Angular Mechanics Contents Review Linear and angular Qtys

Angular Mechanics Contents: • Review • Linear and angular Qtys • Tangential Relationships • Angular Kinematics • Rotational KE • Example | Whiteboard • Rolling Problems • Example | Whiteboard

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

Angular Mechanics - Angular kinematics Linear: u + at = v ut + 1/2 at 2 = s u 2 + 2 as = v 2 (u + v)t/2 = s ma = F 1/ mv 2 = E 2 kin Fs = W Angular: = o + t = ot + 1/2 t 2 2 = o 2 + 2 = ( o + )t/2* = I Ek rot = 1/2 I 2 W = * *Not in data packet

Angular Mechanics - Useful Substitutions = I = r. F so F = /r = I /r s = r, so = s/r v = r, so = v/r a = r, so = a/r

Angular Mechanics - Rotational Ke Two types of kinetic energy: Translational: Ekin = 1/2 mv 2 Rotational: Ek rot = 1/2 I 2 Of course a rolling object has both Demo – ring and cylinder

Example: A 23. 7 kg 45 cm radius cylinder is rolling at 13. 5 m/s at the bottom of a hill. What is its translational kinetic energy? What is its rotational kinetic energy? What is the total kinetic energy? What was the height of the hill?

Angular Mechanics – Rolling with energy I = 1/2 mr 2 h = v/r mgh = 1/2 mv 2 + 1/2 I 2 An 11. 0 g, 0. 0130 m radius cylinder rolls down an incline that is 2. 90 m long, and loses 0. 340 m of elevation. What is its acceleration down the plane, and its velocity at the bottom of the plane? mgh = 1/2 mv 2 + 1/2 I 2 mgh = 1/2 mv 2 + 1/2(1/2 mr 2)(v/r)2 mgh = 1/2 mv 2 + 1/4 mv 2 = 3/4 mv 2 4/ gh = v 2 3 v = (4/3 gh)1/2 0. 767 m/s/s 2. 11 m/s

Whiteboards: Rolling 1

A marble (a solid sphere) has a mass of 23. 5 g, a radius of 1. 20 cm, and rolls 2. 75 m down an incline that loses 0. 650 m of elevation. 1. Solve for v in terms of g and h 2. Plug in and get v 3. suvat for a mgsin = ma + I /r, I = 2/5 mr 2, = a/r mgsin = ma + (2/5 mr 2)(a/r)/r mgsin = ma + 2/5 ma = 7/5 ma gsin = 7/5 a a = 5/7 gsin 3. 02 m/s 1. 66 m/s/s

Whiteboards: Rotational KE 1 -6

What is the rotational kinetic energy of an object with an angular velocity of 12. 0 rad/s, and a moment of inertia of 56. 0 kg m 2? Ek rot = 1/2 I 2 Ek rot = 1/2(56 kgm 2)(12 rad/s)2 Ek rot = 4032 J = 4. 0 x 103 J

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? Ek rot = 1/2 I 2, I = 1/2 mr 2 Ek rot = 1/2(1/2 mr 2) 2 = 1/4 mr 2 2 2 = 4(Ek rot)/mr 2 =(4(Ek rot)/mr 2)1/2=(4(10000 J)/(22. 4 kg)(. 54 m)2)1/2 = 78. 25 rad/s = 78 rad/s

What is the total kinetic energy of a 2. 5 cm diameter 405 g sphere rolling at 3. 5 m/s? I=2/5 mr 2, = v/r, Ek rot=1/2 I 2 , Ekin=1/2 mv 2 Ek total= 1/2 mv 2 +1/2 I 2 Ek total= 1/2 mv 2 +1/2(2/5 mr 2)(v/r)2 Ek total= 1/2 mv 2 +2/10 mv 2 = 7/10 mv 2 Ek total= 7/10 mv 2 = 7/10(. 405 kg)(3. 5 m/s)2 Ek total= 3. 473 J = 3. 5 J

What work do you do if you exert 14. 0 m. N of torque through 3. 10 rotations? solution 273 J

If you exert 14. 0 m. N of torque through 3. 10 rotations on a potter’s wheel that is a 26. 0 kg, 68. 0 cm diameter uniform cylinder, what will be the final angular velocity? solution 19. 1 rad/s

A tangential friction force of 2. 30 N slows down a 26. 0 kg 68. 0 cm diameter potter’s wheel. (That is a uniform cylinder) If it was originally going at 71. 0 RPM, in how many rotations will it stop? solution 8. 45 rot

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

Angular Mechanics – Pulleys and such r Find velocity of impact, and acceleration of system r = 12. 5 cm m 1 = 15. 7 kg m 2 = 0. 543 kg h = 0. 195 m m 1 m 2 h Set up dynamics, do RKE 0. 498 m/s 0. 635 m/s/s

Angular Mechanics – Pulleys and such r m 1 Find acceleration of system r = 46 cm m 1 = 55 kg m 2 = 15 kg m 3 = 12 kg h = 1. 0 m m 2 m 3 h = (made up)

Whiteboards: Rolling with Energy 1|2|3

A 4. 5 kg ball with a radius of 0. 12 m rolls down a 2. 78 m long ramp that loses 0. 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 + 1/2 I 2 mgh = 1/2 mv 2 + 1/2(2/5 mr 2)(v/r)2

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

A 4. 5 kg ball with a radius of 0. 12 m rolls down a 2. 78 m long ramp that loses 0. 345 m of elevation. What is the ball’s velocity at the bottom? (v = (10/7 gh)1/2) v = (10/7(9. 8 m/s/s)(. 345 m))1/2 = 2. 1977 m/s v = 2. 20 m/s

A 4. 5 kg ball with a radius of 0. 12 m rolls down a 2. 78 m long ramp that loses 0. 345 m of elevation. What was the rotational velocity of the ball at the bottom? (v = 2. 1977 m/s) = v/r = (2. 1977 m/s)/(. 12 m) = 18. 3 s-1 = 18 s-1 18 rad/s

A 4. 5 kg ball with a radius of 0. 12 m rolls down a 2. 78 m long ramp that loses 0. 345 m of elevation. What was the linear acceleration of the ball down the ramp? (v = 2. 1977 m/s) v 2 = u 2 + 2 as v 2/(2 s) = a (2. 1977 m/s)2/(2(2. 78 m)) = 0. 869 m/s/s

Angular Mechanics – Pulleys and such r m 1 Find acceleration of system r = 46 cm m 1 = 55 kg m 2 = 15 kg m 3 = 12 kg h = 1. 0 m m 2 m 3 h = (made up)

Angular Mechanics – yo yo ma Find acceleration of system (assume it is a cylinder) r 1 r 2 h = 1. 0 m r 1 = 6. 720 cm r 2 = 0. 210 cm m = 273 g