Unit 5 Notes Torque

τ = r x F Or for those who may not know cross-products, τ = rF sin (Ө) τ (tau) stands for torque. It is equal to the radius from the fulcrum times the force times the sine of the angle between the radius and the torque. Units for torque are Nm Clockwise is negative and counterclockwise is positive (just like angles)

Torque Example You apply a force of 20N to a 10cm long wrench at a30 degree angle. Solve for the torque. τ = rF sin (Ө) τ = (0.1m)(20N) sin (60) τ = 1.73 N*m

Torque and Force ∑F = F net For a non-accelerating system, ∑F = F net = 0 At the same time, ∑ τ = τ net For a non-rotating system, ∑ τ = τ net = 0

Example A meter stick has a fulcrum placed at the 50cm mark. An object that weighs 20N is hung from it 20cm from the center. What force would you have to apply on the other side of the meter stick 30cm from the fulcrum to keep the meter stick from rotating? ∑ τ = τ person - τ weight = 0 F(0.3m)-(0.2m)(20N)=0 F=13.3N

Rotational Dynamics The equation describing most situations in this section is: ∑ τ = I α The sum of torques is equal to the moment of inertia times angular acceleration.

Moment of Inertia Moment of inertia of an object tells you how difficult it is to make the object rotate. In general, I = mr 2 That is true for each individual mass, but an entire system of masses can be more complicated.

Some commonly used moments of inertia are: I = mr 2 for a point mass I = ½ mr 2 for a disc I = 1/3 mr 2 for a rod rotating around its end

You exert a 60N force at the end of a 3m long rod. The rod has a mass of 20kg. Assuming the rod is rotating about its end (the circle in the picture), a)Calculate the moment of inertia of the rod. b)Calculate the net torque on the rod. c)What is the angular acceleration of the rod?

a) This is a rod rotating about its end, so I = 1/3 mr 2. = 1/3(20kg)(3m) 2 = 60 kg*m 2 b) τ = rF sin (Ө) = (3m)(60N)sin(90) =180Nm c) ∑ τ = I α 180Nm = 60(kgm 2 ) α α = 30 rad/s 2

Angular Momentum Angular momentum is another quantity that is conserved in physics. Always: L = I ω For each individual point, L = mrv. Depending on the shape of the entire rotating object, there may be a coefficient.

Example A ball is tied to a pole with a string that is 2m long and is orbiting the pole. The initial speed of the ball is 6m/s. What is the distance of the ball from the pole when its speed is 20m/s? L 1 = L 2 I 1 ω 1 = I 2 ω 2 or mr 1 v 1 = mr 2 v 2 mr 1 2 (v 1 /r 1 ) = mr 2 2 (v 2 /r 2 ) r 1 v 1 = r 2 v 2 (2m)(6m/s) = r 2 (20m/s) r 2 = 0.6m/s

Linear Kinematics: x = ½at 2 + v 0 t + x 0 v = at + v 0 v 2 = 2a∆x + v 0 2 ∑ F = ma p = mv E k = ½ mv 2 Rotational Kinematics: θ = ½αt 2 + ω 0 t + θ 0 ω = αt + ω 0 ω 2 = 2α∆ θ + ω 0 2 ∑ τ = I α L = I ω E k = ½ I ω 2