Physics 101 Lecture 11 Torque Dr. Ali ÖVGÜN EMU Physics Department www.aovgun.com
Force vs. Torque q Forces cause accelerations q What cause angular accelerations? q A door is free to rotate about an axis through O q There are three factors that determine the effectiveness of the force in opening the door: n n n The magnitude of the force The position of the application of the force The angle at which the force is applied
Torque Definition q Torque, τ, is the tendency of a force to rotate an object about some axis q Let F be a force acting on an object, and let r be a position vector from a rotational center to the point of application of the force, with F perpendicular to r. The magnitude of the torque is given by τ = rf
Torque Units and Direction q The SI units of torque are N. m q Torque is a vector quantity q Torque magnitude is given by τ = rf sinθ = Fd q Torque will have direction n n If the turning tendency of the force is counterclockwise, the torque will be positive If the turning tendency is clockwise, the torque will be negative
Net Torque q The force F r will tend to 1 cause a counterclockwise rotation about O q The force F r will tend to 2 cause a clockwise rotation about O q Στ = τ 1 + τ 2 = F 1 d 1 F 2 d 2 q If Στ 0, starts rotating q If Στ = 0, rotation rate does not change q Rate of rotation of an object does not change, unless the object is acted on by a net torque
General Definition of Torque q The applied force is not always perpendicular to the position vector q The component of the force perpendicular to the object will cause it to rotate q When the force is parallel to the position vector, no rotation occurs q When the force is at some angle, the perpendicular component causes the rotation
General Definition of Torque q Let F be a force acting on an object, and let r be a position vector from a rotational center to the point of application of the force. The magnitude of the torque is given by τ = rf sinθ q θ = 0 or θ = 180 : torque are equal to zero q θ = 90 or θ = 270 :magnitude of torque attain to the maximum
Understand sinθ q The component of the force (F cos θ ) has no tendency to produce a rotation q The moment arm, d, is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force d = r sinθ τ = rf sinθ = Fd
Ex: The Swinging Door q Three forces are applied to a door, as shown in figure. Suppose a wedge is placed 1.5 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied won t open the door? Assume F 1 = 150 N, F 2 = 300 N, F 3 = 300 N, θ = 30 F 3 F 2 θ 2.0m F 1
Newton s Second Law for a Rotating Object q When a rigid object is subject to a net torque ( 0), it undergoes an angular acceleration Σ τ = I α q The angular acceleration is directly proportional to the net torque q The angular acceleration is inversely proportional to the moment of inertia of the object q The relationship is analogous to F = ma
Strategy to use the Newton 2 nd Law Draw or sketch system. Adopt coordinates, indicate rotation axes, list the known and unknown quantities, Draw free body diagrams of key parts. Show forces at their points of application. Find torques about a (common) axis May need to apply Second Law twice, once to each part Ø Translation: Ø Rotation: F! τ net net = F! = =! ma Make sure there are enough (N) equations; there may be constraint equations (extra conditions connecting unknowns) Simplify and solve the set of (simultaneous) equations. Find unknown quantities and check answers i! τ i! = Iα Note: can have F net = 0 but τ net 0
Ex: The Falling Object q A solid, frictionless cylindrical reel of mass M = 2.5 kg and radius R = 0.2 m is used to draw water from a well. A bucket of mass m = 1.2 kg is attached to a cord that is wrapped around the cylinder. q (a) Find the tension T in the cord and acceleration a of the object. q (b) If the object starts from rest at the top of the well and falls for 3.0 s before hitting the water, how far does it fall?
Newton 2nd Law for Rotation q Draw free body diagrams of each object q Only the cylinder is rotating, so apply Σ τ = Iα q The bucket is falling, but not rotating, so apply Σ F = ma q Remember that a = αr and solve the resulting equations a mg r
Cord wrapped around disk, hanging weight Cord does not slip or stretch à constraint Disk s rotational inertia slows accelerations Let m = 1.2 kg, M = 2.5 kg, r =0.2 m For mass m: T y T mg FBD for disk, with axis at o : N Mg F = ma= mg T y T = m( g a) Unknowns: T, a τ0 = + Tr = Iα 1 2 2 So far: 2 Equations, 3 unknowns àneed a constraint: Substitute and solve: 2mgr 2mα r α = 2 2 Mr Mr I = 1 2 Mr Tr m( g a) r α = = Unknowns: a, α I Mr 2 m α 1+ 2 = M 2mg Mr 2 a a mg = αr support force at axis O has zero torque r from no slipping assumption α = mg ( 24 rad / s 2 ) rm ( + M/2) =
Cord wrapped around disk, hanging weight Cord does not slip or stretch à constraint Disk s rotational inertia slows accelerations Let m = 1.2 kg, M = 2.5 kg, r =0.2 m For mass m: y T mg F = ma= mg T y T = m( g a) Unknowns: T, a α = mg ( 24 rad/s 2 ) rm ( + M/2) = mg a = ( 4.8 m/s 2 ) ( m+ M /2) = a r support force at axis O has zero torque T = m( g a) = 1.2(9.8 4.8) = 6 N mg 1 2 1 2 2 xf xi = vt i + at = 0 + (4.8 m/s )(3 s) = 21.6m 2 2
Torque! τ =! r! F q The torque is the cross product of a force vector with the position vector to its point of application τ = rf sinθ = r r F F = q The torque vector is perpendicular to the plane formed by the position vector and the force vector (e.g., imagine drawing them tail-to-tail) q Right Hand Rule: curl fingers from r to F, thumb points along torque. q q Superposition:!!! τ net = τ i = ri F! i all i all i (vector sum) Can have multiple forces applied at multiple points. Direction of τ net is angular acceleration axis
Ex3: Calculate torque given a force and its location Solution:!! F = ( 2ˆ i + 3 ˆ) j N r = (4ˆ i + 5 ˆj ) m!!! τ = r F = (4ˆ i + 5 ˆ) j (2ˆ i + 3 ˆ) j = 4ˆ i 2ˆ i + 4ˆ i 3 ˆj + 5 ˆj 2ˆ i + 5 ˆj 3 ˆj = 0 + 4ˆ i 3 ˆj + 5 ˆj 2ˆ i + 0 = 12kˆ 10kˆ = r r A B= 2kˆ (Nm) iˆ ˆj kˆ 4 5 0 2 3 0 i j k
P1: P2:
P3:
P4: P5: