Physics 121 March 20 2008 Frank L H Wolfs Department of Physics and Astronomy University of Rochester Physics 121 March 20 2008 Course Information Quiz Topics to be discussed today Rotational Variables Review Torque Rolling Motion Review for Exam 2 Frank L H Wolfs Department of Physics and Astronomy University of Rochester Physics 121 March 20 2008 Homework set 6 is now available on the WEB and will be due on Saturday morning March 22 at 8 30 am There will be no homework due on March 29 Exam 2 will take place on Tuesday March 25 at 8 am in Hubbell It will cover the material discussed in Chapters 7 8 and 9 There will be no workshops or office hours on Tuesday Friday next week Extra office hours will be scheduled for Sunday 3 23 and Monday 3 24 Frank L H Wolfs Department of Physics and Astronomy University of Rochester Physics 121 Quiz lecture 16 The quiz today will have 3 questions Frank L H Wolfs Department of Physics and Astronomy University of Rochester Rotational variables A quick review The variables that are used to describe rotational motion are Angular position Angular velocity d dt Angular acceleration d dt The rotational variables are related to the linear variables Linear position l R Linear velocity v R Linear acceleration a R Frank L H Wolfs Department of Physics and Astronomy University of Rochester Rotational variables A quick review Things to consider when looking at the rotation of rigid objects around a fixed axis Each part of the rigid object has the same angular velocity Only those parts that are located at the same distance from the rotation axis have the same linear velocity The linear velocity of parts of the rigid object increases with increasing distance from the rotation axis Frank L H Wolfs Department of Physics and Astronomy University of Rochester Rotational variables A quick review Note the acceleration at r is only one of the two component of the acceleration of point P The two components of the acceleration of point P are The radial component this component is always present since point P carried out circular motion around the axis of rotation The tangential component this component is present only when the angular acceleration is not equal to 0 rad s2 Frank L H Wolfs Department of Physics and Astronomy University of Rochester Rotational variables A quick review Angular velocity and acceleration are vectors They have a magnitude and a direction The direction of is found using the right hand rule The angular acceleration is parallel or antiparallel to the angular velocity If increases parallel If decreases anti parallel Frank L H Wolfs Department of Physics and Astronomy University of Rochester Rotational Kinetic Energy A quick review Since the components of a rotating object have a non zero linear velocity we can associate a kinetic energy with the rotational motion 1 1 1 1 2 2 2 2 2 K mi vi mi ri mi ri I 2 i 2 i 2 i 2 The kinetic energy is proportional to square of the rotational velocity Note the equation is similar to the translational kinetic energy 1 2 mv2 except that instead of being proportional to the the mass m of the object the rotational kinetic energy is proportional to the moment of inertia I of the object Note units of I kg m2 I mi ri 2 or I r 2 dm i Frank L H Wolfs Department of Physics and Astronomy University of Rochester Torque Consider a force F applied to an object that can only rotate The force F can be decomposed into two two components A radial component directed along the direction of the position vector r The magnitude of this component is Fcos This component will not produce any motion A tangential component perpendicular to the direction of the position vector r The magnitude of this component is Fsin This component will result in rotational motion Frank L H Wolfs F r A Department of Physics and Astronomy University of Rochester Torque If a mass m is located at the position on which the force is acting and we assume any other masses can be neglected it will experience a linear acceleration equal to Fsin m The corresponding angular acceleration is equal to Fsin mr Since in rotational motion the moment of inertia plays an important role we will rewrite the angular acceleration in terms of the moment of inertia rFsin mr2 rFsin Frank L H Wolfs F r A Department of Physics and Astronomy University of Rochester Torque Consider rewriting the previous equation in the following way rFsin The left hand side of this equation is called the torque of the force F This equation looks similar to Newton s second law for linear motion F ma Note linear rotational mass m moment I force F torque Frank L H Wolfs F r A Department of Physics and Astronomy University of Rochester Torque In general the torque associated with a force F is equal to rF sin r F The arm of the force also called the moment arm is defined as rsin The arm of the force is the perpendicular distance of the axis of rotation from the line of action of the force If the arm of the force is 0 the torque is 0 and there will be no rotation The maximum torque is achieved when the angle is 90 Frank L H Wolfs Department of Physics and Astronomy University of Rochester Rotational motion Sample problem Consider a uniform disk with mass M and radius R The disk is mounted on a fixed axle A block with mass m hangs from a light cord that is wrapped around the rim of the disk Find the acceleration of the falling block the angular acceleration of the disk and the tension of the cord M R m Expectations Linear acceleration should approach g when M approaches 0 kg Frank L H Wolfs Department of Physics and Astronomy University of Rochester Rotational motion Sample problem Start with considering the forces and torques involved Define the sign convention to be used The block will move down and we choose the positive and we choose the positive y axis in the direction of the linear acceleration The net force on mass m is equal to T R a T mg ma mg T Frank L H Wolfs Department of Physics and Astronomy University of Rochester Rotational motion Sample problem The net torque on the pulley is equal to RT The resulting acceleration is equal to angular T RT 2T 1 I MR 2 MR 2 Assuming the cord is not slipping we can determine the linear acceleration R a T mg 2T a M Frank L H Wolfs Department of Physics and Astronomy University of Rochester Rotational motion Sample problem We now have two expressions for a 2T a M T T a g M R a Solving these equations we find M T mg M 2m 2m a g M 2m Frank L H Wolfs T mg Note a