2. Angular Mechanics - Rolling using dynamics Contents: Review Linear and angular Qtys Tangential Relationships Useful Substitutions Force causing  Rolling | WhiteboardWhiteboard Strings and pulleys Example | WhiteboardExampleWhiteboard

3. 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) TOC  - Torque I - Moment of inertia

4. Angular Mechanics - Tangential Relationships Linear: (m) s (m/s) v (m/s/s) a Tangential: (at the edge of the wheel) =  r - Displacement =  r - Velocity =  r - Acceleration* TOC *Not in data packet

5. 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

6. F r Rolling objects accelerate linearly and angularly: Force causing   = I   = rF so F =  /r = I  /r

7. Rolling: TOC  m r - cylinder mgsin  F = ma +  /r mgsin  = ma + I  /r (  = I  ) mgsin  = ma + ( 1 / 2 mr 2 )(a/r)/r (  =a/r) mgsin  = ma + 1 / 2 ma = 3 / 2 ma gsin  = 3 / 2 a a = 2 / 3 gsin  I = 1 / 2 mr 2

8. Whiteboards: Rolling 11 | 2 | 323 TOC

9. A marble (a solid sphere) has a mass of 23.5 g, a radius of 1.2 cm, and rolls 2.75 m down a 21 o incline. Solve for a in terms of g and  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  5 / 7 gsin  W

10. A marble (a solid sphere) has a mass of 23.5 g, a radius of 1.2 cm, and rolls 2.75 m down a 21 o incline. Plug in and get the actual acceleration. ( a = 5 / 7 gsin  ) a = 5 / 7 gsin  = 5 / 7 (9.81 m/s 2 )sin(21 o ) a = 2.5111 m/s 2 = 2.5 m/s 2 2.5 m/s/s W

11. A marble (a solid sphere) has a mass of 23.5 g, a radius of 1.2 cm, and rolls 2.75 m down a 21 o incline. What is its velocity at the bottom of the plane if it started at rest? ( a = 2.5086 m/s 2 ) v 2 = u 2 + 2as v 2 = 0 2 + 2(2.5111 m/s 2 )(2.75 m) v = 3.716 m/s = 3.7 m/s 3.7 m/s W

12. Angular Mechanics – Pulleys and such TOC r m1m1 m2m2 For the cylinder:  = I  rT = ( 1 / 2 m 1 r 2 )(a/r) (Where T is the tension in the string) For the mass: F = ma m 2 g - T = m 2 a

13. Pulleys and such: TOC r m1m1 m2m2 So now we have two equations: rT = ( 1 / 2 m 1 r 2 )(a/r) or T = 1 / 2 m 1 a and m 2 g - T = m 2 a

14. Angular Mechanics – Pulleys and such TOC r m1m1 m2m2 T = 1 / 2 m 1 a m 2 g - T = m 2 a Substituting: m 2 g - 1 / 2 m 1 a = m 2 a Solving for a: m 2 g = 1 / 2 m 1 a + m 2 a m 2 g = ( 1 / 2 m 1 + m 2 )a m 2 g/( 1 / 2 m 1 + m 2 ) = a

15. Whiteboards: Pulleys 11 | 2 | 3 | 4234 TOC

16. A string is wrapped around a 12.0 cm radius 4.52 kg cylinder. A mass of 0.162 kg is hanging from the end of the string. Set up the dynamics equation for the hanging mass. (m 2 ) m 2 g - T = m 2 a figure it out for yourself W r m1m1 m2m2

17. A string is wrapped around a 12.0 cm radius 4.52 kg thin ring. A mass of 0.162 kg is hanging from the end of the string. Set up the dynamics equation for the thin ring (m 1 )  = I , I = m 1 r 2,  = a/r,  = rT rT = (m 1 r 2 )(a/r) T = m 1 a figure it out for yourself W r m1m1 m2m2

18. A string is wrapped around a 12.0 cm radius 4.52 kg thin ring. A mass of 0.162 kg is hanging from the end of the string. Solve these equations for a: m 2 g - T = m2am2a T = m1am1a - m 1 a = m2am2a m 2 g = m 1 a + m2am2a m 2 g = a(m 1 + m2)m2) a = m 2 g/(m 1 + m2)m2) figure it out for yourself W r m1m1 m2m2

19. A string is wrapped around a 12.0 cm radius 4.52 kg thin ring. A mass of 0.162 kg is hanging from the end of the string. Plug the values in to get the acceleration: a = m 2 g/(m 1 + m 2 ) a = m 2 g/(m 1 + m 2 ) a = (0.162 kg)(9.80 N/kg)/(4.52 kg + 0.162 kg) a = 0.339 m/s/s 0.339 m/s/s W r m1m1 m2m2