Lecture 15 Torque and Center of Gravity

LECTURE 15 TORQUE AND CENTER OF GRAVITY

7.3 Torque What is unusual about this door? Net torque 7.4 Gravitational torque and the center of gravity Calculating the position of the center of gravity Learning objectives

! Compute torque for a force applied to extended objects. ! Identify moment arm for a force ! Compute net torque for forces applied to extended objects. ! Calculate center of gravity of an object or a collection of objects. 7.3 Torque

! Torque (!) is a measure of twisting, and the magnitude of torque is defined as $⃗

! = #$ sin ( = $)# = $#) ( ⃗ ! #) is called moment arm, or lever arm of $⃗. $) ! By convention: ! A torque that tends to rotate the object in a counter-clockwise direction is positive. #⃗ ! A torque that tends to rotate the object in a clockwise direction is negative. Pivot ! The net torque is the sum of the torques due to the applied forces.

!+,- = . !

Line of action Quiz: 7.3-1

! Suppose a ladder on a rough floor is leaning against a frictionless !*" wall as shown, and you are trying to calculate the torque on the ladder about the pivot through where the ladder touches the floor due to the weight of the ladder. What is the moment arm of this force? A. 2, B. , h C. ℎ $%" D. ℎ. + 2, .

E. ℎ. + ,. !"# 0 . F. + ,. . ⃗ '(,#" r r Quiz: 7.3-1 answer

! ! $-%

! The moment arm is the shortest distance between the line of action and the axis. h ! Follow-up: What is the sign of this torque? Line of action '(%

$%&

⃗ *+,&% r r !" = ! Quiz: 7.3-2

! Suppose a ladder on a rough floor is leaning against a frictionless !*" wall as shown, and you are trying to calculate the torque on the ladder about the pivot through where the ladder touches the floor due to the normal force by the wall. What is the moment arm of this force? A. 2, B. , h C. ℎ $%" D. ℎ. + 2, .

E. ℎ. + ,. !"# 0 . F. + ,. . ⃗ '(,#" r r Quiz: 7.3-2 answer

Line of action ! ℎ %.&

! The moment arm is the shortest distance between the line of action and the axis. " = ℎ h ! Follow-up: What is the sign of this # torque? ()&

%&'

⃗ +,,'& r r Quiz: 7.3-3

! The four forces shown have the same strength. Rank the force in the order of the effectiveness in in opening the door, smallest first. Quiz: 7.3-3 answer / demo

$ ! The four forces shown have the same strength. %+ Rank the force in the order of the effectiveness in $%) in opening the door, smallest first.

! The greater the torque a force can apply on the

door about its hinges, the more easily you can $%* open the door. Line of action for #⃗* ! ! = #$%

! 0 = $%' < $%) < $%* < $%+

! 0 = !' < !) < !* < !+ ! 2 < 4 < 3 < 1 Quiz: 7.3-4

! A 2.20-m flagpole is held to the side of a building by a string. The flagpole has a mass of 1.15 kg. A weight force acts at the center of the String flagpole.

! About the pivot point, what is the magnitude of Flagpole the torque in N " m applied by the weight force?

35.0°

pivot point Quiz: 7.3-4 answer

! About the pivot point, what is the magnitude of the torque in N " m applied by the weight force?

! $ = &'( &⃗) ! $ = &),'+ = &) cos 35.0° 45 35.0° = +BC ! $ = (1.10 m) cos 35.0° (1.15 kg)(9.80 ) >? pivot point ! $ = 10.2 N " m

! Follow-up: What is the sign of this torque? &⃗D &⃗),∥ 35.0° +BC &⃗),' Quiz: 7.3-5

! A 2.20-m flagpole is held to the side of a building by a string. The magnitude of the torque applied by the weight force about the String pivot point is 10.2 N " m.

! A tension force is applied by the string at a Flagpole point 1.80 m from the wall in the direction along the flagpole. 35.0° ! What is the magnitude of the tension force in N?

pivot point Quiz: 7.3-5 answer

! What is the magnitude of the tension force

in N? )*+ 35.0° ! !"#$ = !& − !( = )*+,&,. − !( = ,⃗& )*+ ,& sin 35.0° − !( = 0 78 @?.A BCD 35.0° ! )*+ = = = 9: ;<" =>.?° @.E? D ;<" =>.?° 9.88 N pivot point

,⃗&,∥ )*+ 35.0° ,⃗&,. ,⃗& Quiz: 7.4-1

! Where is the center of gravity of the lifebuoy? Quiz: 7.4-1 answer

! Where is the center of gravity of the lifebuoy?

! The gravitational torque can be calculated by assuming that the net force of gravity (the object’s weight) acts at the center of gravity.

! For a highly symmetric object, the center of gravity is at the center of symmetry. The center of gravity is at the geometric center of a uniform object even if there is no mass at that location. 7.4 Gravitational torque and the center of gravity

! An object that is free to rotate about a pivot will come to rest with the center of gravity below the pivot point.

! An object is balanced if its center of gravity is directly above the base on which it is supported. Carrying water / good posture

!17

! Why is it easier to carry the same amount of water in two buckets, one in each hand, than in a single bucket in one hand? ! With two buckets, the center of gravity will be in the center of the support base provided by one’s feet, so there is no need to lean.

! Good posture places the upper body’s center of gravity over the pivots in the hips. Poor posture requires exertion by the back muscles to counteract the torque produced around the pivot by the upper body’s weight. 7.4 Gravitational torque and the center of gravity / demo

! The torque due to gravity when the pivot is at the center of gravity is zero.

! The location of the center of gravity is:

!%&% + !(&( !"# = &% + &(

! The center of gravity tends to lie closer to the heavier objects or particles that make up the object.

! Demos: meter stick, see-saw Quiz: 7.4-2

! Two uniform thin beams are joined end-to-end as shown to make a single object. The left beam is 10.0 kg and the 1.00 m long and the right side is 40.0 kg and 2.00 m long.

! How far from the left end of the left beam is the center of gravity of the object? Quiz: 7.4-2 answer

! Two uniform thin beams are joined end-to-end as shown to make a single object. The left beam is 10.0 kg and the 1.00 m long and the right side is 40.0 kg and 2.00 m long.

! How far from the left end of the left beam is the center of gravity of the object? %&'&(%)') ! !"# = '&(') *., - .*.* /# ( 0.* - 1*.* /# ! ! = "# .*.* /#(1*.* /#

! !"# = 1.7 m Fosbury flop

! In the 1968 Olympic games, University of Oregon jumper Dick Fosbury Introduced a new technique of high jumping.

! It raised the world record by ~30 cm, and is used by nearly every world-class jumper.

! The jumper goes over the bar facing up while arching their back, placing their center of gravity outside of their body, below their back.