Chapter 8 Rotational Motion
The Action of Forces and Torques on Rigid Objects In pure translational motion, all points on an object travel on parallel paths. The most general motion is a combination of translation and rotation.
The Action of Forces and Torques on Rigid Objects According to Newton s second law, a net force causes an object to have an acceleration. What causes an object to have an angular acceleration? TORQUE Torque tells you how effective a given force is at rotating something about some axis.
The Action of Forces and Torques on Rigid Objects The amount of torque depends on where and in what direction the force is applied, as well as the location of the axis of rotation.
The Action of Forces and Torques on Rigid Objects DEFINITION OF TORQUE Magnitude of Torque = (Magnitude of the force) x (Lever arm) τ = F The Lever arm, l, for some force acting about some rotation axis is defined as the distance between the rotation axis and a perpendicular to the force. Direction: The torque is positive when the force tends to produce a counterclockwise rotation about the axis. SI Unit of Torque: newton x meter (N m)
The Action of Forces and Torques on Rigid Objects Example: The Achilles Tendon The tendon exerts a force of magnitude 720 N with a geometry as shown in the figure. Determine the torque (magnitude and direction) of this force about the ankle joint.
The Action of Forces and Torques on Rigid Objects τ = F cos55 = 2 3.6 10 m First, calculate the magnitude of τ 720 N τ = ( 720 N) ( 3.6 10 2 m)cos55 =15 N m Since the force rotates the foot about the ankle joint in a clockwise direction --> τ negative τ = - 15 N m, if the direction is included
Consider a powered model airplane on a massless string. F = ma T T τ = F T r a T = rα m τ = ( mr 2 )α Moment of Inertia, I
Torques acting on a general rigid body è torques due to internal forces cancel out in pairs τ = ( mr 2 ) α Net external torque Moment of inertia τ 1 = ( 2 m 1 r 1 )α τ 2 = ( 2 m 2 r 2 )α τ N = ( 2 m N r N )α
ROTATIONAL ANALOG OF NEWTON S SECOND LAW FOR A RIGID BODY ROTATING ABOUT A FIXED AXIS Net external torque & Moment of = $ % inertia #! " & Angular $ % acceleration #! " τ = I α Requirement: Angular acceleration must be expressed in radians/s 2. I = ( mr 2 ) SI unit: kg m 2
Example: The Moment of Inertial Depends on Where the Axis Is. Two particles have identical mass and are fixed at the ends of a thin rigid rod. The length of the rod is L. Find the moment of inertia when this object rotates relative to an axis that is perpendicular to the rod at (a) one end and (b) the center.
(a) I = ( mr 2 ) = m 1 r 2 1 + m 2 r 2 2 = m( 0) 2 + m( L) 2 = ml 2 m m = 1 = 2 m r r = 1 = 0 2 L (b) I = ( mr 2 ) = m 1 r 2 1 + m 2 r 2 2 = m( L 2) 2 + m( L 2) 2 = 1 2 ml2 r = 1 = L 2 r2 L 2
Flat plate with axis along one edge I = 1 3 Ma2 parallel to axis shown:
Example: Hoisting a Crate A motor is used to lift a crate with the dual pulley system shown below. The combined moment of inertia of the dual pulley is 46.0 kg m 2. The crate has a mass of 451 kg. A tension of 2150 N is maintained in the cable attached to the motor. Find the angular acceleration of the dual pulley and the tension in the cable connected to the crate.
equal " F y = T2 mg = ma y 1 1 2 2 τ = T T = Iα a y = a T = l 2 α T = mg + 2 ma y T 1 1 ( mg + ma y ) 2 = Iα
T 1 1 ( mg + ma y ) 2 = Iα a y = 2 α T 1 1 ( mg + m 2 α) 2 = Iα α = T 1 l 1 mgl 2 I + ml 2 2 ( )( 0.600 m) ( 451 kg) ( 9.80m s 2 )( 0.200 m) = 6.34rad s 2 46.0 kg m 2 + ( 451 kg) ( 0.200 m) 2 = 2150 N
Tension in the cable connected to the crate: T 2 = mg + ma y = mg + ml 2 α = (451 kg)(9.80 m/s 2 ) + (451 kg)(0.200 m)(6.34 rad/s 2 ) = 4990 N
Maximum mass that could be lifted with this system: α = T 1l 1 m max gl 2 I + m max l 2 2 = 0, i.e. m < m max to lift the crate T 1 l 1 m max gl 2 = 0 m max = T 1 l 1 gl 2 = ( 2150 N ) 0.600 m 9.80m s 2 ( ) ( ) ( ) 0.200 m T 2 = m max g + m max l 2 α = m max g + 0 = ( 658 kg) ( 9.80m s 2 ) = 6450 N = 658 kg