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Phys 2101 Gabriela González

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⇒= amF

tt

rr

maFmaF

=

=

rmFrmTF

αφ

ωφ

=

=−

sincos 2

ατ

αφτ2

2sinrm

rmFr=

==

ατ I=

2

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We know all of that already!!

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The figure shows the massive shield door at a neutron test facility at Lawrence Livermore Laboratory; this is the world's heaviest hinged door. The door has a mass of 44,000 kg, a rotational inertia about a vertical axis through its huge hinges of 8.7x104 kg·m2, and a (front) face width of 2.4 m. Neglecting friction, what steady force, applied at its outer edge and perpendicular to the plane of the door, can move it from rest through an angle of 90° in 30 s?

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In the figure, a block has mass M = 500 g, the other has mass m = 460 g, and the pulley, which is mounted in horizontal frictionless bearings, has a radius of 5.00 cm. When released from rest, the heavier block falls 75.0 cm in 5.00 s (without the cord slipping on the pulley).

(a) What is the magnitude of the blocks' acceleration?

What is the tension in the part of the cord that supports (b) the heavier block and (c) the lighter block?

(d) What is the magnitude of the pulley's angular acceleration? (e) What is its rotational inertia?

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(a)  What is the magnitude of the blocks' acceleration?

What is the tension in the part of the cord that supports (b) the heavier block and

(c) the lighter block?

(Tensions are not equal any more!!!) (d) What is the magnitude of the pulley's angular acceleration?

(e) What is its rotational inertia?

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y =12at 2 ⇒ a = 2y / t 2 = 2 × 0.75m /25s2 = 0.06m /s2

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Mg−T1 = Ma⇒ T1 = M(a + g) = 0.5kg × 9.86m /s2 = 4.93N

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T2 −mg = ma⇒ T2 = m(a − g) = 0.496kg × 9.794m /s2 = 4.86N

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α = a /R = 0.06m /s2 /0.05m =1.2rad /s2

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τ = −T1R + T2R = IαI = (T2 −T1)R /α = 0.07N × 0.05m /1.2rad /s2 = 0.0029kgm2

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A meter stick is held vertically with one end on the floor and is then allowed to fall. Assuming that the end on the floor does not slip… (a) What forces are acting on the stick? (b) What forces are doing work? (c) Find the speed of the other end when it hits the floor.

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A meter stick is held vertically with one end on the floor and is then allowed to fall. Assuming that the end on the floor does not slip… (a) What forces are acting on the stick? Gravity (at the center of mass), static friction and normal force (at bottom end) (b) What forces are doing work? Only gravity does (positive) work.

(c) Find the speed of the other end when it hits the floor. The motion is pure rotation about the lower end. Use conservation of energy:

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Ibot = Icom + M(L /2)2 =1

12ML2 +

14ML2 =

13ML2

12Ibotω

2 −Mg L2

= 0⇒ω =MgLIbot

= 3 gL

vtop =ωL = 3gL = 5.4 m/s

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Assume a 50 kg woman is in a rotor, with µs=0.5, R=2m.

•  What’s the minimum angular velocity that will let the woman be stuck to the wall without a floor?

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Assume a 50 kg woman is in a rotor, with µs=0.5, R=2m.

What’s the minimum angular velocity that will let the woman be stuck to the wall without a floor?

We need enough static friction to cancel gravity. Maximum value of static friction is µsN. The normal force is the centripetal force, or mω2R.

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mg ≤ µsN = µsmω2R⇒ω ≥ g /(µsR) = 3.13 rad/s ~ 30 rpm

Answer is independent of the mass, but very much dependent on coefficient of friction!

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If we sit at the center of the wheel If we sit on the road, following the bike

It depends on the reference frame!

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2212

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comcom MvIKE += ω

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A solid ball starts from rest at the upper end of the track shown in the figure, and rolls without slipping until it rolls off the right-hand end. If H = 6.0 m and h = 2.0 m and the track is horizontal at the right-hand end, how far horizontally from point A does the ball land on the floor?

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A solid ball starts from rest at the upper end of the track shown in the figure, and rolls without slipping until it rolls off the right-hand end. If H = 6.0 m and h = 2.0 m and the track is horizontal at the right-hand end, how far horizontally from point A does the ball land on the floor?

From H to h: use conservation of energy to get speed. Notice that kinetic energy has both translational and rotational components:

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ΔK + ΔU = 012mvcom

2 +12Icomω

2 −mg(H − h) = 0

12mvcom

2 +12

25mR2 vcom

R

2

= mg(H − h)

710mvcom

2 = mg(H − h)

vcom= (10 /7)g(H − h) = 7.48 m/s

Notice that this is a smaller velocity than a point mass: rolling bodies fall slower than point masses, and the larger the rotational inertia, the slower they fall). The velocity of the center of mass at h is horizontal. From h on, com motion is a projectile motion (and the ball keeps rotating, but with a constant angular velocity, since gravity does not produce a torque)

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h =12gt 2 ⇒ t = 2h /g = 0.64 s

x = vt = 4.78 m

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•  No slipping! Friction is static; and does no work. •  There is a net force, so the center of mass is accelerated. •  If the center of mass is accelerated, there is also angular

acceleration.

comnet amF =

ατ comnet I=€

Mgsinθ − f s = Macom

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R fs = Icomα = Icomacom /R

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acom =gsinθ

1+ Icom /R2

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A constant horizontal force of magnitude 10 N is applied to a wheel of mass 10 kg and radius 0.30 m as shown in the figure. The wheel rolls smoothly on the horizontal surface, and the acceleration of its center of mass has magnitude 0.60 m/s2. (a) What are the magnitude and direction of the frictional force on the wheel? (b) What is the rotational inertia of the wheel about the rotation axis through its center of mass?

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A constant horizontal force of magnitude 10 N is applied to a wheel of mass 10 kg and radius 0.30 m as shown in the figure. The wheel rolls smoothly on the horizontal surface, and the acceleration of its center of mass has magnitude 0.60 m/s2.

(a) What are the magnitude and direction of the frictional force on the wheel?

(b) What is the rotational inertia of the wheel about the rotation axis through its center of mass?

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Fnet = ma⇒ F − fs = macom ⇒ f s = F −macom = 4N

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τ net = Iα ⇒ f sR = Iacom /R⇒ I = f sR2 /acom = 0.6kgm2