Advanced Physics Term 3 - Notes

E0T for Eleventh Advanced Physics Term 3

• 4 writing questions (40 marks total) - 4 Questions of 10 marks each
• 15 MCQ questions (60 marks total) - 4 marks each
• Total: 100 marks

Determining the Center of Mass

• The center of mass is an imaginary point where the entire mass of the system can be concentrated.

• Methodology

  • Imagine a coordinate axis (if not given). If it is given already then you can use that coordinate access. If it is not given make important, make a coordinate axes.
  • Write the coordinates of every mass.
  • For two masses m1 and m2 at coordinates (x1, y1) and (x2, y2), the x-coordinate of the center of mass (x{CM}) is: x{CM} = \frac{m1x1 + m2x2}{m1 + m2}
  • Generalization for n masses:
    x{CM} = \frac{\sum mi xi}{\sum mi}
  • The center of mass resides closer to the heavier mass.

• Example

  • Three identical balls of mass m at (0,0), (a,0) and (0,a).
  • x_{CM} = \frac{m \cdot 0 + m \cdot a + m \cdot 0}{3m} = \frac{a}{3}
  • y_{CM} = \frac{m \cdot 0 + m \cdot 0 + m \cdot a}{3m} = \frac{a}{3}
  • The center of mass can be represented by vector \vec{r_{CM}} = \frac{a}{3}\hat{x} + \frac{a}{3}\hat{y}

Center of Mass for the Extended Objects

• Given center of mass coordinates for an object and the mass distribution, determine the coordinates of an unknown mass.

• Example:

  • System of masses: 4 kg at (0,0), 3 kg at (L,0), and 2 kg at (x,y).
  • Given xCM = L/4.
  • x_{CM} = \frac{4 \cdot 0 + 3 \cdot L + 2 \cdot x}{4 + 3 + 2} = \frac{L}{4}
  • \frac{3L + 2x}{9} = \frac{L}{4}
  • 12L + 8x = 9L
  • 8x = -3L
  • x = -\frac{3}{8}L
  • You can assume L=1 to solve for x and then substitute back as needed.

Polar Coordinates

• Polar coordinates represent a point by its distance from the origin (r) and the angle it makes with the x-axis ([theta]).

  • x = r \cos(\theta)
  • y = r \sin(\theta)
  • r = \sqrt{x^2 + y^2}
  • \theta = \tan^{-1}(\frac{y}{x})

• Example

  • A point on a circle of radius 8 cm has an x-component of 4 cm.
  • \sin(\theta) = \frac{y}{8}
  • \cos(\theta) = \frac{4}{8}
  • \theta = \cos^{-1}(\frac{4}{8})

Radians

• One radian is the angle subtended at the center of a circle by an arc equal in length to the radius of the circle.
• 180^\circ = \pi \text{ radians}
• Converting degrees to radians: multiply by \frac{\pi}{180}
• Dividing a circle (360 degrees) into 12 parts gives each part as 30 degrees.
• Example: Converting 40 degrees to radians involves multiplying \frac{4\pi}{30} = \frac{4\pi}{180}

Circular Arc Length

• The relationship between arc length (s), radius (r), and angle in radians (\theta) is:
  \theta = \frac{s}{r}

• s = r\theta

• Example

  • A diameter of 12 meters has an angular displacement of \pi radians. So, r = 6
  • s = 6\pi

• Example

  • A boy travels a distance of 20 meters along the edge of a circle with a radius of 5 meters.
  • \theta = \frac{20}{5} = 4 \text{ radians}

Angular Velocity and Linear Velocity

• Points on a rotating object travel different linear distances in the same time, thus having different linear speeds.
• Angular speed (\omega) is the same for all points on a rotating object.
• The relationship between linear speed (v) and angular speed is:
  v = r\omega
• When two objects are in continuous contact (e.g., gears), their linear speeds at the point of contact are the same, but their angular speeds may differ.
• Angular acceleration (\alpha) is the rate of change of angular speed.
• The relationship between linear acceleration (a) and angular acceleration is:
  a = r\alpha
• RPM (revolutions per minute) can be converted to radians per second using:
  \omega = \frac{2\pi \cdot RPM}{60}
• The vector relationship is actually \vec{v} = \vec{\omega} \times \vec{r}, and it is often used as magnitude not vector because V & Omega have completely different directions.

Example Problems: Angular and Linear Motion

• Three Discs - No Slipping: a small disc of radius 0.100 m, driven by a motor, turns a larger disc this two of a radius 0.5 hundred meter.
  • Disk 3 completes one revolution every 30 seconds.
• When there is no slipping, the linear speeds at contact points are equal (v1 = v2 = v_3).
• For disc 3 T = 30 \text{ seconds}
  • \omega_3 = \frac{2\pi}{30} \approx 0.209 \text{ rad/s}
  • v = r3 \omega3 = 1 \cdot 0.209 = 0.209 \text{ m/s}
  • \omega2 = \frac{v}{r2} = \frac{0.209}{0.5} = 0.418 \text{ rad/s}
  • If disc 1 has an angular acceleration of \alpha_1 = 0.118 \text{ rad/s}^2, then the linear acceleration is the equal!
  • a = r1 \alpha1 = 0.1 \cdot 0.118 = 0.0118 \text{ m/s}^2

Angular Kinematics

• Equations are analogous to linear kinematics.
  • \omegaf = \omegai + \alpha t
  • \theta = \omega_i t + \frac{1}{2} \alpha t^2
  • $\omegaf^2 = \omegai^2 + 2\alpha \theta
• Example: A discus thrower with an arm of length 1.2m starts from rest and rotates with \alpha = 2.5 \text{ rad/s}^2
  • Time to reach \omega_f = 4.7 \text{ rad/s}
    \4.7 = 0 + 2.5t \implies t = \frac{4.7}{2.5} = 1.88 \text{ s}
  • Linear speed: v = r \omega = 1.2 \cdot 4.7 = 5.64 \text{ m/s}
  • Linear acceleration: a = r \alpha = 1.2 \cdot 2.5 = 3 \text{ m/s}^2

Centripetal Acceleration

• Centripetal acceleration (a_c) is the acceleration directed towards the center of a circle, necessary to keep an object moving in a circular path.
• a_c = \frac{v^2}{r}
• If an object is speeding up or slowing down in a circle, there is also tangential acceleration (a_t).
• Net acceleration is a = \sqrt{ac^2 + at^2}
• Example: A Particle is moving clockwise in a circle with its net acceleration is 25 m/s^2 and having position vector 50° angled.
  a{net} = \sqrt{ac^2 + at^2} ac = a{net} \cos(50) at = a{net} \sin(50) ac = \frac{v^2}{r}
  v = \sqrt{a_c * r}

Gears and Rotations

• Gears that do not slip have the same linear speed at their point of contact.
  vA = vB
• Example
  Two gears is in contact each other mA = 1kg \RA = 55 cm, mB = 0.5kg \RB = 30 cm and not slipping.
  A rate rotates 120 RPM slows uniformly to 60 RPM in 2.5 Sec.
  First to find Angular acceleration
  \w = \frac{2\pi}{60}RMP \[0.5ex] \alpha = change in angular velocity/TIME then using \a = r\alpha \alphab = \frac{Ra\alphaa}{Rb}
  then we can find omega for A using \wFinal= \winitial + \alpha_atime
  Using it this for all the parameters can be calculated rotation speed or angular momentum.

Centripetal Force

• Centripetal force (F_c) is the force required to keep an object moving in a circular path.
• It is directed towards the center of the circle.
• Fc = m ac = m \frac{v^2}{r}

Examples for Centripetal Forces

• Example
  Average value is 33.3 *\frac{2\pi}{60}. Initial point angular velocity is 0 with the tire, and the final is what we found. So $\alpha = \frac{w}{t}. Once we have alpha, we can find $\Delta\theta = wat + 1/2 a^2t = (1/2a^2)^{\theta}$.
• Loop the Loops
  So there is always one force acting toward the center in the circle. MG has to equal \frac{mv^2}{r}.
  So minimum velocity will be \sqrt{gr}
• U-Turn Car problem constant Speed And having constant friction
  Constant friction is represented. This is the Maximum speed it can handle. mgs = \frac{mv^2}{r}.
• At a slope is what a bank the is doing so that a car can go faster.

Vertical circle Motion

• MG has to equal \frac{mv^2}{r}.
• Therefore, B comes out the root RB
• Constant circular speed: When ball is on top of it, bottom tensions is more higher. In general. Tensions is 0. When is the point MG downward. So something happens and there is a force. Very important equation here is: the minimum force here is the root five GR at this force root GR you have to never remember it, but it needs to know. If you don't, the example won't work.
• A Phono record is Initially a 33 RPM. How much rotation happened on 0 and after it reached its final speed
  find angular displacement after that we can find Number of Angular Displacement Using the Equation
• A tire is running under uniform acceleration and the tires are .58 cm How many rotations did the car have.

Calculating the moment of Inersia

• Rotation mass date is called movement of initia. Moment of inertia by I is equal to Sigma Mir_i with the radius distance away form rotation

• You have to remember that. Some in Asia for example for a ring when the axis changes the movement of ratio changes. If it's a ring and it's rotating around the center moments of pressure's MS Square. If what's the ring or it is a hollow cylinder that is saying hollow cylinder like this. If it's a disk, the moment of nature's MS Square by two. And if it's a solid cylinder, then also it is MS Square by two. These are important and the one spear if it's solid spear inside two by five MS Square is fallow sper 2 by 3 MS Square. Everything is exactly the same whatever the hard it in linear for example in linear we have displacement here we have angular displacement

• Rotation translational energy is called rolling energy. uniform solid cylinder of mass M is equal 5 kilos is rolling without sleeping whenever there is a term called rolling when we say rolling it's basically moving. So, this system is moving translationally that means it is moving forward as well as there is an element of rotation. So it is rotation plus translation. Translation means forward movement and rotation is the movement along the access. So rolling without sleeping. LOCO horizontal surfaces so velocity is Center of mass. Is 30 meters per seconds? So center Mass is moving at the speed of 30 meter. Calculate its energy when we say calculate its energy we mean rolling energy? That means rotation energy and translation energy so total kinetic energy and rolling is given by rational kinetic energy plus translational kinetic energy. If I know the velocity of the center of mass basically it is moving with the or Omega. So Omega comes out to be B Square by R square. Because Omega is B by R. So I also remember the formula anything like that. All you have to do.

• Example 1s sitting on the corners: radius is 12 meters 60 kilos 45 kilos 80 kilos. \sum mr^2 to determine moment if innertia