Problem 31 Find the moments of inertia about the x & y axes: m = 1.8 kg, M = 3.1 kg.

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Transcript of Problem 31 Find the moments of inertia about the x & y axes: m = 1.8 kg, M = 3.1 kg.

Page 1: Problem 31 Find the moments of inertia about the x & y axes: m = 1.8 kg, M = 3.1 kg.
Page 2: Problem 31 Find the moments of inertia about the x & y axes: m = 1.8 kg, M = 3.1 kg.

Problem 31Find the moments of inertia about the x & y axes:

m = 1.8 kg, M = 3.1 kg

Page 3: Problem 31 Find the moments of inertia about the x & y axes: m = 1.8 kg, M = 3.1 kg.

Problem 40m each blade = 160 kg. Moment of inertia I? Starts from

rest, torque τ to get ω = 5 rev/s (10π rad/s) in t = 8 s?

Page 4: Problem 31 Find the moments of inertia about the x & y axes: m = 1.8 kg, M = 3.1 kg.

Example 8-11

M = 4 kg, FT = 15 N

Frictional torque: τfr = 1.1 m N

t = 0, ω0 = 0; t = 3 s, ω = 30 rad/s

I = ?

Calculate the angular acceleration:

ω = ω0+ αt, α = (ω - ω0)/t

= 10 rad/s2

N’s 2nd Law: ∑τ = Iα

FTR - τfr = Iα

I = [(15)(0.33) -1.1]/10

I = 0.385 kg m2

Page 5: Problem 31 Find the moments of inertia about the x & y axes: m = 1.8 kg, M = 3.1 kg.

The same pulley, connected to

a bucket of weight mg = 15 N

(m = 1.53 kg). M = 4 kg

I = 0.385 kg m2; τfr = 1.1 m N

a) α = ? (pulley)

a = ? (bucket)

b) t = 0, at rest.

t = 3 s, ω = ? (pulley)

v = ? (bucket)

a

Example 8-12

Page 6: Problem 31 Find the moments of inertia about the x & y axes: m = 1.8 kg, M = 3.1 kg.

Translation-Rotation Analogues & Connections

Translation Rotation

Displacement x θ

Velocity v ω

Acceleration a α

Force (Torque) F τ

Mass (moment of inertia) m I

Newton’s 2nd Law: ∑F = ma ∑τ = Iα

Kinetic Energy (KE) (½)mv2 ?

CONNECTIONS

v = rω, atan= rα, aR = (v2/r) = ω2 r

τ = rF, I = ∑(mr2)

Page 7: Problem 31 Find the moments of inertia about the x & y axes: m = 1.8 kg, M = 3.1 kg.

Section 8-7: Rotational Kinetic Energy

• Translational motion (Ch. 6): (KE)trans = (½)mv2

• Rigid body rotation, angular velocity ω. Rigid

Every point has the same ω. Body is made of particles, masses m.

• For each m at a distance r from the rotation axis:

v = rω. The Rotational KE is:

(KE)rot = ∑[(½)mv2] = (½)∑(mr2ω2) = (½)∑(mr2)ω2

ω2 goes outside the sum, since it’s the same everywhere in the body

– As we just saw, the moment of inertia, I ∑(mr2)

(KE)rot = (½)Iω2

(Analogous to (½)mv2)

Page 8: Problem 31 Find the moments of inertia about the x & y axes: m = 1.8 kg, M = 3.1 kg.

Translation-Rotation Analogues & Connections

Translation Rotation

Displacement x θ

Velocity v ω

Acceleration a α

Force (Torque) F τ

Mass (moment of inertia) m I

Newton’s 2nd Law ∑F = ma ∑τ = Iα

Kinetic Energy (KE) (½)mv2 (½)Iω2

CONNECTIONS

v = rω, atan= rα, aR = (v2/r) = ω2 r

τ = rF, I = ∑(mr2)

Page 9: Problem 31 Find the moments of inertia about the x & y axes: m = 1.8 kg, M = 3.1 kg.

Sect. 8-7: Rotational + Translational KE

• Rigid body rotation: (KE)rot = (½)Iω2 • Now, consider a rigid body, mass M, rotating (angular velocity ω)

about an axis through the CM. At the same time, the CM is translating with velocity vCM

– Example, a wheel rolling without friction. For this, we saw earlier that vCM = rω.

The KE now has 2 parts: (KE)trans & (KE)rot

Total KE = translational KE + rotational KE

KE = (KE)trans + (KE)rot

or KE = (½)M(vCM)2 + (½)ICMω2 where: ICM = Moment of inertia about an axis through the CM

Page 10: Problem 31 Find the moments of inertia about the x & y axes: m = 1.8 kg, M = 3.1 kg.

Example 8-13

v = ?

y = 0

A sphere rolls down an

incline (no slipping or sliding).

KE+PE conservation:(½)Mv2 + (½)Iω2 + MgH = constant,

or

(KE)1 +(PE)1 = (KE)2 + (PE)2

where KE has 2 parts!!

(KE)trans = (½)Mv2

(KE)rot = (½)Iω2

v = 0, ω = 0

Page 11: Problem 31 Find the moments of inertia about the x & y axes: m = 1.8 kg, M = 3.1 kg.

Conceptual Example 8-14: Who Wins the Race?

MgH = (½)Mv2 + (½)ICMω2

Gravitational PE is Converted to Translational + Rotational KE!

Hoop: ICM = MR2 Cylinder: ICM = (½)MR2

Sphere: ICM = (2/5)MR2 (also, v = ωR)

v increases as I decreases! Demonstration!

Page 12: Problem 31 Find the moments of inertia about the x & y axes: m = 1.8 kg, M = 3.1 kg.

Friction: Necessary for

objects to roll without

slipping. Example: work

done by friction hasn’t

been included (used

KE +PE =const). WHY?

Because Ffr Motion

Ffr does no work!