Lecture 14 Rotational dynamics Torque Give me a lever long enough and a fulcrum on which to place it, and I shall move the world. Archimedes, 87 1 BC
EXAM Tuesday March 6, 018 8:15 PM 9:45 PM
Today s Topics: Torque Static Equilibrium Center of Gravity (Center of Mass) Torque and Angular Acceleration Moment of Inertia
Rotational Kinetic Energy and the Moment of Inertia Imagine a mass at the end of a light rod, KE = 1 mv = 1 m ( rω ) = 1 ( mr )ω We can also write the kinetic energy as KE = 1 Iω Where I, the moment of inertia, is given by I = mr
Torque According to Newton s second law, a net force causes an object to have a linear acceleration. What causes an object to have an angular acceleration? A net torque about an axis 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.
Defining Torque d θ magnitude of torque = (magnitude of the force) x (lever arm) τ = Fl 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)
ACT: Two Forces Two forces produce the same torque. Does it follow that they have the same magnitude? a) yes b) no Because torque is the product of force times distance, two different forces that act at different distances could still give the same torque. Follow-up: If two forces are identical, does that mean their torques are identical as well?
ACT: Using a Wrench You are using a wrench to loosen a rusty nut. Which arrangement will be the most effective in loosening the nut? a b Because the forces are all the same, the only difference is the lever arm. The arrangement with the largest lever arm (case b) will provide the largest torque. c d e) all are equally effective Follow-up: What is the difference between arrangement a and d?
ACT: Garden hose When you pull on a garden hose on a reel, there is a tension in the hose that applies a torque to the reel. Assuming the tension remains constant when you pull, how does the applied torque vary as the reel becomes empty? a) torque increases b) torque decreases c) torque remains constant As the reel empties, the lever arm decreases because the radius of the reel (with hose on it) is decreasing. Thus, as your pull continues, the applied torque decreases.
Example A string is tied to a doorknob 0.79 m from the hinge as shown in the figure. At the instant shown, the force applied to the string is 5.0 N. What is the torque about the hinge? F (a) 3.3 Nm (b). Nm (c) 1.1 Nm (d) 0.84 Nm (e) 0.40 Nm 33 0.79 m l 57
Static Equilibrium In the absence of rotation about some axis: å x F = 0 å = 0 Now we also have to consider rotation about an axis: F y å F y = 0 å Fx = 0 å t = 0 a x = a y = 0 a = 0 A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration. In equilibrium, the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero.
Example A woman whose weight is 530 N is poised at the right end of a diving board with length 3.90 m. The board has negligible weight and is supported by a fulcrum 1.40 m away from the left end. Find the forces that the bolt and the fulcrum exert on the board.
å t = F!! -W W = 0 F F = W! W! ( 530 N)( 3.90 m) = = 1.40 m 1480 N å F y = -F 1 + F -W = 0 - F1 + 1480N-530 N = 0 F 1 = 950 N
What if the weight of the board was not negligible? We treated the diver, fulcrum and bolt as point-like objects, but the board is an extended object! DEFINITION OF CENTER OF GRAVITY The center of gravity of a rigid body is the point at which its weight can be considered to act when the torque due to the weight is being calculated. Center of gravity = Center of mass When an object has a symmetrical shape and its weight is distributed uniformly, the center of gravity lies at its geometrical center.
So, if the board has a nonnegligible weight 1.95 m W B
ACT: Balancing Rod A 1-kg ball hangs at the end of a rod 1-m long. If the system balances at a point on the rod 0.5 m from the end holding the mass, what is the mass of the rod? a) ¼ kg b) ½ kg c) 1 kg d) kg e) 4 kg The total torque about the pivot must be zero!! The CM of the rod is at its center, 0.5 m to the right of the pivot. Because this must balance the ball, which is the same distance to the left of the pivot, the masses must be the same!! same distance X 1 kg CM of rod
Finding the Center of Mass/Gravity DEMO: Iowa CM In equilibrium, the CM will be along a vertical line through the suspension point. Use two suspension points (P 1 and P ) for the object whose CM you want to determine, mark the vertical lines on the object, and find the point where the two lines cross.
What happens if å t ¹ 0? å t = Ia where I is the moment of inertia about the axis of rotation F = ma T T t = F r T a T = ra t ( mr )a = DEMO: Aluminum rods
Moment of Inertia The moment of inertia depends on where the axis is! I = å( mr ) Two particles each have 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 = å mr = m1r 1 + mr = m 0 + m L = ml (b) ( ) ( ) mr = m r + m r = m L + m( L ) I = å ml 1 1 =
Moments of inertia of various continuous objects: