Angular Mechanics - Torque and moment of inertia Contents: Review Linear and angular Qtys Tangential Relationships Angular Kinematics Rotational KE Example | WhiteboardExampleWhiteboard Rolling Problems Example | WhiteboardExampleWhiteboard

Angular Mechanics - Angular Quantities Linear: (m) s (m/s) u (m/s) v (m/s/s) a (s) t (N) F ( kg) m Angular:  - Angle (Radians)  o - Initial angular velocity (Rad/s)  - Final angular velocity (Rad/s)  - Angular acceleration (Rad/s/s) t - Uh, time (s)  - Torque I - Moment of inertia TOC

Angular Mechanics - Angular kinematics Linear:  s/  t = v  v/  t = a u + at = v ut + 1 / 2 at 2 = s u 2 + 2as = v 2 (u + v)t/2 = s ma = F 1 / 2 mv 2 = E kin Fs = W Angular:  =  /  t  =  /  t*  =  o +  t  =  o t + 1 / 2  t 2  2 =  o   = (  o +  )t/2*  = I  *Not in data packet TOC E k rot = 1 / 2 I  2 W =  *

Angular Mechanics - Useful Substitutions  = I   = rF so F =  /r = I  /r s =  r, so  = s/r v =  r, so  = v/r a =  r, so  = a/r TOC

Angular Mechanics - Rotational Ke TOC Two types of kinetic energy: Of course a rolling object has both IP Demo Rolling.ip Translational: E kin = 1 / 2 mv 2 Rotational: E k rot = 1 / 2 I  2

Example: What Energy does it take to speed up a 23.7 kg 45 cm radius cylinder from rest to 1200 RPM? TOC

Whiteboards: Rotational KE 11 | 2 | 323 TOC

What is the rotational kinetic energy of an object with an angular velocity of 12 rad/s, and a moment of inertia of 56 kgm 2 ? E k rot = 1 / 2 I  2 E k rot = 1 / 2 ( 56 kgm 2 )( 12 rad/s ) 2 E k rot = 4032 J = 4.0 x 10 3 J 4.0 x 10 3 J W

What must be the angular velocity of a flywheel that is a 22.4 kg, 54 cm radius cylinder to store 10,000. J of energy? (hint) E k rot = 1 / 2 I  2, I = 1 / 2 mr 2 E k rot = 1 / 2 ( 1 / 2 mr 2 )  2 = 1 / 4 mr 2  2  2 = 4(E k rot )/ mr 2  =(4(E k rot )/mr 2 ) 1/2 =(4(10000J)/(22.4kg)(.54m) 2 ) 1/2  = rad/s = 78 rad/s 78 rad/s W

What is the total kinetic energy of a 2.5 cm diameter 405 g sphere rolling at 3.5 m/s? (hint) I= 2 / 5 mr 2,  = v/r, E k rot = 1 / 2 I  2, E kin = 1 / 2 mv 2 E k total = 1 / 2 mv / 2 I  2 E k total = 1 / 2 mv / 2 ( 2 / 5 mr 2 )(v/r) 2 E k total = 1 / 2 mv / 10 mv 2 = 7 / 10 mv 2 E k total = 7 / 10 mv 2 = 7 / 10 (.405 kg)(3.5m/s) 2 E k total = J = 3.5 J 3.5 J W

Angular Mechanics – Rolling with energy TOC m r - cylinder I = 1 / 2 mr 2  = v/r h mgh = 1 / 2 mv / 2 I  2 mgh = 1 / 2 mv / 2 ( 1 / 2 mr 2 )(v/r) 2 mgh = 1 / 2 mv / 4 mv 2 = 3 / 4 mv 2 4 / 3 gh = v 2 v = ( 4 / 3 gh) 1/2

Whiteboards: Rolling with Energy 11 | 2 | 323 TOC

A 4.5 kg ball with a radius of.12 m rolls down a 2.78 m long ramp that loses.345 m of elevation. Set up the energy equation without plugging any of the knowns into it. Make substitutions for I and , but don’t simplify. I = 2 / 5 mr 2,  = v/r mgh = 1 / 2 mv / 2 I  2 mgh = 1 / 2 mv / 2 ( 2 / 5 mr 2 )(v/r) 2 W

Solve this equation for v: mgh = 1 / 2 mv / 2 ( 2 / 5 mr 2 )(v/r) 2 mgh = 1 / 2 mv / 10 mr 2 v 2 /r 2 mgh = 1 / 2 mv / 10 mv 2 = 7 / 10 mv 2 10 / 7 gh = v 2 v = ( 10 / 7 gh) 1/2 W

A 4.5 kg ball with a radius of.12 m rolls down a 2.78 m long ramp that loses.345 m of elevation. What is the ball’s velocity at the bottom? ( v = ( 10 / 7 gh) 1/2 ) v = ( 10 / 7 (9.8m/s/s)(.345 m)) 1/2 = m/s v = 2.20 m/s 2.20 m/s W

A 4.5 kg ball with a radius of.12 m rolls down a 2.78 m long ramp that loses.345 m of elevation. What was the rotational velocity of the ball at the bottom? ( v = m/s ) 18 rad/s W  = v/r = ( m/s)/(.12 m) = 18.3 s -1  = 18 s -1

A 4.5 kg ball with a radius of.12 m rolls down a 2.78 m long ramp that loses.345 m of elevation. What was the linear acceleration of the ball down the ramp? ( v = m/s ).869 m/s/s W v 2 = u 2 + 2as v 2 /(2s) = a ( m/s) 2 /(2(2.78 m)) =.869 m/s/s

In General: I tend to solve all rotational dynamics problems using energy. 1.Set up the energy equation 2.(Make up a height) 3.Substitute linear for angular:  = v/r I = ?mr 2 4.Solve for v 5.Go back and solve for accelerations TOC

Angular Mechanics – Pulleys and such TOC r m1m1 m2m2 h Find velocity of impact, and acceleration of system r = 12.5 cm m 1 = 15.7 kg m 2 =.543 kg h =.195 m

Angular Mechanics – Pulleys and such TOC r m1m1 m2m2 h = (made up) m3m3 Find acceleration of system r = 46 cm m 1 = 55 kg m 2 = 15 kg m 3 = 12 kg h = 1.0 m

Angular Mechanics – yo yo ma TOC Find acceleration of system (assume it is a cylinder) r 1 = cm r 2 =.210 cm m = 273 g r1r1 r2r2 h = 1.0 m