Applications of torque Announcements: CAPA Set #10 due on Friday. Exam results are available on D2L Check your scores and make sure they are consistent with how you think you marked the test! Drop deadline for A&S Students is this Friday, March 23! W http://www.colorado.edu/physics/phys1110/phys1110_sp12/
Long Answer Exam 2 Note 25% of students got all of the points possible
Exam 2 Total Points
Sum of Exams 1+2 This is 30% of the points for determining the final grade.
Grading 3 midterm exams, 15% each: 45% Final exam: 23% CAPA homework 15% Tutorial participation + homeworks 10% In-lecture clicker participation: 4% SmartPhysics Prelecture participation 3% --------------------------------------- --------- Total 100%
Summary of torque so far Torque is the cross product (or vector product) of the vector which is the distance from the axis to the force application and the force vector : In many cases we only care about the magnitude where is the angle between the and vectors SI units of torque are newton meter (N m). This has the same dimensions as energy but only use Joules for energy Newton s 2 nd law can be rewritten as for cases of rotational motion.
Massive pulleys A 2 kg weight is attached to the end of a rope coiled around a pulley with mass of 6 kg and radius of 0.1 m. What is the acceleration of the weight? Weight free body diagram: For the pulley, rope (and therefore the tension force) comes off at 90 relative to R so. Disk moment of inertia is so becomes 2 kg Acceleration along outside of pulley equals acceleration of 2 kg weight because connected by rope so becomes
Massive pulleys A 2 kg weight is attached to the end of a rope coiled around a pulley with mass of 6 kg and radius of 0.1 m. What is the acceleration of the weight? We have two equations Solving gives rearranges to and 2 kg
Clicker question 1 Set frequency to BA For 2 kg weight attached to 6 kg pulley with R=0.10 m, what is the kinetic energy of the pulley after dropping for 1 second. Remember v=4 m/s for the weight after 1 second and K=½mv 2 and K=½Iω 2 and I disk =½MR 2. A. 4 J B. 8 J C. 16 J D. 24 J E. 48 J 2 kg
Clicker question 1 Set frequency to BA For 2 kg weight attached to 6 kg pulley with R=0.10 m, what is the kinetic energy of the pulley after dropping for 1 second. Remember v=4 m/s for the weight after 1 second and K=½mv 2 and K=½Iω 2 and I disk =½MR 2. A. 4 J B. 8 J C. 16 J D. 24 J E. 48 J The weight velocity and the outside pulley velocity are the same since they are connected by the rope. Can get angular velocity from rim velocity and then find kinetic energy 2 kg
Massive pulleys A 2 kg weight is attached to the end of a rope coiled around a pulley with mass of 6 kg and radius of 0.1 m. The weight acceleration is found to be 4 m/s 2. After 1 second the total kinetic energy was found to be After 1 second the weight will descend losing gravitational potential energy 2 kg Conservation of energy still holds
Possible pitfalls Sometimes more than one R in the problem. R is found in several places (choose correctly) 1. Torque calculation 2. Moment of inertia calculation 3. Conversion between rope linear velocity and pulley angular velocity If pulley has a hole in the center could add another R which affects moment of inertia 2 kg 2 kg
Notes on gravity The potential energy due to gravity for an extended body is measured from the center of mass. When calculating torque due to gravity, the location of the force of gravity can be assumed to be the center of mass. Note the torque will change as the rod falls due to the angle changing
A few more tips If torques and forces aren t helping, try conservation of energy (and vice versa) Everything you have already learned continues to work Newton s 2 nd law is always true (for inertial reference frames) Work and conservation of energy still apply.
Rotation & Translation Problems so far had a stationary axis. Other problems have a moving axis like a boomerang, bowling ball, or yo-yo. This is a combination of translational motion and rotational motion Split motion into two parts 1. Translational motion of the center of mass 2. Rotational motion around the center of mass
Separate kinetic energy terms 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
Clicker question 2 Set frequency to BA A mass m hangs from string wrapped around a pulley of radius R. The pulley has a moment of inertia I and its pivot is frictionless. Because of gravity the mass falls and the pulley rotates. The magnitude of the torque on the pulley is A. greater than mgr B. less than mgr R C. equal to mgr m
Clicker question 2 Set frequency to BA A mass m hangs from string wrapped around a pulley of radius R. The pulley has a moment of inertia I and its pivot is frictionless. Because of gravity the mass falls and the pulley rotates. The magnitude of the torque on the pulley is A. greater than mgr B. less than mgr R C. equal to mgr (Hint: Is the tension in the string = mg?) Free body diagram of mass gives T=mg-ma when a is positive so T<mg so m
Yo-yo A yo-yo of mass M and radius R starts from rest and is dropped, attached by an unrolling string. What is the center of mass velocity after falling a height h? We know that falling a height loses gravitational potential energy which in this case is This energy goes into kinetic energy: Since this is rolling without slipping Setting
Other aspects of translational + rotational motion Sometimes conservation of energy doesn t tell us everything we want to know. In these cases, you probably need both versions of Newton s 2 nd law and You can read the book to see how the previous yo-yo example works when investigating the forces and torques (Section 11.5 on page 281). We are going to try a slightly harder yo-yo problem.
Pulling the yo-yo Yo-yo with inner radius R 1 and outer radius R 2 is on a table. String is pulled with force F as shown. There is enough static friction to prevent the yo-yo from sliding. Which way does the yo-yo move? Rolling without slipping: and but you need to make sure you get the signs right! so Substituting gives Rearranging: