PHYS10101 - Dynamics

How can unit vectors of polar coordinates be written as Cartesian coordinates?

• both r^ and \theta^ are functions of \theta: they change as angle changes

• r^ is in the direction of r (where r =r\left(\cos\theta i+\sin\theta j\right) due to x=r\cos\theta and y=r\sin\theta)

• i.e it is a vector from the centre to a point on the circumference of a unit circle. Therefore (see diagram):

r^=\cos\theta i+\sin\theta j

• the direction can also simply be written as dr/dr, i.e rate of change of total vector r with respect to rate of change of magnitude of vector r

• the unit vector of \theta (\theta^) must correspond to the effect that a change in angle \theta has on r, therefore

\theta^ \cdot r = \frac{dr}{d\theta}=r\left(-\sin\theta i+\cos\theta j\right)

\theta^ =-\sin\theta i+\cos\theta j

• note this proves mathematically that they are always perpendicular to each other

Velocity in polar coordinates

• v = dr/dt, written in polar coordinates this becomes v = d/dt (rr^)

• using product rule we get:

v =\frac{dr}{dt} r^ + r\frac{d}{dt}(r^)

• therefore we get: (see diagram)

• the first term is the radial component of velocity, the second term is the tangential component (in direction of \theta^) and can be easily found by differentiating \frac{d}{dt}(r^) and showing it is \theta^ multiplied by \frac{d\theta}{dt}

Acceleration in polar coordinates

• a=\frac{dv}{dt}

(from now R will be used to represent r^ and \Theta to represent \theta^)

• using equation for velocity in polar, we can write this as:

=\frac{d}{dt}\left(\frac{dr}{dt}R+r\frac{d\theta}{dt}\Theta\right)

=\frac{d^2r}{dt^2}R+\frac{dr}{dt}\frac{dR}{dt}+\left(\frac{dr}{dt}\frac{d\theta}{dt}+r\frac{d^2\theta}{dt^2}\right)\Theta+r\frac{d\theta}{dt}\frac{d\Theta}{dt}

• from finding velocity we know \frac{dR}{dt}=\frac{d\theta}{dt}\Theta. by using the definition of \Theta

  in a similar way to how we found this expression, we can also prove \frac{d\Theta}{dt}=-\frac{d\theta}{dt}R. Therefore, we can finally conclude: (see diagram)

• in this equation, part 1 gives the linear acceleration in the radial direction (results in change in radial component of particle’s velocity), part 2 gives the centripetal acceleration directed radially inwards (note the negative!! this is VERY key for vertical circular motion questions), part 3 gives the Coriolis acceleration - a fictitious force that appears to act on a particle only if it is in motion with respect to a rotating reference frame, and part 4 gives the linear acceleration in the tangential direction (results in change in magnitude of angular velocity)

Static and dynamic friction

• Static friction is the frictional force that prevents an object experiencing another force from sliding along a rough surface. It will be equal and opposite to the force trying to slide the object until the force becomes too large and the object begins to slide. The max value is called limiting static friction, Fs, and is given by Fs=\mu sN, where \mu s is the coefficient of static friction

• Dynamic friction is the frictional force experienced by an object that is sliding/moving over a rough surface. It is given by Fk=\mu kN

• For the same object sliding over the same surface, the coefficient of dynamic friction will be LESS than the coefficient of static friction. This means that the dynamic frictional force experienced when the object is moving can actually be less than the limiting static frictional force that has to be overcome to make the object move to begin with!

Explanation of friction: Why is it proportional to the normal reaction force, and why is the coefficient for dynamic friction less than the one for static friction?

• Surfaces are never truly ‘flat’ - on a microscopic scale they are rough and bumpy.

• When one surface is placed on another, the area of true contact is therefore actually very small - as little as 1/10,000th the area of the surfaces that are supposedly in contact! This means the pressure at contact points is very high leading to the contact points becoming flattened until the weight of the upper solid is supported.

• at these contact points, the atoms/molecules of the two surfaces are so close together that the strong force can actually act between them, and in some cases they even become ‘cold welded’ together. These strong cohesive forces are responsible for friction.

• Increasing the weight/force pushing the objects together (i.e increasing the normal reaction force!!) will increase the pressure experienced at these contact points. This means they are flattened even further and the true area of contact is increased. This means more atoms/molecules are held together by the strong force - and means that more ‘cold welding’ may take place, leading to even stronger cohesive forces = larger frictional force!

• When you try to move one surface across another, you have to overcome the tiny contact points that have been ‘cold welded’ together. However, once broken, the bumps will skim over one another while one surface moves across the other and won’t ‘cold weld’ back together, hence dynamic friction is less than the limiting static friction!

Motion in a vertical circle: Equation for tension at a given point in the circle

See diagram for visualisation of motion in a vertical circle.

• We can see that here the radial components of the resultant force on the ball is given by Fr=mg\cos\theta-T

• The centripetal acceleration for an object moving in circular motion is given by ar=-r\omega^2=-\frac{v^2}{r} . This can be seen from the polar coordinate equation for acceleration (see flashcard on acceleration in polar coordinates.

• Newton’s second law states F=ma thereforear=\frac{\left(mg\cos\theta-T\right)}{m}=-\frac{v^2}{r}

rearranging this equation gives an expression for the tension required for circular motion at different velocities and angles of rotation:

T=m\left(\frac{v^2}{r}+g\cos\theta\right)

(note due to cos term will be max at bottom of circle when\theta=0,T=m\left(\frac{v^2}{r}+g\right) and will be min at top of circle when \theta=\pi,T=m\left(\frac{v^2}{r}-g\right))

Motion in a vertical circle: Critical velocity at highest point of particle’s motion for it to continue moving in circular motion

See diagram for visualisation of motion in a vertical circle.

• recall the equation for tension/force acting radially inwards for object in vertical circular motion:

T=m\left(\frac{v^2}{r}+g\cos\theta\right)

• at top of path, this radially inwards force (tension in this case) must be greater than 0, otherwise the object will no longer travel along a circular path (in this case string would go slack). Therefore, to find the critical speed, we set T = 0 and \theta=\pi (as at top of path):

0=m\left(\frac{v^2}{r}-g\right)

v=\sqrt{rg}

Work: calculating work done from non-constant force

W=\int_{x0}^{x1}\!F\,dx

i.e, work done by a spring pushing a horizontal block (on smooth surface) from x0 to x1 gives:

W=\int_{x0}^{x1}kxdx

W=\frac12k\left(x1-x0\right)^2 , if x1 = x, x0 = 0, gives

W=\frac12kx^2 - Elastic potential energy!! (as W = total energy)

Work: Calculating work with vectors (3D space etc)

W=\int_{r0}^{r1}F\cdot dr , where F is a vector force and is a function of (vector) r.

Therefore, as a\cdot b=\left|a\Vert b\right|\cos\theta,

W=\int_{r0}^{r1}F\left(r\right)\cos\left(\theta\right)dr , note here that F is a scalar and will give scalar answers when r1 and r0 are subbed in

This is called a line intergral.

Work: Simple pendulum example

• See diagram for sketch of pendulum and free body diagram

• We can see that F=T\sin\theta and mg=T\cos\theta, therefore F=mg\tan\theta, giving us an expression of F as a function of theta.

• From the line integral for work, we know that W=\int_{r0}^{r1}F\left(r\right)\cos\theta dr. This is in terms of r, so we want to find in terms of theta. Therefore, we need expressions linking r to theta and dr to d(theta):

• r to theta: From diagram we can see r is unchanging. Therefore, at r0, we can just say \theta0=0 and at r1, \theta1=\theta (where theta is the final angle reached)

• dr to d(theta): From diagram we can also see s=l\theta (simple circle stuff in radians!). We can also see that dr = extra s added on, meaning
  dr=ld\theta

So, W=\int_{r0}^{r1}F\cos\theta dr=\int_0^{\theta}Fl\cos\theta d\theta

W=\int_0^{\theta}mg\tan\theta\cos\theta ld\theta=mgl\int_0^{\theta}\sin\theta d\theta

W=mgl\left(1-\cos\theta\right)

Work: The Work-Energy Theorem

States that the work done by a resultant force on a particle = the change in kinetic energy of the particle

When resultant force is constant, a = constant so this can easily be proved with SUVAT, W = Fx, and F = ma. When F is not constant in magnitude but is constant in direction, we can use the simple W=\int_{x0}^{x1}F\left(x\right)dx and F=ma=mv\frac{dv}{dx} to get W=\frac12m\left(v1^2-v0^2\right), but when F is not constant in both direction and magnitude, things are a bit more complicated:

General case: (called general case as also works for previous 2 cases)

F=m\frac{dv}{dt} where both F and v are vectors.

We must use line intergral W=\int_{r0}^{r1}F\cdot dr with this which gives:

W=\int_{r0}^{r1}m\frac{dv}{dt}\cdot dr=\int_{t0}^{t1}m\frac{dv}{dt}\cdot vdt

Integration by “looking at it” (reverse product rule in this case!) shows us:

\frac{dv}{dt}\cdot v=\frac12\frac{d\left(v^2\right)}{dt} (see diagram for definition of product rule. What happens if a(t) = b(t) = v ?). Note that v is no longer a vector here (dot product has been used!)

Therefore:

W=\int_{t0}^{t1}\frac12m\frac{d\left(v^2\right)}{dt}dt=\int_{\frac12mv0^2}^{\frac12mv1^2}d\left(\frac12mv^2\right)=\frac12m\left(v1^2-v0^2\right)

Work: What is power?

Power is the total work done over total time taken, i.e work done per unit time:

P=\frac{dW}{dt} ,W=\int_{r0}^{r1}F\cdot dr , so P=\int_{v0}^{v1}F\cdot dv (F is vector function of r/v)

When v is constant, P=F\cdot v

CoM: Finding centre of mass of a rigid body

We consider the rigid body as a system of closely packed particles: We divide it into n small elements, each with mass \Delta mi, located at points ri=\left(xi,yi,zi\right)

The coordinates of the centre of mass for the x coordinate is given by:

\left(\sum_{i=1}^{n}\Delta mi\right)xcm=\sum_{i=1}^{n}\Delta mixi (and same for y and z coordinates)

If we increase n, the number of elements, to infinity, \Delta mi will tend to 0, therefore (for x as example but same for each coordinate):

xcm=\lim_{\Delta mi\to0}\frac{\sum_{i=1}^{n}\Delta mixi}{\sum_{i=1}^{n}\Delta mi}=\frac{\int xdm}{\int dm}=\frac{1}{M}\int xdm

Therefore equivalent vector equation is:

rcm=\frac{1}{M}\int rdm (r as vector)

This centre of mass is useful as we can use it as the point that all external forces act

CoM: Mass per unit length and it’s uses

Mass per unit length is.. the mass of an object.. per, it’s.. unit length

depicted with symbol \lambda where for uniform objects, \lambda=\frac{M}{L} (L is total length) and for non-uniform objects, \lambda will equal some function of r (or x if 1D rod), usually over L (as mass varies with x position along L so M/L becomes f(x)/L). This will be given in question (hopefully)

It is useful as it allows us to rewrite the rigid body equation, rcm=\frac{1}{M}\int rdm (or it’s version for individual coordinates, i.e x), as an integral of dr:

\int dm=\int_0^{L}\lambda\left(x\right)dx when working with rods with length along x axis (note when uniform, lambda(x) will simply equal lambda, as it is not a function of x as mass per unit length is same at all points of x)

therefore xcm=\frac{1}{M}\int xdm becomes xcm=\frac{1}{M}\int_0^{L}x\lambda\left(x\right)dx

CoM: Proof that all external forces can be thought of acting on the centre of mass

We know Mrcm=m1r1+m2r2+\cdots . Differentiating this gives us M\frac{d}{dt}rcm=m1\frac{d}{dt}r1+m2\frac{d}{dt}r2+\cdots=Mvcm=m1v1+m2v2+\cdots

Differentiating again will give us:
Macm=m1a1+m2a2+\cdots+mnan . Using F = ma, we see:

Macm=F1+F2+\cdots Fn

i.e, the total mass of the system * by the acceleration of system at CoM is equal to the vector sum of all the forces acting on the group of particles

Among these forces, some will be internal forces, i,e if system is system of planets and stars, gravity, or if system is a solid rock, intermolecular forces between the atoms of the rock. N3 shows us these internal forces occur in equal and opposite pairs, so cancel each out.

Therefore,

Macm=\sum_{i=1}^{n}Fi=Fext , where Fext is the vector sum of external forces acting on the system, and can be thought as all acting at the CoM.

CoM: The kinetic energy of a system of particles - theorem and proof

Theorem:

“The kinetic energy of a system of particles equals the kinetic energy associated with the motion of the centre of mass PLUS the kinetic energy associated with the motion of the particles relative to the centre of mass”

K=Kcm+Krel

Kcm=\frac12Mvcm^2

Krel=\sum_{i}\frac12miui^2 (where ui is velocity of ith particle relative to CoM)

Proof:

KE of system = sum of KE of individual particles, i.e K=\sum_{i}\frac12mivi^2

As ui is the velocity of ith particle relative to CoM velocity, vi can be written as vi=vcm+ui

Therefore:

K=\frac12\sum_{i}\operatorname{mi}\left(vcm+ui\right)^2=\frac12vcm^2\sum_{i}\operatorname{mi}+vcm\cdot\sum_{i}\operatorname{mi}ui+\frac12\sum_{i}\operatorname{mi}ui^2

As Mvcm=\sum_{i}mivi and vi=vcm+ui, \sum^{i}mivi must equal 0 (to keep definition of vcm true)

Therefore:

KK=\frac12Mvcm^2+\frac12\sum_{i}\operatorname{mi}ui^2

CoM: The reduced mass

For two particles,

vcm=\frac{m1v1+m2v2}{m1+m2}

vrel=v1-v2

Rearranging gives expressions for v1 and v2:

v1=vcm+\frac{m2}{M}vrel

v2=vcm-\frac{m1}{M}vrel

Expressing Kinetic energy (\frac12m1v1^2+\frac12m2v2^2) in terms of vcm and vrel using these equations gives:

K=\frac12Mvcm^2+\frac12\mu vrel^2

where mu is the reduced mass and is given by:

\mu=\frac{m1m2}{m1+m2}

Reduced mass describes the relative motion of two particles. This is useful for looking at forces that only depends on separation (i.e electron “orbiting” proton, Earth orbiting Sun), as the associated Ek is the relative kinetic energy, governed by the reduced mass.

If one mass is significantly larger than the other, i.e m1 « m2, we find that \mu\approx m1, so the reduced mass approximately equals the smaller of the two masses (i.e in electron-hydrogen example, the reduced mass would be almost the mass of the electron)

Energy: Using Work-Energy principle to find work done by other forces (than wd by potential energies)

If other forces than gravity and elastic forces do work on a body,

Wtotal=Wgrav+Wel+Wother

Total work done = change in kinetic energy (Work-Energy theorem):

Wgrav+Wel+Wother=\Delta K=K2-K1

Wgrav/Wel is equal to the change in their respective potential energies, so:

\left(Gpe1-Gpe2\right)+\left(Epe1-Epe2\right)+Wother=K2-K1

rearranging gives:

Wother=\left(K2+Gpe2+Epe2\right)-\left(K1+Gpe1+Epe1\right)

Wother=\left(K2+U2\right)-\left(K1+U1\right)

Wother= Final total energy - Initial total energy

Wd by all other forces than gravitational and elastic ones equals the change in total mechanical energy of the system

Continue energy from 8.4