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Physics Chapter 8-9
| Question | Answer |
|---|---|
| Angular Displacement (theta) | Given by theta = s/r, where r is the radius and s is the length of an arc in a circle. |
| Angular Velocity (w) | Measured in radians per second. The average angular velocity is measured by dividing the angular displacement by the time required to travel through the angular displacement. |
| Angular Acceleration (a) | Angular acceleration (a) is measured in radians/s^2. The average angular acceleration is defined as the rate of change of the angular velocity (w) in time. |
| Linear Velocity (special v) | As an object travels in a circle, the motion of a point (a) and distance (r) form the center of the circle and can be described in tangential quantities. v = rw and aT = r * alpha |
| Linear Velocity Equation | v = rw Linear velocity = radius * angular velocity |
| Tangential Acceleration Equation | aT = ra Tangential Acceleration = radius * angular acceleration |
| Total Linear Acceleration of a Particle | The total linear acceleration of a particle traveling in a circle is a = aT + ac. Where ac is the centripetal acceleration known as ac = w^2r. |
| What is each symbol analogous to? | x = theta w = v alpha = a |
| Perform Chapter 8 Lecture Problem 1 | -- |
| Perform Chapter 8 Lecture Problem 2 | -- |
| Perform Chapter 8 Recitation Problem 1 | -- |
| Perform Chapter 8 Recitation Problem 2 | -- |
| Miscellaneous Chapter 8 Tips | To turn revs to radians, multiply by 2 pi. To turn rpm to radians/s, multiply by 2 pi and DIVIDE by secs. Also remember that ac = w^2r! This won't be given to you. If they ask for the angle a wheel rotates through, they're just asking for theta. |
| Torque | A measure of effectiveness of a force in producing rotation of an object about an axis. The torque is measured by the product of the force and the perpendicular distance from the axis of rotation to the line along which. the force acts. |
| Level Arm | Perpendicular distance from the axis of rotation to the line along which the force acts. |
| Major Tip to remember torque | To rotate an object, you don't need a certain amount of force. You need a certain amount of torque. |
| Torque Equation | t = Fl = Frsin(theta) L stands for lever arm The SI unit is meter-newtons |
| Perform Chapter 9 Lecture Problem 1 | -- |
| Perform Chapter 9 Lecture Problem 2 (VERY IMPORTANT) | -- |
| Tip for Positive and Negative Values | Try to keep counterclockwise as your positive and clockwise as your negative. |
| Tip for ball hanging of a string... | At static equilibrium, the tension of the ball on the rope would be equal to its weight. |
| Types of forces associated with ladder problems | Fs (running along the bottom), force of floor on ladder (vertical from the bottom), m1g (weight of ladder), m2g (top down, force of the man on the ladder), and force of the wall on the ladder (top horizontal). |
| Relationship between ladder forces | As the man goes up the ladder, the push on the wall increases. The reaction force from the wall increases -> if Fnwall is increasing then fs is increasing. |
| Introduction to Rotational Inertia | When an unbalanced torque acts on an object, it tends to change an object's state of rotation. The magnitude of change depends on the object's moment of inertia & magnitude of torque. |
| Rotational Inertia Equation | Net torque = rotational inertia * angular acceleration |
| Linear & Moment of Inertia | Linear inertia is tendency to resist any change in linear motion; moment of inertia is tendency to resist rotational motion. |
| Moment of Inertia Equation | Inertia = sum of masses * r^2 |
| Thin-wall hollow cylinder or hoop | I = MR^2 |
| Solid cylinder or disk | I = 1/2MR^2 (notice that the inertia is smaller because the mass is closer to the axis of rotation!) |
| Solid sphere, axis through center | I = 2/5MR^2 |
| Thin rod, axis perpendicular to rod and passing through center | I = 1/12ML^2 |
| Thin rod, axis perpendicular to rod and passing through one end | I = 1/3ML^2 |
| When the object's pulled by the rope are not massless... | You cannot say that the two tensions are equal. |
| Perform Chapter 9 Lecture Problem 3 (VERY IMPORTANT) | -- |
| When using I * alpha in a disk equation, remember | I = whatever the formula of the shape's inertia is and alpha = a/r This applies to when you're trying to find the acceleration (tangential acceleration). |
| Sliding Block Vs. Rolling Disk Accelerating Down an Incline | Both objects are going to have the same motion (sliding motion) and the same acceleration. |
| Sliding Block Specifics | Work of non conservative forces is zero, so you can use conservation of energy to find velocity. |
| Solid Disk Specifics | Work of non conservative forces is zero, so you can use conservation of energy to find velocity. KE NEEDS to be split into horizontal and rotational parts. |
| Horizontal kinetic energy | 1/2mv^2 |
| Rotational kinetic energy | 1/2Iw^2 |
| Which has a higher velocity? The sliding block or the solid disk? | The sliding block, because the energy for the disk needs to be split into translational and rotational energies. |
| When trying to find velocity a disk equation, remember | The rotational kinetic energy tends to translate to 1/2Iw^2 -> 1/2 inertia equation^2 * (v^2/r^2) |
| Solid Disk Vs Hollow Disk Main Takeaway | The speed of the solid disk is going to be larger than the hollow disk at the bottom of the incline. The larger the moment of inertia, the slower the objects are going to roll. |
| Forces acting on sliding objects | Fn going straight up, mg going straight down, fs going back from where they're headed |
| Complete Chapter 9 Lecture Problem 4 (VERY IMPORTANT) | -- |
| Angular Momentum | L = Iw L relates to P I relates to m W relates to v |
| Net Torque Equation | Change in L / change in t |
| The law of conservation of angular momentum states,,, | If the net torque acting on an object is zero, the object's angular momentum must remain consistent in both magnitude and direction. If change in L equals 0, then net torque equals 0. |
| Complete Chapter 9 Recitation Problem 1 | -- |
| Complete Chapter 9 Recitation Problem 2 (Very important!) | -- |
| Complete Chapter 9 Recitation Problem 3 | -- |
| Complete Chapter 9 Recitation Problem 4 | -- |
| Complete Chapter 9 Recitation Problem 5 | -- |
| When you have a pulley problem, that involves torques, here are the steps... | Make FBD. Set sum of forces for each mass to respective ma, find equations. Find the value of (ex: tension) using torqs; remember that lever arm of a string wrapped around a cyl is r! Use sum of torque = I*alpha, then add equations together and solve. |
Created by:
smurtab