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 lever arm, L Axle Force, F

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 Torque To 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

Torque Period 1 stopped here Hewitt video

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 CG If the CG is above the edge, the object will not fall

when does it fall over? STABLE NOT STABLE CG CG If the vertical line extending down from the CG is inside the edge the object will return to its upright position  the torque due to gravity brings it back. STABLE NOT STABLE

Stable and Unstable stable unstable torque due to gravity pulls object back torque due to gravity pulls object down

Stable structures Structures are wider at their base to lower their center of gravity

Playing with your blocks CG If the center of gravity is supported, the blocks do not fall over

Object with low CG Stay low to the ground! 300 lb fullback who is 4 ft, 10 inches tall and runs a 4-40 Stay low to the ground!

As more and more stuff is loaded into a High Profile Vehicles wind As more and more stuff is loaded into a semi, its center of gravity moves upward. It could be susceptible to tipping over.

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 Lever arm length (m) t = r x F Torque (N.m) Force (N)

Sign Convention for Torque By convention, counterclockwise torques are positive and clockwise torques are negative. ccw Positive torque: Counter-clockwise, out of page cw 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 these two forces, but will not rotate.

Balancing torques 20 N 10 N 1 m 0.5 m Left torque = 10 N x 1 m = 10 n m Right torque = 20 N x 0.5 m = 10 N m

Torque Lab

Torque example For the same force, you get more torque What is the torque on a bolt applied with a wrench that has a lever arm of 30 cm with a force of 10 N? F Torque = F  L = 30 N  0.30 m = 9 N m L For the same force, you get more torque with a bigger wrench  the job is easier!

Extend line of action, draw, calculate r. Example 1: An 80-N force acts at the end of a 12-cm wrench as shown. Find the torque. Extend line of action, draw, calculate r. r = 12 cm sin 600 = 10.4 cm t = (80 N)(0.104 m) = 8.31 N m

Example 2: Find resultant torque about axis A for the arrangement shown below: Find t due to each force. Consider 20-N force first: 300 6 m 2 m 4 m 20 N 30 N 40 N A negative r r = (4 m) sin 300 = 2.00 m The torque about A is clockwise and negative. t = Fr = (20 N)(2 m) = 40 N m, cw t20 = -40 N m

Example 2 (Cont.): Next we find torque due to 30-N force about same axis A. Find t due to each force. Consider 30-N force next. 300 6 m 2 m 4 m 20 N 30 N 40 N A negative r = (8 m) sin 300 = 4.00 m The torque about A is clockwise and negative. t = Fr = (30 N)(4 m) = 120 N m, cw t30 = -120 N m

Example 2 (Cont.): Finally, we consider the torque due to the 40-N force. Find t due to each force. Consider 40-N force next: 300 6 m 2 m 4 m 20 N 30 N 40 N A r positive r = (2 m) sin 900 = 2.00 m The torque about A is CCW and positive. t = Fr = (40 N)(2 m) = 80 N m, ccw t40 = +80 N m

Example 2 (Conclusion): Find resultant torque about axis A for the arrangement shown below: 300 6 m 2 m 4 m 20 N 30 N 40 N A Resultant torque is the sum of individual torques. tR = t20 + t20 + t20 = -40 N m -120 N m + 80 N m tR = - 80 N m Clockwise

Practice Problem #1 A 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 exerted (a) perpendicular to the door, and (b) at a 600 angle to the face of the door? 

Practice Problem #1 Solution A 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 exerted (a) perpendicular to the door, and (b) at a 600 angle to the face of the door?  The formula for torque is: r x F = rFsin So for the 60o angle:  = (.84 m)(45 N)sin60oNm = 33 Nm If the force is applied at a 90o angle to the radius, the factor sinbecomes 1, and really the torque is: rF = (.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 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.  If each wheel supports an equal share of the weight, then F = ma F = (1080 kg)(9.80 N/kg) = 10584 N is divided into four equal parts, so the normal force at each tire is FN = (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 = rFsin Since the road must be tangential to the tire,  the force is applied at a 90o angle to the radius, so the factor sinbecomes 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 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.

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 = rFsin Since force is applied at a 90o angle to the radius, so the factor sinbecomes 1, and really the torque is: rF  = 80 Nm r = .30 m F = (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 Nm r = 7.5x10-3 m F = (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