F1 F2 If F1 = F2… …no change in motion (by Newton’s 1st Law)

However… F1 F2 If F1 = F2 under this condition… …the block will twist So the location of the forces is important

Torque… …is the rotational form of the word Force. Think of it as a twisting Force Torque equals the amount of force multiplied by its perpendicular distance away from the axis of rotation. Τ = r x F

F1 Fg Suppose F1 = 100 N. The Fridge is 3m tall and 1 m wide. It is just about to tip. What is the weight of the fridge?

A light cable is wrapped around a large solid cylinder several times A light cable is wrapped around a large solid cylinder several times. When the cable is pulled out with a constant force, the cylinder spins without slipping. It starts from rest. ω2 = ? F M = 50 kg R = 0.7 m x = 6 m F = 100 N α = ? a = ?

Campbell - Physics Notes solid cylinder Campbell - Physics Notes Find the acceleration of the block and the tension in the string 8/31/2018 M r M = 6 kg m = 2 kg m

solid cylinder Find the acceleration of the yo-yo and the tension in the string m = 2 kg m r

Campbell - Physics Notes solid cylinder Find the velocity of the yo-yo after it has fallen 0.5 m. (starting from rest) 8/31/2018 m = 2 kg m r

solid cylinder Find the acceleration of the cylinder and the friction on the ramp m = 3 kg  = 30 m r 

Pulley is a Thin-Shell cylinder w/ mass and radius Table is frictionless. Given: m1 m2 m3 r =3 kg =2 kg =0.5 kg = 0.25 m a1 = a2 = 3.6 m/s2 α = 14.3 rad/s2 a1 T1=? m3 m1 α r T2=? m2 a2 Calculate: a1= a2= α = T1= T2= Free Body Diagrams T1= 10.7 N T2= 12.5 N

P = P’ (mv)a + (mv)b = (mv)’a + (mv)’b Conservation of Angular Momentum P = P’ (mv)a + (mv)b = (mv)’a + (mv)’b

Campbell - Physics Notes solid cylinder A child stands on the edge of a small merry-go-round that is rotating at a rate of 2 rad/s. If the child walks in toward the center of the “disk” and stops 0.5 m for the center, what will happen to ω of the merry-go-round? 8/31/2018 M =100kg m = 30kg R = 2.5 m r = 0.5 m M R r

Angular Momentum P = mv P = mv = F.t …so P = F.t And P = F t

Front view Angled view Side view Identify: Direction of ω Direction of L Direction of Torque Same Perpendicular to L …Use Right Hand Rule RHR’s

Campbell - Physics Notes Angled view 8/31/2018 Side view

Angled view Side view

Free Body Diagrams m1 m2 T1 a T1 T2 T1 = m1a  = (T2 r) – (T1 r) = Iα r(T2– T1)= (m3r2)(a/r) substitute T2 T2– T1 = m3a m2 a m2g – m2a - m1a = m3a m2g m2g – T2 = m2a Solve for “a”… substitute m2g – m2a = T2