Rotational Dynamics

Consider a bicycle wheel to be a ring of radius 30 cm and mass 1. 5 kg Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of 20 N is applied tangentially to a sprocket of radius 4.0 cm for 4.0 s, what linear speed does the wheel achieve, assuming it rolls without slipping? a. 3.0 m/s b. 5.9 m/s c. 7.1 m/s d. 24 m/s

M = 1.5 kg, R = 0.30 m, F = 20 N, r = 0.04 m t = 4 s Thin hoop, I = MR2 = 0.135 kg m2 t = Ia = t/I = 20 N * 0.04 m/0.135 kg m2 = 5.9 rad/s2 = at = 5.9 * 4 = 23.6 rad/s v = wr = 23.6 rad/s * 0.3 m = 7.1 m/s

A wheel of moment of inertia of 5 A wheel of moment of inertia of 5.00 kgm2 starts from rest and accelerates under a constant torque of 3.00 Nm for 8.00 s. What is the wheel's rotational kinetic energy at the end of 8.00 s? a. 57.6 J b. 64.0 J c. 78.8 J d. 122 J

I = 5 kgm2, t = 3.00 Nm , t = 8.00 s Krot = ??? Krot = ½ I w2 = 1/2(5)w2 To find omega, w, use w = wo + at. How do we find alpha, a? a = t/I= 3/5 = 0.6 rad/s2. So, w = wo + at = 0 + 0.6 * 8 = 4.8 rad/s So, Krot = ½ I w2 = 1/2(5)w2 = 57.6 J

A solid sphere of mass 1. 0 kg and radius 0 A solid sphere of mass 1.0 kg and radius 0.010 m is released from the top of a 1.0-m high 370 inclined plane. What is the speed of the sphere when it reaches the bottom of the inclined plane? a. 3.7 m/s b. 4.4 m/s c. 5.6 m/s d. 6.3 m/s

M =1.0 kg, R=0.010 m, y = 1.0 m V = ??? First, let’s set up the linear and rotational dynamics equations for this sphere. F = Ma Mgsinq – f = Ma Mgsinq –Ma = f = Ia rf = I a/r f = Ia/r2 = 2/5MR2 (a/R2) = 2/5 Ma Combining, Mgsinq – Ma = f= 2/5 Ma Gives a = 5/7 g sinq = 4.2 m/s2

v = vo + at = 0 + 4.2 m/s2 * t. Find t from, x = ½ a t2 , t= (2x/a)1/2 = (2*1.66 m/ 4.2 m/s2)1/2 = 0.89 s So, v = 3.7 m/s.

Rotational Dynamics Torque, τ =I α Work, W = τΘ Kinetic Energy of Rotation, K = ½ Iω2 Conservation of Energy Angular Momentum, L = Iω Conservation of Angular Momentum

Newton’s 2nd Law Linear motion Fnet = ma Linear motion τnet = Iα

Fnet = m1 a Fnet = T1 - µN N = m1g T1 - µm1g=m1a T1 = m1a -µm1g τnet = Iα τnet =R(T2 – T1) I = ½ MR2 R(T2 – T1) =½MR2α N Accel. = a Ang. Accel = α f= µN T1 M,R T2 Fnet = m2 a Fnet = m2g – T2 m2g –T2 =m2a T2 = -m2a +m2g Accel = a m1g m2g

-(m2 + m1)a +g(m2 +µm1) =½MRa/R T1= m1a -µm1g;T2 – T1 =½MRα;T2= -m2a +m2g -m2a +m2g – m1a +µm1g =½MRα -(m2 + m1)a +g(m2 +µm1) =½MRa/R (m2 + m1 +½M)a = g(µm1+ m2) a = g(µm1+ m2)/(m2 + m1 +½M) N Accel. = a Ang. Accel = α f= µN T1 M,R T2 Accel = a m1g m2g

What if we ignore pulley? a = 3.92 m/s2, 6.7% higher! a = g(µm1+ m2)/(m2 + m1 +½M) Let M =0.2 kg, m1=1 kg, m2=0.5 kg, µ=0.1 a = 3.7 m/s2, What if we ignore pulley? a = 3.92 m/s2, 6.7% higher! N Accel. = a Ang. Accel = α f= µN T1 M,R T2 Accel = a m1g m2g

It took work to rotate the pulley The pulley goes from zero angular speed to non-zero angular speed, ω = αt. This means that the pulley’s final kinetic energy, Kf , is ½ Iω2 Kf = ½(½ MR2)ω2 The work-energy theorem, W = ΔK W = τΘ=¼MR2ω2

Torque and Angular Momentum Recall, F = ma = mΔv/Δt = Δp/Δt, so making the correspondence to rotational motion, τ = Iα = IΔω/Δt = Δ(Iω)/Δt = ΔL/Δt, where L = Iω = angular momentum. τ = ΔL/Δt When no external forces are acting, torque = 0, so ΔL/Δt = 0 which means, Lf = Lo or Ifωf = Ioωo