M1=100 Kg Adult, M2=10 Kg Baby. the Seesaw Starts from Rest. Which Direction Will It Rotates?

m1 m2

m1=100 kg adult, m2=10 kg baby. The seesaw starts from rest. Which direction will it rotates?

(a) Counter-Clockwise (b) Clockwise ()(c) NttiNo rotation (d) Not enough information

Effect of a Constant Net Torque

2.3 A constant non-zero net torque is exerted on a wheel. Which of the following quantities must be changing?

1. angular position 2. angular velocity 3. angular acceleration 4. moment of inertia 5. kinetic energy 6. the mass center location

A. 1, 2, 3 B. 4, 5, 6 C. 1,2, 5 D. 1, 2, 3, 4 E. 2, 3, 5

1 Example: second law for rotation PP10601-50: A torque of 32.0 N·m on a certain wheel causes an angular acceleration of 25.0 rad/s2. What is the wheel's rotational inertia?

Second Law example: α for an unbalanced bar

Bar is massless and originally horizontal Rotation axis at fulcrum point L1 N L2 Î N has zero torque +y Find angular acceleration of bar and the linear m1gmfulcrum 2g acceleration of m1 just after you let go

τnet Constraints: Use: τnet = Itotα ⇒ α = Itot 2 2 Using specific numbers: where: Itot = I1 + I2 = m1L1 + m2L2 Let m1 = m2= m L =20 cm, L = 80 cm τnet = ∑ τo,i = + m1gL1 − m2gL2 1 2 gL − gL g(0.2 - 0.8) What happened to sin(θ) in moment arm? 1 2 α = 2 2 = 2 2 L1 + L2 0.2 + 0.8 2 net = − 8.65 rad/s Clockwise torque m gL − m gL a ==+ -α L 1.7 m/s2 α = 1 1 2 2 11 2 2 Accelerates UP m1L1 + m2L2 total I about pivot What if bar is not horizontal?

2 See Saw

3.1. Suppose everything is as it was in the preceding example, but the bar is NOT horizontal. Assume both masses are equal. Which of the following is the correct equation for the angular acceleration?

L (L − L ) 1 N θ A) α = 1 2 g L2 + L2 1 2 m1g fulcrum L2 (L − L ) g m g B) 1 2 2 α = 2 2 L1 + L2 cos(θ) (L − L ) C) 1 2 gsin( ) α = 2 2 θ L1 + L2 (L − L ) D) 1 2 gcos( ) α = 2 2 θ L1 + L2 (L − L ) g E) 1 2 α = 2 2 L1 + L2 sin(θ)

τnet = Itotα

See Saw - Solution

L1 N θ α = τ / I net tot m1g fulcrum L2 m2g τnet = ∑ τo,i = + m1gL1 cos(θ) − m2gL2 cos(θ)

2 2 Itot = I1 + I2 = m1L1 + m2L2 [m L − m L ] 1 1 2 2 g cos( ) α = 2 2 θ m1L1 + m2L2

For equal masses m1 –m2 = m (L − L ) If bar is balanced when it’s ∴ α = 1 2 gcos(θ) L2 + L2 horizontal, does that change 1 2 when it is not horizontal?

3 Method for solving (complex) problems using the Second Law Many components in the system means several (N) unknowns…. … need an equal number of independent equations

Method: • Draw or sketch system. Adopt coordinates, name the variables, indicate rotation axes, list the known and unknown quantities, … • Draw free bbdody diagrams of key parts. Show forces at thhieir poi nts of application. find torques about a (common) axis Note: can have

• May need to apply Second Law twice to each part Fnet .eq. 0 G G but τ .ne. 0 ¾ Translation: net Fnet = ∑Fi = ma G G G ¾ Rotation: τnet = ∑ τi = Iα • Make sure there are enouggqh (N) equations; there may be constraint equations (extra conditions connecting unknowns) • Simplify and solve the set of (simultaneous) equations. • Interpret the final formulas. Do they make intuitive sense? Refer back to the sketches and original problem • Calculate numerical results, and sanity check anwers (e.g., right order of magnitude?)

Example: Torque and Angular Acceleration of a Wheel

Analysis approach:

Break into two sub-systems wheel is accelerated angularly by tension T block is accelerated linearly by weight mg, with tension opposing

Free body diagrams shown r

The wheel is rotating and so we apply Στ = Ια a The tension supplies the torque via tangential force

The mass is moving in a straight line, so apply Newton’ s Second Law

ΣFy = may = mg -T How to connect the two problems above? mg Need a constraint linking linear acceleration to α a = αr

4 Application of Method using the Second Law • Cord wrapped around disk, hanging weight • Cord does not slip or stretch Æ constraint • Disk’s rotational inertia slows accelerations • Let m = 1.2 kg, M = 2.5 kg, r =0.2 m •ForFind mass acceleration m: of mass m, find α for disk F = ma = mg − T r T ∑ y y mg T = m (g − a) Unknowns: T, a a support force at axis “O” has FBD for disk, with axis at “o”: zero torque 1 2 N τ = + Tr = Iα I = Mr ∑ 0 2 Tr m(g − a)r α = = 1 Unknowns: a, α mg T Mg I Mr2 2 from “no So far: 2 Equations, 3 unknowns ÆNeed a constraint: a = + αr slipping” Substitute and solve: assumption 2 mg 2mgr 2mαr 2 α = - α = (= 24 rad/s ) Mr2 Mr2 r(m + M/2) α proportional to g m 2mg α(1 + 2 ) = mg 2 a = (= 4.8 m/s ) M Mr (m + M/2)

Example: Heavy pulley Atwood’s machine application of Newton’s 2nd Law Given the numerical values:

• m1= 5.0 kg, • m2= 5.5 kg r • r = 0.2 m, • M=60kgM = 6.0 kg O • I = 0.20 kg-m2 M, I

a) Find the accelerations of the hanging weights. Are they up or down? b) Find the angular acceleration of the pulley. Is it CW or CCW? m2 c) Find the tensions in each cord. d) How long d oes it ittk take the 55k5.5 kg wei ght htt to m fall 0.6 m from rest? 1

Strategy: Apply Newton’s Second Law (linear and rotational form) to all three bodies, connected by constraints. Make sure there are as many independent equations as there are unknowns

5 Heavy pulley Atwood’s machine: solution Apply Second Law: F = F = ma N (I) net ∑ i (each component) τ = τ = I α α (II) net ∑ i (pulley) r

If I = 0 problem simplifies O M, I For m1 use I: T1 m a = T - m g y 1 1 1 1 y a1 Mg m g 1. T1 = m1 (a1 + g) 1 m2

• positive α Æ negative a1 (down) m1 For m2 use I again: T • choose y positive up for m 2a 2 = T2 - m 2g 2 both masses a 2. T2 = m 2 (a 2 + g) 2 • CCW rotation Æ positive α m2g • positive α Æ positive a2 (up) • Free fall if T1 or T2 = 0

Constraints: cord cannot stretch or slip Note: if I = 0, then T1 = T2. Result using 1 & 2 becomes: a 2 = −a1 only 2 of these are independent m1 - m 2 a 2 = g a 2 = + αr a1 = −αr m1 + m 2

Atwood’s solution, continued For the Pulley use II: N NOTE: τnet = + T1r − T2ri = I α α r T ≠ m g (T − T ) r 1 1 3. α = 1 2 O T ≠ m g I M, I 2 2 T T • positive α Æ T > T 1 2 1 2 Mg Check ability to solve:

• 5 unknowns: a1, a2, T1, T2, α • Have 5 independent equations: 1,2,3 and 2 constraints OK! Solution: Begin eliminating unknowns (tension equations)

I α = T1r − T2ri = (m1g + m1a1 - m 2a 2 - m2g)r = (m 1 - m 2 )gr + (m1 + m 2 )a 1r a 2 = −a1 2 I α - (m1 + m2 )a 1r = (m 1 - m2 )gr I α + (m1 + m2 )αr = (m 1 - m2 )gr a1 = −αr

Result: (m1 − m 2 ) g r Net torque of hanging weights interpretation α = 2 Total moment of Inertia is simple I + (m1 + m 2 ) r Limiting case: (m1 − m 2 ) g m1 - m2 Let I Æ 0: α → a1 = −αr = − g massless (m1 + m2 ) r m1 + m 2 pulley Negative if m1 > m2:

6 Atwood’s solution, numerical evaluation

(5.0 − 5.5) (9.8) (0.2) α = (b) α = - 1.58 rad/s 2 (CW ) 0.20 + (5.0 − 5.5) 0.2 2

2 (a) a1 = - αr = + 0.32 m/s (up) 2 a2 = - a1 = − 0320.32 m/s (down)

T1 = 5.0 (9.8 + 0.32) (c) T1 = 506 N. (a1 increases tension) T2 = 5.5 (9.8 − 0.32) T2 = 521 N. (a 2 decreases tension)

How long does m2 take to fall 0.6 m from rest?

1 2 2d d = a 2t (d) t = = 1.94 s 2 a 2