chapter8_slides_updated

chapter8_slides_updated - 10/24/2009 Chapter 8 Rotational...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
10/24/2009 1 Chapter 8 Rotational Equilibrium and Dynamics ) sin( rF 0 0 i i F Equilibrium: I L f f i i I I Even opening the door needs PHYSICS??? !!! Yes, you already know it! But may not have realized. Three factors that determine the effectiveness of the force in opening the door: The magnitude of the force The position of the application of the force The angle at which the force is applied Combining three factors - Torque
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
10/24/2009 2 Torque ( ) The effectiveness of a force to rotate an object about some axis SI unit is N . m r : magnitude of the position vector of the point where force is applied, measured from the rotational axis If the force is perpendicular to the position vector, 1 ) sin( ) rF It is sometime convenient to introduce the concept of lever arm so that dF ) ) r d
Background image of page 2
10/24/2009 3 Forces cause (linear) accelerations Torques cause angular accelerations Like force, torque is a vector quantity If the turning tendency of the force is counterclockwise , the torque is positive If the turning tendency is clockwise , the torque is negative When two or more torques are acting on an object, the torques are added as vectors ( can be positive or negative ) If the net torque is zero, the object’s rate of rotation doesn’t change: It does not rotate, or does constant speed rotation Two people are pushing a revolving door at the same time. Both forces are perpendicular to the position vectors. Calculate the net torque on the revolving door. Analysis: Torque can be positive (CCW) or negative (CW) Solution: Torque 1: negative (CW) Torque 2: positive (CCW) Total torque: m 2 . 1 ; N 625 1 1 r F m 8 . 0 ; N 850 2 2 r F 1 1 1 F r m N 750 625 2 . 1 2 2 2 F r m N 680 850 8 . 0 m N 70 680 750 2 1 The lady wins!
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
10/24/2009 4 If the magnitude of the applied force is the same, in which case the torque is the smallest? (1) (2) (3) A man applies a force of 300 N at an angle of 60 o to the door. The position of the hinges is the axis of rotation. (a) Find the torque on the door. Solution: (b) Another person applies a force F 2 perpendicular to the door. What must be F 2 so that the man in part (a) won’t open the door? ) sin( 1 rF 2 F 520 2 2 m N 520 ) 60 300 0 . 2 o 1 2 2 1 0 N 260 / 520 2 r F
Background image of page 4
10/24/2009 5 Still remember this “amazing” photo? How can a small car overturns a big truck? Torque Suppose the arm of the crane truck extends 5 m. The mass of the car is 1000 kg. The mass of the truck is 3000 kg. Net torque: m N 49000 5 8 . 9 1000 car car car car mgr r F mg F car m 5 m 5 . 1 m N 44100 5 . 1 8 . 9 3000 truck truck truck truck Mgr r F m N 4900 44100 49000 truck car Mg F truck We have discussed the condition for equilibrium: the net force applied on the object is zero.
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/02/2010 for the course PHY 101 taught by Professor Pralle during the Spring '08 term at SUNY Buffalo.

Page1 / 22

chapter8_slides_updated - 10/24/2009 Chapter 8 Rotational...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online