu^2=\tfrac12mv^2+mgr(1-\cos\theta)$, where $u$ is bottom speed.
• Therefore $v^2=u^2-2gr(1-\cos\theta)$. This formula applies while the particle follows the circle. It finds speed, not tension; and $v^2\geq0$ alone does not prove the circular path is possible.

Take inward components of weight and tension

• For a string pulling towards the centre, the inward equation is $T-mg\cos\theta=mv^2/r$. At the bottom, weight acts outwards; at the top, weight acts inwards. The changing sign is already in $\cos\theta$.
• Combining with energy gives $T/m=u^2/r-2g+3g\cos\theta$. At the bottom $T=mu^2/r+mg$; at the top $T=m(u^2/r-5g)$. These tensions use different speeds at the two heights.

Work from a specified bottom speed

• Suppose $u^2=6gr$. At $\cos\theta=1/2$, energy gives $v^2=5gr$, and the inward equation gives $T=5mg+mg/2=11mg/2$.
• At the bottom the tension is $7mg$; at the top it is $mg$. The particle moves fastest and needs the greatest tension at the bottom. In between, gravity also changes its speed, so the full acceleration has a tangential component.

A complete loop needs a feasible top force

• A string can pull, not push: a taut 绷紧 string needs $T\geq0$. The top has the least tension. There $T+mg=mv_{\rm top}^2/r$, so $v_{\rm top}^2\geq gr$. Energy then gives the complete-loop condition $u^2\geq5gr$.
• Equality is the limiting complete loop, with zero tension at the top. Merely reaching top height by energy would require $u^2\geq4gr$; that weaker test allows a negative calculated tension before the top. The string would already have gone slack 松弛.

Start at a point other than the bottom

• If the starting radius makes angle $\phi$ with the upward vertical, its height above the bottom is $r(1+\cos\phi)$. The climb to the top is $r(1-\cos\phi)$.
• A complete loop requires $v^2\geq gr+2gr(1-\cos\phi)=gr(3-2\cos\phi)$. At $\cos\phi=1/4$, the minimum starting squared speed is $5gr/2$. At equality the bottom squared speed is $5gr$ and the greatest tension is $6mg$.

Use a tension ratio with two energy states

• In a complete loop, bottom tension exceeds top tension by $6mg$. If their ratio is $9:1$, write $T_{\rm bottom}=9T_{\rm top}$. Then $T_{\rm top}=3mg/4$, so $v_{\rm top}^2=7gr/4$ and bottom squared speed is $23gr/4$.
• If the speed at angle $\theta$ from downward vertical is $\sqrt{4gr}$, energy gives $4gr=23gr/4-2gr+2gr\cos\theta$. Thus $\cos\theta=1/8$. Equal top and bottom speeds would violate energy conservation.

A turning point and a slack point are different

• If bottom squared speed is $gr$, the equation $v=0$ gives $\cos\theta=1/2$: the particle turns back at $60^\circ$, with tension $mg/2>0$. It never reaches the horizontal level.
• If bottom squared speed is $3gr$, the equation $T=0$ instead gives $\cos\theta=-1/3$, about $109.5^\circ$. Here $v^2=gr/3>0$: the string slackens while the particle is still moving. Stop using the string's circular-motion equation beyond this point.

An inner surface can push only inwards

• On the smooth inner surface of a spherical shell, the normal reaction, or normal contact force 法向接触力 $N$, points towards the sphere centre. Its radial equation is $N-mg\cos\theta=mv^2/r$ with $\theta$ measured from downward vertical.
• The same energy and complete-loop results apply, replacing $T$ by $N$. Require $N\geq0$; a negative answer means the surface would have to pull. For loss of contact 脱离接触, test the boundary $N=0$: the particle leaves if continuing the circle would require $N<0$. A limiting complete loop can have $N=0$ at the top without separating.

A reaction ratio can locate a rim

• A particle starts vertically downwards from the side of a smooth inner shell with squared speed $u^2=2gr$. A rim is at height $kr$ above the centre. At the start, $N_A=mu^2/r=2mg$; at the rim, energy gives $v_B^2=(2-2k)gr$.
• The inward weight component there is $mgk$, so $N_B/m=v_B^2/r-gk=(2-3k)g$. If $N_B=N_A/4$, then $k=1/2$. Keeping contact as far as this rim requires $u^2\geq3gr/2$, not just the energy test $u^2\geq gr$.

An outer surface has the opposite normal direction

• A particle starts at the top of a smooth fixed sphere with tangential squared speed $gr/2$. Let $\phi$ be its angle from upward vertical. Energy gives $v^2=gr/2+2gr(1-\cos\phi)$.
• On the outside, the normal points outwards. The inward equation is $mg\cos\phi-N=mv^2/r$, giving $N=mg(3\cos\phi-5/2)$. It loses contact at $\cos\phi=5/6$, with $v^2=5gr/6>0$. Do not copy the inner-surface sign.

Release gives a tangent velocity, then a curved path

• At release, position and velocity remain continuous. The initial free-flight velocity is tangent to the circle. Gravity still acts, so the later path is a projectile parabola, not a straight tangent line.
• For the string that slackens at $\cos\theta=-1/3$ with bottom squared speed $3gr$, take centre $O$ as origin. At release, $(x,y)=(2\sqrt2r/3,r/3)$ and speed is $\sqrt{gr/3}$. On the upward trip along the right side, velocity components are $(-v/3,2\sqrt2v/3)$.

Join circular motion to a flight across a rim

• Cut an inner shell at height $r/2$ above its centre. To fly from the right rim to the left rim at the same height, let speed at release be $v$. The horizontal gap is $\sqrt3r$; velocity components are $(-v/2,\sqrt3v/2)$.
• The non-zero return time is $t=\sqrt3v/g$. Horizontal motion gives $\sqrt3r=(v/2)t$, so $v^2=2gr$. If the launch began at the side at centre height, energy gives side squared speed $u^2=3gr$. The rim reaction just before exit is $3mg/2>0$: the surface ends before contact is lost.

Check the model before accepting a root

• Draw forces and identify the orbit radius. For steady horizontal motion, resolve vertically and inwards. For vertical motion, combine height-based energy with one inward force equation. Check $v^2\geq0$ and the sign of tension or reaction along the route.
• State which event happens first: a turning point, slack string, loss of contact, or the end of the surface. Then use the next model. A collision changes the motion through an impulse; do not carry the no-loss circular energy equation through an impact without a separate collision argument.

Keep the direction and constraint together