Plan for Today (AP Physics 2)

Plan for Today (AP Physics 2) • B Testers • Thermo AP Problems • C Testers • Go over Parallel Axis Problems • Example Problems Using Moment of Inertia • Finish Lab, AP Problems

Torque • I have a wheel with a diameter of 26 cm that weighs 50 kg that is spinning at 30 m/s....How much torque is needed to accelerate the wheel to 35 m/s in 2.0 seconds

Torque Example Problem • R = .13 m • m = 50 kg • Vi = 30 m/s • vf= 35 m/s • t = 2.0 s • τ = ? • τ = I ɑ • ɑ = (wf-wi)/t = • (35/.13 – 30/.13)/2 • = 19.23 rad/s2 • I = ½ mR2 • = .5 * 50 *.132 • = .4225 kgm2 • τ = I ɑ = 19.23 *.4225 • = 8.1 N-m

Attwood Machine • An Attwood machine is constructed by mounting a well-lubricated pulley (not necessarily a ring/disc), whose radius is 6.00cm on an axle and passing a rope over the pulley. Each end of the rope is attached to a different mass of 3.0 kg and 2.0kg so they both hang vertically. when released from rest, a motion detector measures the descending mass's acceleration as 0.70m/s2. The mass of the pulley is 4.00kg. • a.) Find the rotational inertia of the pulley. • b) find the kinetic energy of the ENTIRE system when the mass has fallen 3.5 m.

Attwood Solution • Lets call the larger mass M and the smaller mass m • Mg - mg - F = (M + m) a • F = g(M - m) - a(M + m) = • 9.81 m/s2(3 kg - 2 kg) - 0.70 m/s2(3 kg + 2 kg) • F = 6.31 N • But also F = Torque/r = I * α/r = I * a/r2 • I = F * r2/a • I = (6.31 N) * (0.06m)2/(0.70 m/s2) • I = 0.032 kg-m2

Attwood Solution • Total KE = translational KE plus rotational KE • KEt= (1/2) (M + m) v2 • KEr= (1/2) I w2 • w = v/r • KE total = (1/2) (M + m) v2+ (1/2) I (v2/r2) • KE total = (1/2) v2[(M + m ) + (I/r2)] • Now we just need to find the velocity when distance = 3.5 m • vf2=vi2+ 2 a x • vi=0 • vf2= 2 a x • v = • v = • v = 2.21 m/s • KE = (1/2) (2.21 m/s)2 [(3 kg + 2 kg) + (0.032 kg-m2/(0.06 m)2] • KE = 6.08 J

YoYo Problem • Ayoyo has mass M and moment of inertia I. The string is wrapped around the axle whose radius is R. It is released with one end of the string held at rest. Find its acceleration. Neglect the width of the string

Yoyo answer • moment of inertia of disk: I = ½ MR² • Torque: T = FR = MgR = Iα • where, α: angular acceleration, and thus; • α = MgR / I = MgR / ½ MR² = 2g/R • The linear acceleration is: • a = Rα = 2g = 2*9.81 = 19.6 m/s²

Falling Rod • A thin uniform rod is initially positioned in the vertical direction, with its lower end attached to a frictionless axis that is mounted on the floor. The rod has a length of 3.00 m and is allowed to fall, starting from rest. Find the tangential speed of the free end of the rod, just before the rod hits the floor after rotating through 90°. The moment of inertia for a rod with axis at the end : I = 1/3 * m * L2

Falling Rod Solution • moment of inertia for a rod with axis at the end : I = 1/3 * m * L2 • Take the total mass m to be at its center of mass which is at L/2. • That mass has potential energy of mgh = mg(L/2). • When it falls to the ground this PE is converted to rotational kinetic energy = • Ker= ½ I w2 • So mg(L/2) = ½ I w2 • mgL= I w2 • but I = (1/3) m L2 • mgL= (1/3) m L2 w2 • g = (1/3)Lw2 • w2= 3g/L • w = • w = • w= 3.13049517 • w= v/L • so v = L (w) = 3 (3.13049517)= 9.39148551 m/s • v= 9.39 m/s

Plan for Today (AP Physics 2)