Review for 3 rd Midterm

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Blaise Norman
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1 Review for 3 rd Midterm Midterm is on 4/19 at 7:30pm in the same rooms as before You are allowed one double sided sheet of paper with any handwritten notes you like. The moment-of-inertia about the center-of-mass for any necessary geometries are on page 2 of the exam. The main material covered will be Chapters 10,11,12,13. You also need to know previous material like collisions and forces. Parts of the book explicitly NOT included in this exam/class are: 12.6, 12.7, 13.9 Don t forget to contact Professor Dessau if you have an exam conflict

2 Exam questions Multiple choice questions 70% statics questions gravity questions conservation of angular momentum questions rotational energy problems torque & force type problems Short answer questions 30%

3 Study suggestions (approximately in order) 1. CAPA: Make sure you understand and can do the CAPA problems. Think about possible modifications to the problems and what you would need to do different to solve it. 2. Lecture: Make sure you understand the lecture notes and can do all of the clicker questions. Think about modifications to the clicker questions that could be asked. 3. Tutorials: Review the tutorials including the homework to make sure you understand. For questions you got wrong, analyze where your thinking went astray and make sure you have corrected that aspect. For ungraded questions, go over them again to see if you have changed your mind about your answers.

4 Study suggestions (approximately in order) 4. End of chapter problems: These problems should help you setup and solve problems. Try to solve the problems using only your formula sheet and calculator as in the exam. 5. Old exams: The exams this time are usually quite representative. Try to solve the problems using only your formula sheet and calculator and impose a time limit. 6. Read the text: There are examples worked out, suggestions for how to solve problems, and interesting relationships explored that might help you remember the essential bits.

5 Chapter 10 angular kinematics Introduced rigid body rotation. Before we only dealt with point particles. Now we can deal with extended bodies as long as they don t flop around (every part has to be rigidly connected). We found rotational analogues for all of the linear equations: Constant linear acceleration Constant angular acceleration Related angular and linear terms: Moment of inertia and rotational kinetic energy Look at the Lecture clicker question examples of a CD rotating

6 Clicker question 1 Set frequency to BA A bike has a front wheel of radius 0.25 m. It starts from rest, and accelerates down a hill with linear acceleration a= 2.0 m/s^2. At t= 1.0 sec, what is the angular velocity ω of the bike wheel? A) 1.0 rad/s B) 4.0 rad/sec C) 8.0 rad/s D) 8 π rad/s E) 16 π rad/s

7 Clicker question 1 Set frequency to BA A bike has a front wheel of radius 0.25 m. It starts from rest, and accelerates down a hill with linear acceleration a= 2.0 m/s^2. At t= 1.0 sec, what is the angular velocity ω of the bike wheel? A) 1.0 rad/s B) 4.0 rad/sec C) 8.0 rad/s D) 8 π rad/s E) 16 π rad/s α z = a /r ω z = 0 + a r t = 2m /s2.25m (1s) = 8rad /s

8 Chapter 10 other stuff Can write rotational versions of work and power: becomes becomes Torque and angular momentum are vectors. This can be used to understand how gyroscopes and bicycles work.

9 Chapter 11: Rotation We can separate the total kinetic energy into two parts, translational kinetic energy and rotational kinetic energy. Sometimes center of mass velocity & angular velocity are related. Rolling without slipping: translational rotational total

10 Chapter 11 torque & angular momentum Torque is the rotational equivalent to force. It is determined by the force and lever arm. and F Newton s 2 nd law F=ma has a r rotational equivalent:. Add up the known torques by drawing an extended free body diagram and using. Set the result equal to. Remember the signs (usually counterclockwise is positive). Rotational equivalent to momentum is angular momentum and just as no net force means momentum is conserved, no net torque means angular momentum is conserved. Angular momentum depends on object and the axis. (usually for particle) and (usually for rigid body).

11 Torque and energy problem A cylinder (r = 0.14 m, I cm = kg m 2, M = 1.5 kg) starts at rest and rolls without slipping down a plane with an inclination angle of θ=15 o. Find the time it takes to travel 1.4 m. Remember rolling without slipping means! Problem diagram Free body diagram!!! Use Newton s 2 nd law: Torque about center of mass From torque: Solve: substitute into force: Use to find the time:

12 Torque and energy problem A cylinder (r = 0.14 m, I cm = kg m 2, M = 1.5 kg) starts at rest and rolls without slipping down a plane with an inclination angle of θ=15 o. Find the time it takes to travel 1.4 m. Remember rolling without slipping means! Can also use conservation of energy to solve this. Remember, rolling without friction is static friction no energy is lost! Set equal and use : Solve for final velocity: Use to get time:

13 Clicker question 2 Set frequency to BA

14 Clicker question 2 Set frequency to BA K = K trans + K rot = 1 2 mv Iω 2

15 What are some variations? Variations that use the forces and torques approach What is the frictional force? Minimum coefficient of static friction to prevent slipping? Given a coefficient of static friction, what is the maximum angle to roll without slipping? Given a coefficient of static friction, angle, and mass, what is the maximum moment of inertia that still rolls without slipping? Variations that use energy approach What is the final velocity? What is the final angular velocity?

16 Chapter 12 - Statics Statics deals with situations in which nothing is moving. Therefore, there is no acceleration or angular acceleration. In two dimensions, the static equilibrium equations are: To solve problems, draw an extended free body diagram, pick an origin, and setup the three equations. If you have three unknowns or less then solve the equations; otherwise check the problem for additional information.

17 Clicker question 3 Set frequency to BA A ladder leans against a frictionless wall at an angle θ. The floor exerts an upward normal force of magnitude N and a horizontal friction force of magnitude f. A) f/n = sin θ B) f/n = cos θ / sin θ C) f/n = cos θ D) f/n = 2 sin θ / cos θ E) f/n = 0.5 cos θ / sin θ θ

18 Clicker question 3 Set frequency to BA A ladder leans against a frictionless wall at an angle θ. The floor exerts an upward normal force of magnitude N and a horizontal friction force of magnitude f. A) f/n = sin θ B) f/n = cos θ / sin θ C) f/n = cos θ D) f/n = 2 sin θ / cos θ E) f/n = 0.5 cos θ / sin θ θ F x = 0 = f s F w F y = 0 = F N W = 0 τ = 0 = F W, L W (L /2) F W, = F W sinθ = f s sinθ f s sinθ = W cosθ /2 W = W cosθ

19 Clicker question 4 Set frequency to BA A uniform hanging meter stick is supported by tension T 1 in a string at 0 cm and tension T 2 in a string at 80 cm. The ratio T 2 /T 1 is A) 5/3 B) 5/4 C) 3/5 D) 4/5 E) 4/3 T 1 T 2 0 cm 80 cm

20 Clicker question 4 Set frequency to BA A uniform hanging meter stick is supported by tension T 1 in a string at 0 cm and tension T 2 in a string at 80 cm. The ratio T 2 /T 1 is A) 5/3 B) 5/4 C) 3/5 D) 4/5 E) 4/3 T 1 T 2 0 cm 80 cm F = 0 = T 1 + T 2 mg = 0 Pick pivot axis at 0.5m τ = 0 = 0.3T 2 0.5T 1 T 2 T 1 = 5 3

21 Chapter 13:Universal Gravitation Newton s Law of Gravitation: Any two masses, anywhere in the universe, are attracted to each other by this force. Can still use Newton s 2 nd law to find acceleration If you (with mass m) are a distance r away from a planet with mass M P then the force on you is You will be accelerated at (independent of your mass) The planet feels the exact same force and will be accelerated at As you fall, you accelerate at 9.8 m/s 2 toward the ground The earth accelerates toward you at about 1 x m/s 2!

22 Universal Gravitation: Potential energy Gravity is a conservative force and therefore has a potential energy associated with it Can use conservation of energy to determine things like velocity at a given height or the escape velocity. Total energy Often one mass is much more massive and is basically motionless and so the total energy is is conserved Negative total energy is a bound system (closed orbits or objects eventually colliding). Total energy 0 is an unbound system.

23 Universal Gravitation: Escape velocity Escaping a gravitational field implies total energy 0. Escape velocity is speed at which you have enough kinetic energy to balance out the negative potential energy so total energy is 0. That is, means For a small mass m escaping a big mass M:

24 Universal Gravitation: Orbits Applying Newton s 2 nd law with the gravitational force and radial acceleration gives M This gives a speed of m Can relate orbital period to velocity and derive Using the speed we determined All this was for circular orbits only! For elliptical orbits, r is not constant so speed, kinetic energy and potential energy all change (total energy is conserved) and

= o + t = ot + ½ t 2 = o + 2

= o + t = ot + ½ t 2 = o + 2 Chapters 8-9 Rotational Kinematics and Dynamics Rotational motion Rotational motion refers to the motion of an object or system that spins about an axis. The axis of rotation is the line about which the