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

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 - 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* *Not in data packet TOC

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

Angular Mechanics - Force causing   = rF so F = /r = I/r TOC

Angular Mechanics - Rolling mgsin I = 1/2mr2 m r - cylinder  F = ma + /r mgsin = ma + I/r ( = I) mgsin = ma + (1/2mr2)(a/r)/r ( =a/r) mgsin = ma + 1/2ma = 3/2ma gsin = 3/2a a = 2/3gsin TOC

Whiteboards: Rolling 1 | 2 | 3 TOC

mgsin = ma + I/r, I = 2/5mr2,  = a/r mgsin = ma + (2/5mr2)(a/r)/r 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 21o incline. Solve for a in terms of g and  mgsin = ma + I/r, I = 2/5mr2,  = a/r mgsin = ma + (2/5mr2)(a/r)/r mgsin = ma + 2/5ma = 7/5ma gsin = 7/5a a = 5/7gsin 5/7gsin W

a = 5/7gsin = 5/7 (9.8m/s2)sin(21o) a = 2.5086 m/s2 = 2.5 m/s2 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 21o incline. Plug in and get the actual acceleration. (a = 5/7gsin) a = 5/7gsin = 5/7 (9.8m/s2)sin(21o) a = 2.5086 m/s2 = 2.5 m/s2 W 2.5 m/s/s

A marble (a solid sphere) has a mass of 23. 5 g, a radius of 1 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 21o incline. What is its velocity at the bottom of the plane if it started at rest? (a = 2.5086 m/s2) v2 = u2 + 2as v2 = 02 + 2(2.5086 m/s2)(2.75 m) v = 3.714 m/s = 3.7 m/s W 3.7 m/s

Angular Mechanics – Pulleys and such For the cylinder:  = I rT = (1/2m1r2)(a/r) (Where T is the tension in the string) r m1 m2 For the mass: F = ma m2g - T = m2a TOC

Angular Mechanics – Pulleys and such So now we have two equations: rT = (1/2m1r2)(a/r) or T = 1/2m1a and m2g - T = m2a TOC

Angular Mechanics – Pulleys and such T = 1/2m1a m2g - T = m2a Substituting: m2g - 1/2m1a = m2a Solving for a: m2g = 1/2m1a + m2a m2g = (1/2m1 + m2)a m2g/(1/2m1 + m2) = a TOC

Whiteboards: Pulleys 1 | 2 | 3 | 4 TOC

A string is wrapped around a 12 cm radius 4. 52 kg cylinder. A mass of A string is wrapped around a 12 cm radius 4.52 kg cylinder. A mass of .162 kg is hanging from the end of the string. Set up the dynamics equation for the hanging mass. (m2) r m1 m2 m2g - T = m2a W figure it out for yourself

A string is wrapped around a 12 cm radius 4.52 kg thin ring. A mass of .162 kg is hanging from the end of the string. Set up the dynamics equation for the thin ring (m1) r m1 m2  = I, I = m1r2,  = a/r,  = rT rT = (m1r2)(a/r) T = m1a W figure it out for yourself

A string is wrapped around a 12 cm radius 4.52 kg thin ring. A mass of .162 kg is hanging from the end of the string. Solve these equations for a: m2g - T = m2a T = m1a m2g - m1a = m2a m2g = m1a + m2a m2g = a(m1 + m2) a = m2g/(m1 + m2) r m1 m2 W figure it out for yourself

A string is wrapped around a 12 cm radius 4.52 kg thin ring. A mass of .162 kg is hanging from the end of the string. Plug the values in to get the acceleration: a = m2g/(m1 + m2) r m1 m2 a = m2g/(m1 + m2) a = (.162 kg)(9.80 N/kg)/(4.52 kg + .162 kg) a = .339 m/s/s W .339 m/s/s