Post on 19-Dec-2015
Chapter 8
Rotational Motion
Objectives
• state the meaning of the symbols used in the kinematics equations for uniformly accelerated angular motion.
• describe torque.
• solve word problems related to angular kinematics.
Torque
• A torque is an action that causes objects to rotate.
• Torque is not the same thing as force.
• For rotational motion, the torque is what is most directly related to the motion, not the force.
TORQUE
• To make an object rotate, a force must be applied in the right place.
• the combination of force and point of application is called TORQUE
Force, F
lever arm, L
Axle
Torque = force times lever arm
Torque = F L
Torque
• Motion in which an entire object moves is called translation.
• Motion in which an object spins is called rotation.
• The point or line about which an object turns is its center of rotation.
• An object can rotate and translate.
8-4 TorqueTo make an object start rotating, a force is needed; the position and direction of the force matter as well.
The perpendicular distance from the axis of rotation to the line along which the force acts is called the lever arm.
8-4 Torque
A longer lever arm is very helpful in rotating objects.
Torque
• Torque is created when the line of action of a force does not pass through the center of rotation.
• The line of action is an imaginary line that follows the direction of a force and passes though its point of application.
Torque
• To get the maximum torque, the force should be applied in a direction that creates the greatest lever arm.
• The lever arm is the perpendicular distance between the line of action of the force and the center of rotation
Why things fall over• Every object has a special point called the
center of gravity (CG). The CG is usually right smack in the center of the object.
• if the center of gravity is supported, the object will not fall over.
• You generally want a running back with a low CG then it’s harder to knock him down.
• The lower the CG the more stable an object is. stable not easy to knock over!
Condition for stability
If the CG is above the edge, the objectwill not fall
CG
when does it fall over?
CG CG
STABLE NOT STABLE
If the vertical lineextending down fromthe CG is inside theedge the object willreturn to its uprightposition the torquedue to gravity bringsit back.
Stable and Unstable
stable unstable
torque due to gravitypulls object back
torque due to gravitypulls object down
Stable structures
Structures arewider at their
base to lower theircenter of gravity
If the center of gravityis supported, the blocksdo not fall over
Playing with your blocks
CG
Object with low CG
Stay low to the ground!
300 lb fullback who is4 ft, 10 inches talland runs a 4-40
As more and more stuff is loaded into asemi, its center of gravity moves upward.It could be susceptible to tipping over.
High ProfileVehicles
wind
Enrichment
Linear and Rotational Motion of Cars.
8-4 Torque
Here, the lever arm for FA is the distance from the knob to the hinge; the lever arm for FD is zero; and the lever arm for FC is as shown.
Torque
t = r x F
Lever arm length (m)
Force (N)
Torque (N.m)
Sign Convention for Sign Convention for TorqueTorque
By convention, counterclockwise torques are positive and clockwise
torques are negative.Positive torque:
Counter-clockwise, out of
pagecw
ccw
Negative torque: clockwise, into page
8-4 Torque
The torque is defined as:
(8-10a)
(8-10c)
Net Force = 0 , Net Torque ≠ 0
10 N
10 N
• > The net force = 0, since the forces are applied in opposite directions so it will not accelerate.
• > However, together these forces will make the rod rotate in the clockwise direction.
Net torque = 0, net force ≠ 0
The rod will accelerate upward under thesetwo forces, but will not rotate.
Balancing torques
10 N20 N
1 m 0.5 m
Left torque = 10 N x 1 m = 10 n mRight torque = 20 N x 0.5 m = 10 N m
Torque Lab
Torque example
F
L
What is the torque on a boltapplied with a wrench that has a lever arm of 30 cmwith a force of 10 N?
Torque = F L = 30 N 0.30 m = 9 N m
For the same force, you get more torquewith a bigger wrench the job is easier!
Example 1:Example 1: An An 80-N80-N force acts at the end of force acts at the end of a a 12-cm12-cm wrench as shown. Find the torque. wrench as shown. Find the torque.
• Extend line of action, draw, calculate r.
= (80 N)(0.104 m) = 8.31 N m
= (80 N)(0.104 m) = 8.31 N m
r = 12 cm sin 600 = 10.4 cm
r = 12 cm sin 600 = 10.4 cm
Example 2:Example 2: Find resultant torque Find resultant torque about axis about axis AA for the arrangement for the arrangement shown below:shown below:
300300
6 m 2 m4 m
20 N30 N
40 NA
Find due to each force. Consider 20-N force first:
Find due to each force. Consider 20-N force first:
r = (4 m) sin 300 = 2.00 m
= Fr = (20 N)(2 m) = 40 N m, cw
The torque about A is clockwise and negative.
20 = -40 N m20 = -40 N m
r
negative
Example 2 (Cont.):Example 2 (Cont.): Next we find Next we find torque due to torque due to 30-N30-N force about force about same axis same axis AA..
300300
6 m 2 m4 m
20 N30 N
40 NA
Find due to each force. Consider 30-N force next.
Find due to each force. Consider 30-N force next.
r = (8 m) sin 300 = 4.00 m
= Fr = (30 N)(4 m) = 120 N m, cw
The torque about A is clockwise and negative.
30 = -120 N m30 = -120 N m
rnegative
Example 2 (Cont.):Example 2 (Cont.): Finally, we Finally, we consider the torque due to the consider the torque due to the 40-N40-N force.force.
Find due to each force. Consider 40-N force next:
Find due to each force. Consider 40-N force next:
r = (2 m) sin 900 = 2.00 m
= Fr = (40 N)(2 m) = 80 N m, ccw
The torque about A is CCW and positive.
40 = +80 N m40 = +80 N m
300300
6 m 2 m4 m
20 N30 N
40 NA
r
positive
Example 2 (Conclusion):Example 2 (Conclusion): Find Find resultant torque about axis resultant torque about axis AA for the for the arrangement shown below:arrangement shown below:
300300
6 m 2 m4 m
20 N30 N
40 NA
Resultant torque is the sum of individual torques.
Resultant torque is the sum of individual torques.
R = - 80 N mR = - 80 N m Clockwise
R = 20 + 20 + 20 = -40 N m -120 N m + 80 N m
Practice Problem #1Practice Problem #1
A person exerts a force of 45N on A person exerts a force of 45N on the end of a door 84cm wide. the end of a door 84cm wide. What is the magnitude of the What is the magnitude of the torque if the force is exerted (a) torque if the force is exerted (a) perpendicular to the door, and (b) perpendicular to the door, and (b) at a 60at a 6000 angle to the face of the angle to the face of the door? door?
Practice Problem #1 SolutionPractice Problem #1 SolutionAA person exerts a force of 45N on the end of a door 84cm person exerts a force of 45N on the end of a door 84cm wide. What is the magnitude of the torque if the force is wide. What is the magnitude of the torque if the force is exerted (a) perpendicular to the door, and (b) at a 60exerted (a) perpendicular to the door, and (b) at a 6000 angle angle to the face of the door? to the face of the door?
a.a.The formula for torque is:The formula for torque is:r x F = rFsinr x F = rFsinSo for the 60So for the 60oo angle: angle: = (.84 m)(45 N)sin= (.84 m)(45 N)sin6060ooNm = 33 NmNm = 33 Nm
b.b.If the force is applied at a 90If the force is applied at a 90oo angle to the radius, the factor angle to the radius, the factor sinsinbecomes 1, and really the torque is:becomes 1, and really the torque is:rF = (.84 m)(45 N) = 37.8 Nm = 38 NmrF = (.84 m)(45 N) = 37.8 Nm = 38 Nm
Practice Problem 2
If the coefficient of static friction between tires and pavement is 0.75, calculate the minimum torque that must be applied to the 66-cm-diameter tire of a 1080kg automobile in order to "lay rubber" (make the wheel spin, slipping as the car accelerates). Assume each tire supports an equal share of the weight.
Practice Problem 2If the coefficient of static friction between tires and pavement is 0.75, calculate the minimum torque that must be applied to the 66-cm-diameter tire of a 1080kg automobile in order to "lay rubber" (make the wheel spin, slipping as the car accelerates). Assume each tire supports an equal share of the weight.
If each wheel supports an equal share of the weight, thenF = maF = (1080 kg)(9.80 N/kg) = 10584 N is divided into four equal parts, so the normal force at each tire isFN = (10584 N)/4 = 2646 N
This means that the maximum static force of friction tangential to the drive wheels must be:Ffr µsFN = (.75)(2646 N) = 1984.5 N
The formula for torque is:r x F = rFsinSince the road must be tangential to the tire, the force is applied at a 90o angle to the radius, so the factor sinbecomes 1, and really the torque is:rF = (.33 m)(1984.5 N) = 654.9 Nm = 650 Nm(They gave you the diameter - .66 m diameter, .33m radius)
Practice Problem 3The bolts on the cylinder head of an engine require tightening to a torque of 80m.N. If a wrench is 30cm long, what force perpendicular to the wrench must the mechanic exert at its end? If the six-sided bolt head is 15mm in diameter, estimate the force applied near each of the six points by a rocket wrench.
Practice Problem 3
The bolts on the cylinder head of an engine require tightening to a torque of 80m.N. If a wrench is 30cm long, what force perpendicular to the wrench must the mechanic exert at its end? If the six-sided bolt head is 15mm in diameter, estimate the force applied near each of the six points by a rocket wrench.
The formula for torque is:r x F = rFsinSince force is applied at a 90o angle to the radius, so the factor sinbecomes 1, and really the torque is:rF = 80 Nmr = .30 mF = (80 Nm)/(.30 m) = 266.67 N
This force is much greater at the locale of the bolt head itself, with a diameter of 15 mm, or radius of 7.5 mm:rF = 80 Nmr = 7.5x10-3 mF = (80 Nm)/(7.5x10-3 m) = 10666.7 N, which divided by the six sides is:F = 1777.8 N = 1800 N per side
Elaboration
• Torque Magnitude
Homework
• Chapter 8 Problems
• #s 22, 24, and 25
Closure
• Kahoot 8-4