eifueo/docs/1a/ece105.md
2023-10-30 15:47:39 -04:00

7.9 KiB

ECE 105: Classical Mechanics

Motion

Please see SL Physics 1#2.1 - Motion for more information.

Kinematics

Please see SL Physics 1#Kinematic equations for more information.

Projectile motion

Please see SL Physics 1#Projectile motion for more information.

Uniform circular motion

Please see SL Physics 1#6.1 - Circular motion for more information.

Forces

Please see SL Physics 1#2.2 - Forces for more information.

Work

Please see SL Physics 1#2.3 - Work, energy, and power for more information.

Momentum and impulse

Please see SL Physics 1#2.4 - Momentum and impulse for more information.

The change of momentum with respect to time is equal to the average force so long as mass is constant.

dpdt=mdvdt+vdmdt\frac{dp}{dt} = \frac{mdv}{dt} + \frac{vdm}{dt}

Impulse is actually the change of momentum over time.

J=pipfdp\vec J = \int^{p_f}_{p_i}d\vec p

Centre of mass

The centre of mass xx of a system is equal to the average of the centre of masses of its components relative to a defined origin.

xcm=m1x1+m2x2+...+mnxnm1+m2+...+mnx_{cm} = \frac{m_1x_1 + m_2x_2 + ... + m_nx_n}{m_1 + m_2 + ... + m_n}

To determine the centre of mass of a system with a hole, the hole should be treated as negative mass. If the geometry of the system is symmetrical, the centre of mass is also symmetrical in the x and y dimensions.

For each mass, its surface density σ\sigma is equal to:

σ=mAm=σA \sigma = \frac{m}{A} \\ m = \sigma A

Holes have negative mass, i.e., m=σAm = -\sigma A.

For a one-dimensional hole, the linear mass density uses a similar formula:

λ=mLλ=dmdx \lambda =\frac{m}{L} \\ \lambda = \frac{dm}{dx}

This means that a hole in a rod can use a different formula: xcm=1M0Mxdmx_{cm} = \frac{1}{M}\int^M_0 x\cdot dm

For a solid object, the centre of mass can be expressed as a Riemann sum and thus an integral:

rcm=1M0Mrdmr_{cm} = \frac{1}{M}\int_0^M r\cdot dm

In an isolated system, it is guaranteed that the centre of mass of the whole system never changes so long as only rigid bodies are involved.

Rotational motion

Moment of inertia

The moment of inertia of an object represents its ability to resist rotation, effectively the rotational equivalent of mass. It is equal to the sum of each point and distance from the axis of rotation.

I=(mr)2I=\sum(mr)^2

For more complex objects where the distance often changes:

I=0MR2dmI=\int^M_0 R^2 dm

Common moment shapes

  • Solid cylinder or disc symmetrical to axis: I=12MR2I = \frac{1}{2}MR^2
  • Hoop about symmetrical axis: I=MR2I=MR^2
  • Solid sphere: 25MR2\frac{2}{5}MR^2
  • Thin spherical shell: I=23MR2I=\frac{2}{3}MR^2
  • Solid cylinder about the central diameter: I=14MR2+112ML2I=\frac{1}{4}MR^2 + \frac{1}{12}ML^2
  • Hoop about diameter: I=12MR2I=\frac{1}{2}MR^2
  • Rod about center: I=112ML2I=\frac{1}{12}ML^2
  • Rod about end: I=13ML2I=\frac{1}{3}ML^2
  • Thin rectangular plate about perpendicular axis through center: I=13ML2I=\frac{1}{3}ML^2

Rotational-translational equivalence

Most translational variables have a rotational equivalent.

Although the below should be represented as cross products, this course only deals with rotation perpendicular to the axis, so the following are always true.

Angular acceleration is related to acceleration:

α=ar\alpha = \frac{a}{r}

Angular velocity is related to velocity:

ω=vr\omega = \frac{v}{r}

The direction of the tangential values can be determined via the right hand rule. Where rr is the vector from the origin to the mass:

v=r×ωa=r×α \vec v = r\times\omega \\ \vec a = r\times\alpha

And all kinematic equations have their rotational equivalents.

  • θ=12(ωf+ωi)t\theta = \frac{1}{2}(\omega_f + \omega_i)t
  • ωf=ωi+αt\omega_f = \omega_i + \alpha t
  • θ=ωit+12αt2\theta = \omega_i t + \frac{1}{2}\alpha t^2
  • ωf2+ωi2+2αθ\omega_f^2 + \omega_i^2 + 2\alpha\theta

Most translational equations also have rotational equivalents.

Ek rot=12Iω2E_\text{k rot} = \frac{1}{2}I\omega^2

Torque

Torque is the rotational equivalent of force.

τ=Iα\vec\tau=I\vec\alpha τ=r×F\vec\tau=\vec r\times\vec F τ=rFsinθ|\vec\tau=|r||F|\sin\theta

In the general case, especially when the force is variable, the work done is equal to the integral of force over displacement.

W=xixfFxdxW=\int^{x_f}_{x_i}F_xdx

Work is also related to torque:

W=τΔθW=\tau\Delta\theta W=FΔSW=F\Delta S

The total net work from torque from external forces is equivalent to:

W=ΔEk=θiθfτdθW=\Delta E_k = \int^{\theta_f}_{\theta_i}\tau d\theta

Angular momentum

This is the same as linear momentum.

L=r×p\vec L = \vec r\times\vec p L=Iω\vec L = I\vec\omega L=τt\vec L =\vec\tau t

Rolling motion

!!! definition - Slipping is sliding faster than spinning. - Skidding is spinning faster than sliding.

Pure rolling motion is only true if the tangential velocity of the centre of mass is equal to its rotational velocity.

vcm=Rωv_{cm}=R\omega

In pure rolling motion, the point at the top is moving at two times the velocity while the point at the bottom has no tangential velocity.

(Source: Wikimedia Commons)

For any point in the mass:

vnet=vtrans+vrotvnet=vcm+R×ωEkroll=Ektrans+Ekrot v_{net} = v_{trans} + v_{rot} \\ v_{net} = v_{cm} + \vec R \times\vec\omega \\ E_{k roll} = E_{k trans} + E_{k rot}

Alternatively, the rolling can be considered as a rotation about the pivot point between the disk and the ground, allowing rolling motion to be represented as rotational motion around the pivot point. The parallel axis theorem can be used to return it back to the original point.

τb=Ibα\sum\tau_b=I_b\alpha

At least one external torque and one external force is required to initiate pure rolling motion because the two components are separate.

If an object is purely rolling and then it moves to:

  • a flat, frictionless surface, it continues purely rolling
  • an inclined, frictionless surface, external torque is needed to maintain pure rolling motion
  • an inclined surface with friction, it continues purely rolling

Where cc is the coefficient to the moment of inertia (I=cMR2I=cMR^2), while rolling down an incline:

vcm=21+cghacm=gsinθ1+c v_{cm} = \sqrt{\frac{2}{1+c}gh} \\ a_{cm} = \frac{g\sin\theta}{1+c}

Statics

An object at static equilibrium has no rotational or translational velocity with zero net force and torque.

An object at dynamic equilibrium has a constant rotational and translational velocity with zero net force and torque.

F=0τ=0 \sum\vec F = 0 \\ \sum\vec\tau = 0

Whether an object stays at static equilibrium depends on the

  • It is at unstable equilibrium if the object moves away and does not return to equilibrium
  • It is at stable equilibrium if the object returns to its original position and equilibrium
  • It is neutral if the object does not move

Simple harmonic motion

!!! definition - The amplitude AA of a wave is always greater than zero and is equal to the height of the wave above the axis. - The angular frequency ω\omega is the angular velocity, which is dependent only on the restorative force. - The phase constant ϕ\phi is the offset from equilibrium at t=0t=0.

Please see SL Physics 1#Simple harmonic motion for more information.

The position of any periodic motion can be represented as a sine or cosine function (adjust phase as needed).

x(t)=Acos(ωt+ϕ)x(t)=A\cos(\omega t+\phi)

This means that the velocity function has a phase shift of π2\frac{\pi}{2} and the acceleration function has a phase shift of π\pi along with other changes.

SHM is linked to uniform circular motion:

  • ϕ\phi is the angle from the standard axis
  • AA is the radius

The restorative force can be modelled by substituting in a(t)a(t) into F=maF=ma

F=mω2x(t)F=-m\omega^2x(t)

Because restoring force is proportional to the negative position for smaller displacements, F=Cx(t)F=-Cx(t).

Torque is also linear: τ=kθ\tau=-k\theta

!!! warning For small angles, sinθ=θ\sin\theta = \theta.

ω=Cm\omega=\sqrt{\frac{C}{m}}