Physics 2210 Fall 2015

smartPhysics 17-18 Rotational Statics

11/18/2015

Poll 11-18-01

In both cases the beam has the same mass and length and is attached to the wall by a hinge.

In which of the static cases shown is the tension in the supporting wire bigger?

A. Case 1 B. Case 2 C. Same

𝜏𝑔𝑔 = 𝐿

𝑔∙ 𝑀𝑀 ∙ sin +90°

= 1𝑔𝑀𝑀𝑀

𝑴𝑴

+90°

𝑴𝑴

+90° 𝜏𝑔1 = 𝐿

𝑔∙ 𝑀𝑀 ∙ sin +90°

= 1𝑔𝑀𝑀𝑀

𝜏𝑇1 = 𝑀 ∙ 𝑇1 ∙ sin −150° = −1

𝑔𝑇1𝑀

−150° −150°

𝜏𝑇𝑔 = 𝐿𝑔∙ 𝑇𝑔 ∙ sin −150°

= −14𝑇𝑔𝑀

𝜏𝑇1 + 𝜏𝑔1 = 0 𝑇1 = 𝑀𝑀

𝜏𝑇𝑔 + 𝜏𝑔𝑔 = 0 𝑇𝑔 = 2𝑀𝑀

Unit 17

This is exactly the same as 𝝉 = 𝒓 × 𝑭 Some people have found the “lever arm” confusing … here is my take on it The line-of-action of the force is a line parallel to the direction of the force and passing through the point of application of the force) The “Lever arm”, 𝒓⊥, is the perpendicular distance (of closest approach) from the rotation axis to the line-of-action Sometimes makes calculating the cross product easier

line-of-action 𝒓⊥

Lever Arm

3/3

Example 17-1 A traffic light hangs from a beam as shown in the figure. The uniform aluminum beam AB is 7.20 m long and has a mass of 12.0 kg. The mass of the traffic light is 21.5 kg. Determine (a) the tension in the horizontal massless cable CD, and (b) the vertical and horizontal components of the force exerted by the pivot A on the aluminum beam.

Solution This problem is all about balancing all forces and torques on the beam AB. We solve such problems by making a catalogue of all forces acting on the beam, and the torque each exerts. But first we must decide on a rotation axis about which to calculate torque. • It is usually convenient to pick a “pivot” that exerts force in both x and y directions on the

element in question: because then these forces exert NO torque!!! We pick A (1) Force of pivot: 𝐹𝑝𝑝 in the x-direction, 𝐹𝑝𝑝 in the y direction. Notice we are treating them

like two separate forces for convenience. They exert no torque (they act at the chosen rotation axis).

(2) Tension Force in the cable: 𝑇 directed in the –x direction, The line-of-action is horizontal, The lever arm, 𝒓⊥, is given in the diagram to be 3.80m. The torque (it acts CCW so it is POSITIVE) is 𝜏𝑇 = +𝒓⊥𝑇 = + 3.80𝑚 𝑇

𝑇 line-of-action

𝐹𝑝𝑝

𝐹𝑝𝑝

𝒓⊥= Lever Arm

Example 17-1

(3) Weight of the beam itself. 𝑚𝑀 in the –y direction Torque: 𝑚𝑀 acts at radius 𝑟=ℓ 2⁄ =3.60 m. Angle from radial vector to weight is −127° So the torque is 𝜏 𝑚 = ℓ 2⁄ 𝑚𝑀sin −127° : Note the negative orientation of this torque (it wants to deflect beam CW around the pivot) is contained in sin −127° = − sin 127° (4) Weight of traffic light: M𝑀 in the –y direction Acts at distance 𝑟=ℓ=7.20 m, and angle from radial vector to weight is AGAIN −127° 𝜏 𝑀 = ℓ𝑚𝑀 sin −127° Force components in the x- and y-direction independently sum to zero:

𝐹𝑝 = 𝐹𝑝𝑝 − 𝑇 = 0 … (1) 𝐹𝑝 = 𝐹𝑝𝑝 − 𝑚𝑀 −𝑀𝑀 = 0 … 2

(*** note you CANNOT add the magnitudes of the forces) Torques add to zero

𝜏 = 𝜏𝑇 + 𝜏 𝑚 + 𝜏 𝑀 = 3.80 m 𝑇 − 3.60 m 𝑚𝑀 sin 127° − 7.20 m 𝑀𝑀 sin 127° = 0 … (3)

𝐹𝑝𝑝

𝐹𝑝𝑝

𝑇

𝑚𝑀

37° 53° 127°

𝑀𝑀 Uniform beam AB: ℓ=7.20 m, 𝑚 =12.0 kg. (gi09-019) Traffic light 𝑀=21.5 kg. Determine (a) tension 𝑇 (b) components 𝐹𝑝𝑝, 𝐹𝑝𝑝 of the force exerted by the pivot on the beam

Example 17-1 Uniform beam AB: ℓ=7.20 m, 𝑚 =12.0 kg. (gi09-019) Traffic light 𝑀=21.5 kg. Determine (a) tension 𝑇 (b) components 𝐹𝑝𝑝, 𝐹𝑝𝑝 of the force exerted by the pivot on the beam

From equation (3) we have

𝑇 =3.60 m 𝑚𝑀 sin 127° + 7.20 m 𝑀𝑀 sin 127°

3.80 m = 𝑀 sin 127°3.60 m 𝑚 + 7.20 m 𝑀

3.80 m= 9.8 m s𝑔⁄ 0.799

3.60 m 12.0 kg + (7.20 m)(21.5 kg)3.80 m

= 7.827 m s𝑔⁄198 kg ∙ m

3.80 m = 408 N

From equation (1): 𝐹𝑝𝑝 − 𝑇 = 0 → 𝐹𝑝𝑝 = 𝑇 = 408 N

From equation (2): 𝐹𝑝𝑝 − 𝑚𝑀 −𝑀𝑀 = 0

𝐹𝑝𝑝 = 𝑚𝑀 + 𝑀𝑀 = 12.0 kg + 21.5 kg 9.8 m s𝑔⁄ = 328 N

𝐹𝑝𝑝

𝐹𝑝𝑝

𝑇

𝑚𝑀

37° 53° 127°

𝑀𝑀

Example 17-1 synopsis A traffic light hangs from a beam as shown in the figure. The uniform aluminum beam AB is 7.20 m long and has a mass of 12.0 kg. The mass of the traffic light is 21.5 kg. Determine (a) the tension in the horizontal massless cable CD, and (b) the vertical and horizontal components of the force exerted by the pivot A on the aluminum beam.

Choose pivot at A to be rotation axis. (1) Force components 𝐹𝑝𝑝 in the +x direction, 𝐹𝑝𝑝 in the +y direction. They exert no torque (2) Tension Force : 𝑇 directed in the –x direction 𝜏𝑇 = +𝑇 ∙ (3.80m) (3) Weight of the beam: 𝑚𝑀 in the –y direction , torque is 𝜏 𝑚 = − ℓ 2⁄ 𝑚𝑀sin 127° : (4) Weight of traffic light: M𝑀 in the –y direction 𝜏𝑀 = −ℓ𝑚𝑀 sin 127°

𝐹𝑝 = 𝐹𝑝𝑝 − 𝑇 = 0 … (1) 𝐹𝑝 = 𝐹𝑝𝑝 − 𝑚𝑀 −𝑀𝑀 = 0 … 2

𝜏 = 𝜏𝑇 + 𝜏 𝑚 + 𝜏 𝑀 = 3.80 m 𝑇 − 3.60 m 𝑚𝑀 sin 127° − 7.20 m 𝑀𝑀 sin 127° = 0 … 3

𝑇 = 9.8 m s𝑔⁄ 0.7993.60 m 12.0 kg + (7.20 m)(21.5 kg)

3.80 m = 408 N

𝐹𝑝𝑝 − 𝑇 = 0 → 𝐹𝑝𝑝 = 𝑇 = 408 N

𝐹𝑝𝑝 − 𝑚𝑀 −𝑀𝑀 = 0 𝐹𝑝𝑝 = 𝑚𝑀 + 𝑀𝑀 = 12.0 kg + 21.5 kg 9.8 m s𝑔⁄ = 328 N

Unit 18

We already covered this in Unit 16

This statement is somewhat misleading: because 𝑑𝑈𝑔 𝑑𝑌𝐶𝑀⁄ = 𝑀𝑀 is never ZERO JUST IGNORE IT!!!!

𝝋

What we really want to say is 𝑑𝑈𝑔 𝑑𝜑⁄ = 0

To minimize the potential energy

𝑈𝑔 = 𝑀𝑀𝑌𝐶𝑀

= 𝑀𝑀ℓ2

sin𝜑 + 𝑏

𝒃

Unit 18

𝑁𝐴

Poll 11-18-02

Suppose you hang one end of a beam from the ceiling by a rope and the bottom of the beam rests on a frictionless sheet of ice. The center of mass of the beam is marked with a black spot.

Which of the above configurations best represents the equilibrium condition of this setup?

A. A B. B C. C

𝑇𝐴

𝑀𝑀 𝑥

𝑦

𝑭𝒙 > 𝟎 𝑇𝐵

𝑀𝑀 𝑁𝐴

𝑭𝒙 > 𝟎 𝑇𝐵

𝑀𝑀 𝑁𝐴

𝑭𝒙 = 𝟎!!!!

Poll 11-18-03

In the two cases shown above identical ladders are leaning against frictionless walls and are not sliding.

In which case is the force of friction between the ladder and the ground the biggest?

A. Case 1 B. Case 2 C. Same

Unit 19

Unit 19

Law of Conservation of Angular Momentum

If the sum of the external torques on a system is zero, the angular momentum of the system is conserved.

Poll 11-20-01

Consider the two collisions shown above. In both cases a solid disk of mass M, radius R, and initial angular velocity ω0 is dropped onto an initially stationary second disk having the same radius. In Case 2 the mass of the bottom disk is twice as big as in Case 1. If there are no external torques acting on either system, in which case is the final kinetic energy of the system biggest?

A. Case 1 B. Case 2 C. Same

Poll 11-20-02

The angular momentum of a freely rotating disk around its center is Ldisk. You toss a heavy block horizontally onto the disk along the direction shown. Friction acts between the disk and the block so that eventually the block is at rest on the disk and rotates with it. We will choose the initial angular momentum of the disk to be positive.

What is the Ltotal, the magnitude of the angular momentum of the disk-block system?

A. Ltotal > Ldisk B. Ltotal = Ldisk C. Ltotal < Ldisk

Poll 11-20-03

The angular momentum of a freely rotating disk around its center is Ldisk. You toss a heavy block horizontally onto the disk along the direction shown. Friction acts between the disk and the block so that eventually the block is at rest on the disk and rotates with it. We will choose the initial angular momentum of the disk to be positive.

What is the magnitude of the final angular momentum of the disk-block system?

A. Ltotal > Ldisk B. Ltotal = Ldisk C. Ltotal < Ldisk

Law of Conservation of Angular Momentum Definition (1) : Angular Momentum of a rotor (sign of 𝑀 follows sign of 𝜔)

𝑀 ≡ 𝐼𝜔 Definition (2) : Angular Momentum of a particle in linear motion

𝑀 ≡ 𝑚𝑚𝑏 Sign of L is positive if particle misses P to the right Negative if it misses P to the left

P

𝐼 𝜔 𝑀 ≡ 𝐼𝜔

Total angular momentum of a system (rotors and particles defined above) is the sum (with sign: they are vectors!!!) of the angular momenta of all components. In the absence of external torque about a pivot (rotation axis) P, the total angular momentum (sum of individual angular momenta) of a system is conserved.

Example 19-1 putty 𝑚=100 g , 𝑚0=25 m/s, sticks to a stick, 𝑀=1.4 kg , ℓ=1.2 m. 𝑏=0.80 m Find 𝜔 of the stick + putty after the collision.

Solution by Conservation of Angular Momentum The only external force acting on the system is that by the pivot P. But since this force acts at point P, it exerts no torque No net external torque in collision Angular Momentum is conserved Before: (stick at rest)

𝑀𝑖 = 𝑚𝑚0𝑏 (note in this case 𝑏 turns out to be the minimum approach distance) After: rod and putty become joined, with total moment-of-inertia:

𝐼(𝑟𝑟𝑟+𝑝𝑝𝑝𝑝𝑝) = 𝐼𝑟𝑟𝑟 + 𝑚𝑏𝑔 And the system now rotates at angular velocity 𝜔

𝑀𝑓 = 𝐼(𝑟𝑟𝑟+𝑝𝑝𝑝𝑝𝑝)𝜔 = 𝑀ℓ 𝑔 3⁄ + 𝑚𝑏𝑔 𝜔 Total Angular Momentum is conserved:

𝑀𝑓 = 𝑀𝑖 , → 𝑀ℓ 𝑔 3⁄ + 𝑚𝑏𝑔 𝜔 = 𝑚𝑚0𝑏

𝜔 =𝑚𝑚0𝑏

𝑀ℓ 𝑔 3⁄ + 𝑚𝑏𝑔=

0.10 kg 25 m s⁄ 0.80 m1.4 kg 1.2 m 𝑔 3⁄ + 0.10 kg 0.80 m 𝑔 = 2.72 rad/s

Unit 20

Demos In many systems one can change the moment-of-inertia of a system by changing the radial distribution of mass. To conserve angular momentum the angular speed must change to compensate

https://www.youtube.com/watch?v=yAWLLo5cyfE https://www.youtube.com/watch?v=yAWLLo5cyfE

https://www.youtube.com/watch?v=UZlW1a63KZs https://www.youtube.com/watch?v=UZlW1a63KZs

https://www.youtube.com/watch?v=AQLtcEAG9v0 https://www.youtube.com/watch?v=AQLtcEAG9v0

Example 20-1 (gi08-69) A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad/s. Its total moment of inertia is 1360 kg⋅m2. Four people standing on the ground, each of mass 65 kg, suddenly step onto the edge of the merry-go-round. (a) What is the angular velocity of the merry-go-round now? (b) What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)? Answer: (a) 0.43 rad/s; (b) 0.80 rad/s.

Solution: (a) An external force acts at the center of the merry-go-round to keep the disk from moving away from the pivot. But this force exerts no external torque (to the disk + people system) So the total angular momentum is conserved. Before “collision” (jumping on)

𝑀𝑖 = 𝐼𝑀𝑀𝑀𝜔𝑖 Where 𝐼𝑀𝑀𝑀 =1360 kg m2 is the moment-of-inertia of the merry-go-round given, and 𝜔𝑖 =0.80 rad/s is its initial angular velocity. The people were at rest on the ground and do not contribute to the angular momentum. After the four people jump on the rim of the disk, the total moment of inertia changes to

𝐼 = 𝐼𝑀𝑀𝑀 + 4𝑚𝑅𝑔 Where m is the mass of each person, R=2.1m is the radius of the merry-go-round And so after the “collision”:

𝑀𝑓 = 𝐼𝜔𝑓 = 𝐼𝑀𝑀𝑀 + 4𝑚𝑅𝑔 𝜔𝑓

Example 20-1 (gi08-69) A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad/s. Its total moment of inertia is 1360 kg⋅m2. Four people standing on the ground, each of mass 65 kg, suddenly step onto the edge of the merry-go-round. (a) What is the angular velocity of the merry-go-round now? (b) What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)? Answer: (a) 0.43 rad/s; (b) 0.80 rad/s.

Solution: (a) continued: Total angular momentum is conserved: 𝑀𝑖 = 𝑀𝑓 → 𝐼𝑀𝑀𝑀𝜔𝑖 = 𝐼𝑀𝑀𝑀 + 4𝑚𝑅𝑔 𝜔𝑓

𝜔𝑓 =𝐼𝑀𝑀𝑀𝜔𝑖

𝐼𝑀𝑀𝑀 + 4𝑚𝑅𝑔 =1360 kg ∙ m𝑔 0.80 rad s⁄

1360kg ∙ m𝑔 + 4 65 kg 2.1 m 𝑔 = 0.434 rad s⁄

(b) Here we start with all 4 people on board at the rim, so wehad, before the collision: 𝑀𝑖 = 𝐼𝜔𝑖 = 𝐼𝑀𝑀𝑀 + 4𝑚𝑅𝑔 𝜔𝑖

When the people jump off, they pushed radially, so their tangential speeds are unchanged: 𝑚𝑓 = 𝑚𝑖 = 𝑅𝜔𝑖

And so after the collision: 𝑀𝑓 = 𝐼𝑀𝑀𝑀𝜔𝑓 + 4𝑚𝑚𝑓𝑏 = 𝐼𝑀𝑀𝑀𝜔𝑓 + 4𝑚𝑚𝑓𝑅 = 𝐼𝑀𝑀𝑀𝜔𝑓 + 4𝑚𝑅𝑔𝜔𝑖

𝑀𝑖 = 𝑀𝑓 → 𝐼𝑀𝑀𝑀 + 4𝑚𝑅𝑔 𝜔𝑖 = 𝐼𝑀𝑀𝑀𝜔𝑓 + 4𝑚𝑅𝑔𝜔𝑖 → 𝐼𝑀𝑀𝑀𝜔𝑖 + 4𝑚𝑅𝑔𝜔𝑖 = 𝐼𝑀𝑀𝑀𝜔𝑓 + 4𝑚𝑅𝑔𝜔𝑖 → 𝐼𝑀𝑀𝑀 𝜔𝑖 = 𝐼𝑀𝑀𝑀𝜔𝑓

𝜔𝑓 = 𝜔𝑖 = 0.80 rad s⁄

Precessing Gyropscope Demo Video from MIT https://www.youtube.com/watch?v=8H98BgRzpOM https://www.youtube.com/watch?v=8H98BgRzpOM