Today Torque on Extended Object by Gravitational Force → Assume That

Physics 106 Week 3 Rotational Dynamics II SJ 7th Ed.: Chap 10.7

• Newton’s Second Law for Rotation • Problem solving methods • Examples. • Heavy pulley Atwood’s machine Today • Double pulley Atwood’s machine

Torque on extended object by gravitational force Æ Assume that the total gravitational force effectively acts at the center of mass.

Gravitational potential energy of extended object Æ M g H, where H is the height of the center of mass and M is the total mass.

1 iClicker Q

Axis of rotation

HiHorizonta l unif orm rod of length L & mass M

Find the torque by gravitational force.

A. LMg B. (/)(L/2)Mg C. 2LMg D. (3/2)LMg E. None of the above

iClicker Q

Axis of rotation

HiHorizonta l unif orm rod of length L & mass M 1 Find the moment of inertia. I = ML2 cm, rod 12 A. ML^2 B. (1/3)ML^2 C. (1/4)ML^2 D. (1/12)ML^2 E. (13/12)ML^2

2 Moments of Inertia of Various Rigid Objects

iClicker Q

Axis of rotation

HiHorizonta l unif orm rod of length L & mass M

Find the angular acceleration.

A. (2/3)g/L B. 2(g/L^2) C. g/L D. (1/2)(g/L) E. (3/2)(g/L)

3 Example of energy conservation

Axis of rotation

Horizontal uniform rod of length L & mass M is released from rest.

Find its angular speed at the lowest point, assuming no friction between axis of rotation and the rod.

iClicker Q

Axis of rotation

Horizontal uniform massless rod of length L Uniform circular disk of diameter L & mass M

Find the torque by gravitational force. A. LMg B. (L/2)Mg C. 2LMg D. (3/2)LMg E. None of the above

4 Example A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal, frictionless pin through one end of the rod. (The moment of inertia of the rod about this axis is ML2/3.) The rod is released when it makes an angle of 37° with the horizontal. What is the angular acceleration of the rod at the instant it is released?

Example

Axis of rotation

HiHorizonta l unif orm rod of length L & mass M Uniform circular disk of diameter L & mass M

Find the net torque. Find the total moment of inertia Find the angular acceleration.

5 Example: Torque and Angular Acceleration of a Wheel

•Cord wrapped around disk, hanging weight • Cord does not slip or stretch •Let m = 1.2 kg, M = 2.5 kg, r =0.2 m • Find acceleration of mass m, r find angular acceleration α for disk, tension, and torque on the a disk 1 Formula: I = Mr2 2 mg

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

6 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 does 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

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

7 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:

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

8 Example: Two-level-pulley Atwood’s machine

α Assume: • a = +2 m/s2 (CCW) A rA rB • r = 2 cm, r = 5 cm A B O Find α for the pulley:

MP, I a A aB Find aB (mass B linear acceleration): mB

No slipping on double pulley means:

• αrA = aA • αrB = aB • α is the same for A & B

• but aA is different from aB

Example: no-slipping conditions on a double pulley

No slipping on double pulley means: α • αrA = aA • αr = a B B rA rB • α is the same for A & B O • but aA is different from aB Example: Assume: MP, I a 2 A aB • aA = +2 m/s (CCW) • r = 2 cm, r = 5 cm A B mB Find α for the pulley: mA aA 2 2 α = = = 100 rad/s (CCW) rA .02

Find aB (mass B linear acceleration):

aArB 2 x 5 2 aB = αrB = = = 5 m/s rA 2

What if there IS slippage? Can not connect a’s with α

9 Two-level-pulley Atwood’s machine - application of Newton’s 2nd Law α For the “two-level-pulley” Atwood’s machine, r rB two cords connect to the 10 kg pulley, whose A angular acceleration is 2 rad/s2 clockwise, The O tension in the cord attached to mass B is 50 N a) Draw free body diagrams b) and apply Newton’s second law to all three M , I bodies. P aA a c) Calculate the rotational inertia of the pulley B from the dynamics and conditions given mB d) Find the unknown mass mB. mA

Strategy – same as before: Numerical values: m = 2.0 kg, m = ?? Apply Newton’s Second Law A B r = 60 cm, r = 40 cm (linear and rotational) applying it A B MP = 10 kg I = ??? to all three bodies, connected by α= - 2 rad/s2 (CW) constraints TB= 50 N.

mass A:

mass B:

pulley:

TA= ?

aA= ? aB= ?

10 No slipping constraints:

5 equations Find I : P 5 unknowns TA,aA,mB,aB, IP I All is well p Problem can be solved

In the above:

• don’t know MB yet • other quantities are known • first term is just the rotational inertia of the hanging masses

Side remark:

Find MB using equation 2:

Now find the numerical value of IP: