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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: 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 TOC Angular Mechanics - Tangential Relationships Linear: Tangential: (at the edge of the wheel) (m) s = r - Displacement (m/s) v = r - Velocity (m/s/s) a = r - Acceleration* *Not in data packet TOC 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 Rolling objects accelerate linearly and angularly: Force causing F r = I = rF so F = /r = I/r TOC Rolling: 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 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 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 (9.81 m/s2)sin(21o) a = 2.5111 m/s2 = 2.5 m/s2 2.5 m/s/s W 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(2.5111 m/s2)(2.75 m) v = 3.716 m/s = 3.7 m/s 3.7 m/s W Angular Mechanics – Pulleys and such For the cylinder: r m1 = I rT = (1/2m1r2)(a/r) (Where T is the tension in the string) For the mass: m2 F = ma m2g - T = m2a TOC Pulleys and such: r m1 So now we have two equations: rT = (1/2m1r2)(a/r) or T = 1/2m1a and m2g - T = m2a m2 TOC Angular Mechanics – Pulleys and such r m1 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 m2 TOC Whiteboards: Pulleys 1|2|3|4 TOC 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. (m2) m2g - T = m2a r m1 m2 figure it out for yourself W 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 (m1) = I, I = m1r2, = a/r, = rT rT = (m1r2)(a/r) T = m1a r m1 m2 figure it out for yourself W 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: m2g - T = m2a T = m1a m2g - m1a = m2a m2g = m1a + m2a m2g = a(m1 + m2) a = m2g/(m1 + m2) r m1 m2 figure it out for yourself W 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 = m2g/(m1 + m2) a = m2g/(m1 + m2) a = (0.162 kg)(9.80 N/kg)/(4.52 kg + 0.162 kg) a = 0.339 m/s/s r m1 m2 0.339 m/s/s W