Physics 8 — Friday, October 23, 2015

I Turn in HW6. Pick up HW7 handout (due next Friday). HW7will be quite easy in comparison with HW6, to make up foryour having a lot to read next week.

I For Monday, read Chapter 2 (“statics”) of Onouye/Kane. It’sabout 90 pages, but mostly review, from an ARCHperspective, of material we’ve already seen this semester.

I For Wednesday, read Chapter 3 (analyzes cables, trusses,arches, etc.) of Onouye/Kane: about 75 pages — yikes!Carefully read 3.1, 3.2, 3.3, 3.5. Quickly skim 3.4 and 3.6.

I Many of you already bought copies of Onouye/Kane. I havemany more used copies available. You need your own copy.

I Today: we’ll finally discuss torque!

I Warning! Mechanical and structural engineers usually use thephrase “moment of a force” to mean what we call “torque.”

Torque: the rotational analogue of force

Just as an unbalanced force causes linear acceleration

~F = m~a

an unbalanced torque causes rotational acceleration

τ = Iα

Torque is (lever arm) × (force)

τ = r⊥ F

where r⊥ is the “perpendicular distance” from the rotation axis tothe line-of-action of the force.

position

~r = (x , y)

velocity

~v = (vx , vy ) =d~rdt

acceleration

~a = (ax , ay ) =d~vdt

force~F = m~a

rotational coordinate

ϑ = s/r

rotational velocity

ω =dϑdt

rotational acceleration

α =dωdt

torqueτ = Iα

τ = r⊥ F

I wind a string around a coffee can of radius R = 0.05 m. (That’s5 cm.) Friction prevents the string from slipping. I apply a tensionT = 20 N to the free end of the string. The free end of the stringis tangent to the coffee can, so that the radial direction isperpendicular to the force direction. What is the magnitude of thetorque exerted by the string on the coffee can?

(A) 1 N ·m(B) 2 N ·m(C) 5 N ·m(D) 10 N ·m(E) 20 N ·m

Suppose that the angular acceleration of the can is α = 2 s−2

when the string exerts a torque of 1 N ·m on the can. What wouldthe angular acceleration of the can be if the string exerted a torqueof 2 N ·m instead?

(A) α = 0.5 s−2

(B) α = 1 s−2

(C) α = 2 s−2

(D) α = 4 s−2

(E) α = 5 s−2

(F) α = 10 s−2

I apply a force of 5.0 N at a perpendicular distance of 5 cm(r⊥ = 0.05 m) from this rotating wheel, and I observe someangular acceleration α. What force would I need to apply to thissame wheel at r⊥ = 0.10 m (that’s 10 cm) to get the same angularacceleration α?

(A) F = 1.0 N(B) F = 2.5 N(C) F = 5.0 N(D) F = 10 N(E) F = 20 N

Suppose that I use the tension T in the string to apply a giventorque τ = r⊥T to this wheel, and it experiences a given angularacceleration α. Now I increase the rotational inertia I of the wheeland then apply the same torque. The new angular accelerationαnew will be

(A) larger: αnew > α

(B) the same: αnew = α

(C) smaller: αnew < α

I want to tighten a bolt to a torque of 1.0 newton-meter, but Idon’t have a torque wrench. I do have an ordinary wrench, a ruler,and a 1.0 kg mass tied to a string. How can I apply the correcttorque to the bolt?

(A) Orient the wrench horizontally and hang the mass at adistance 0.1 m from the axis of the bolt

(B) Orient the wrench horizontally and hang the mass at adistance 1.0 m from the axis of the bolt

If the wrench is at 45◦ w.r.t. horizontal, will the 1.0 kg masssuspended at a distance 0.1 m along the wrench still exert a torqueof 1.0 newton-meter on the bolt?

(A) Yes. The force of gravity has not changed, and the distancehas not changed.

(B) No. The torque is now smaller — about 0.71 newton-meter— because the “perpendicular distance” is now smaller by afactor of 1/

√2.

(C) No. The torque is now larger — about 1.4 newton-meter.

τ = r⊥F = rF⊥ = rF sin θrF = |~r × ~F |Four ways to get the magnitude of the torque

I (perpendicular component of distance) × (force)

I (distance) × (perpendicular component of force)

I (distance) (force) (sin θ between ~r and ~F )

I use magnitude of “vector product” ~r × ~F (a.k.a.“cross product”)

To tighten a bolt, I apply a force of the same magnitude F atdifferent positions and angles. Which torque is largest?

To tighten a bolt, I apply a force of magnitude F at differentpositions and angles. Which torque is smallest?

I want to apply to this meter stick two torques of the samemagnitude and opposite sense, so that the stick has zero rotationalacceleration. I apply one force of 5 N at a lever arm of 0.5 m. Iwant to apply an opposing force at a lever arm of 0.2 m, so thatthe second torque balances the first torque. How large must thissecond force be?

(A) 1.0 N

(B) 2.0 N

(C) 12.5 N

(D) 25 N

I want to apply to this meter stick two torques of the samemagnitude and opposite sense, so that the stick has zero rotationalacceleration. I apply one force of 10 N at a lever arm of 0.5 m. Itie a second string on the opposite end, 0.5 m from the pivotpoint. The second force is applied at a 45◦ angle w.r.t. thevertical. How large must this second force be?

(A) 5 N

(B) 7 N

(C) 10 N

(D) 14 N

(E) 20 N

If the rod doesn’t accelerate (rotationally, about the pivot), whatforce does the scale read?

(A) 1.0 N

(B) 5.0 N

(C) 7.1 N

(D) 10 N

(E) 14 N

(F) 20 N

If the rod doesn’t accelerate, what force does the scale read?

(A) 1.0 N

(B) 5.0 N

(C) 7.1 N

(D) 10 N

(E) 14 N

(F) 20 N

If the rod doesn’t accelerate, what force does the scale read?

(A) 1.0 N

(B) 5.0 N

(C) 7.1 N

(D) 10 N

(E) 14 N

(F) 20 N

If the rod doesn’t accelerate, what force does the scale read?

(A) 1.0 N

(B) 5.0 N

(C) 7.1 N

(D) 10 N

(E) 14 N

(F) 20 N

If the rod doesn’t accelerate, what force does the scale read?

(A) 1.0 N

(B) 5.0 N

(C) 7.1 N

(D) 10 N

(E) 14 N

(F) 20 N

τ = r⊥F = rF⊥ = rF sin θrF = |~r × ~F |

Four ways to get the magnitude of the torque dueto a force:

I (perpendicular component of distance) × (force)

I (distance) × (perpendicular component of force)

I (distance) (force) (sin θ between ~r and ~F )

I use magnitude of “vector product” ~r × ~F (a.k.a.“cross product”)

To get the “direction” of a torque, use theright-hand rule.

Physics 8 — Friday, October 23, 2015

I Turn in HW6. Pick up HW7 handout (due next Friday). HW7will be quite easy in comparison with HW6, to make up foryour having a lot to read next week.

I For Monday, read Chapter 2 (“statics”) of Onouye/Kane. It’sabout 90 pages, but mostly review, from an ARCHperspective, of material we’ve already seen this semester.

I For Wednesday, read Chapter 3 (analyzes cables, trusses,arches, etc.) of Onouye/Kane: about 75 pages — yikes!Carefully read 3.1, 3.2, 3.3, 3.5. Quickly skim 3.4 and 3.6.

I Many of you already bought copies of Onouye/Kane. I havemany more used copies available. You need your own copy.

I Today: we’ll finally discuss torque!

I Warning! Mechanical and structural engineers usually use thephrase “moment of a force” to mean what we call “torque.”