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Page 1: By the way… α = “alpha” ω = “omega” θ = “theta” τ = “tau”

Extra Practice: 8.1, 8.3, 8.5, 8.7, 8.25, 8.27

By the way…α = “alpha” ω = “omega” θ = “theta” τ = “tau”

I’ve drawn a wrench but there’s a lot of physiotherapy application in this concept. So all you future physiologists, pay attention!

Page 2: By the way… α = “alpha” ω = “omega” θ = “theta” τ = “tau”

Movie project due TODAY!• If you don’t have it now, deliver to my mailbox

in White Hall 111 by 4pm.• Can be hand-written or computer (or both).• Check through rubric before you finish the draft:

http://sarahspolaor.faculty.wvu.edu/home/physics-101

Will be 30% penalty if <week late, 50% penalty if 1-2 weeks late, 75% penalty if up to 2-3 weeks late. If not turned in by April 20 you will receive a 0.

Page 3: By the way… α = “alpha” ω = “omega” θ = “theta” τ = “tau”

Today’s goals.• Calculate torque.

• Rotational equilibrium.

• Find something’s center of gravity.

Page 4: By the way… α = “alpha” ω = “omega” θ = “theta” τ = “tau”

Torque

The ability of a force to rotate an object around an axis.

⊥= rFτUnits: N m

r is a vector starting from axis of rotation pointing to where force is applied

A FORCE at a given RADIUS makes a TORQUE. More torque means more rotational acceleration.[See light board notes for examples]

Page 5: By the way… α = “alpha” ω = “omega” θ = “theta” τ = “tau”

Torque: Notes…Convention:

Counterclockwise spin,torque positive

+ τ

Ex: wheel with two

forces acting on it

F!

F!

r We can have a net torque even though

net force = 0!

Net torque: an object can have a net torque acting on it even though the net force is zero. Fnet = F - F = 0 tau_net = rF + rF = 2rF

Page 6: By the way… α = “alpha” ω = “omega” θ = “theta” τ = “tau”

120 N

Identify:1. Pivot point.2. Line from pivot to force (r).3. Force’s component

perpendicular to r.

The fishing pole in the figure makes an angle of 20.0° with the horizontal. What is the magnitude of the torque exerted

by the fish about an axis through the angler's hand if the fish pulls on the fishing line with a force of 120 N at an

angle 37.0° below the horizontal? The force is applied at a point with distance L = 2 m from the angler's hands.

Many problems will appear ugly (complicated). They’re not! Don’t get intimidated!

[Problem done on light board]

Page 7: By the way… α = “alpha” ω = “omega” θ = “theta” τ = “tau”

Everyone grab something heavy. Is it harder to hold it straight out or with your elbows bent?

Q82

A. Straight outB. With elbows bentC. Same for both

⊥= rFτ

ANSWER: A.

Page 8: By the way… α = “alpha” ω = “omega” θ = “theta” τ = “tau”

Q83

A. B.

C. D.

⊥= rFτ

E. There are 2 best ones (equal torque)

You are using a wrench and trying to loosen a rusty nut. Which of the arrangements shown is most

effective in loosening the nut?

SAY ANSWER: B. Whenever I ask you a question like this, always consider your experience, but also consult the relevant equation!

Page 9: By the way… α = “alpha” ω = “omega” θ = “theta” τ = “tau”

Joe the plumber pushes straight down on the end

of a long wrench as shown. What is the

magnitude of the torque Joe applies about the

pipe?

A. (0.80 m)(900 N) sin 19° B. (0.80 m)(900 N) cos 19°

C. (0.80 m)(900 N) tan 19° D. (0.80 m)(900 N) E. None of the above Q84

19°

⊥= rFτ

Answer: B

Page 10: By the way… α = “alpha” ω = “omega” θ = “theta” τ = “tau”

Equilibrium (again)• Same concept as linear momentum.

• AT REST

• *OR* Rotating at CONSTANT VELOCITY!

ΣFx = 0 ΣFy = 0 Στ = 0

I’m just gonna say that in the next few lectures you’ll notice a common theme: that I’m going to keep saying, “this is just like we did before, only now for rotational motion.” Previously we saw that kinematics equations have rotational equivalents, and here we are seeing that you need a torque to change rotation rate. Suggest breaking this forces into x and y directions (one radial, other perpendicular). Like the fishing problem.THESE EQUATIONS ARE THE DEFINITION OF EQUILIBRIUM!Note, you’ve seen this before (with just forces, not torque) in linear equilibrium!

Page 11: By the way… α = “alpha” ω = “omega” θ = “theta” τ = “tau”

Center of Gravity“Where should I draw my Fg line?”

The center of gravity is the point around which a body’s mass is equally distributed in all directions. Center of gravity answers a simple question but has practical conceptual applications. For instance, if I throw something around what point will it rotate? Or, where should I put the fulcrum of my see-saw? If they were equal weight it’d be in the middle, but since one kid is heavier it’s off to one side.

Page 12: By the way… α = “alpha” ω = “omega” θ = “theta” τ = “tau”

Center of Gravity“Where should I draw my Fg line?”

60 kg

15 kgFind the center of gravity of the system of two blocks above if they are 50 m apart.

xcg = Σmixi Σmi

In a uniform object, COG is at its mid point

Center of gravity can be outside the object

Page 13: By the way… α = “alpha” ω = “omega” θ = “theta” τ = “tau”

Toppling Rule of Thumb

• If the CG of the object is above the area of support, the object will remain upright.

• If the CG is outside the area of support the object will topple. Leaning tower in Pisa, Italy

Try standing with your heels and butt against the wall and try to bend over. Can you keep standing? What’s happening to your center of gravity?

Page 14: By the way… α = “alpha” ω = “omega” θ = “theta” τ = “tau”

Human Center of GravityWithin the body, near the navel

(Generally) higher in men than women

To avoid toppling - stand so that your center of gravity is above area enclosed by your feet!


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