Chapter13 rigid body rotation

13
11B: Rigid Body Rotation

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Transcript of Chapter13 rigid body rotation

Page 1: Chapter13 rigid body rotation

11B: Rigid Body Rotation

Page 2: Chapter13 rigid body rotation

Inertia of Rotation

Consider Newton’s first law for the inertia of rotation to be patterned after the law for translation.

Moment of Inertia or Rotational Inertia is the rotational analog of mass.

Units of I: kg-m2 ; g-cm2 ; slug-ft2

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Common Rotational Inertias

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Example 2: A circular hoop and a disk each have a mass of 3 kg and a radius of 30 cm. Compare their rotational inertias.

R

R

Disk

I = 0.3 kg m2

I = 0.1 kg m2

hoop

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Inertia of RotationConsider Newton’s second law for the inertia of rotation to be patterned after the law for translation.

ΣF = 20 N

a = 4 m/s2

Linear Inertia, m

F = 20 NR = 0.5 ma = 2 rad/s2

Rotational Inertia, I

Force does for translation while torque does for rotation

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Important Analogies

For many problems involving rotation, there is an analogy to be drawn from linear motion.

V

f

R

4 kg

wt wo = 50

rad/s t = 40 N m

A resultant force F produces negative acceleration a for a mass m.

I

A resultant torque tproduces angular acceleration a of disk with rotational inertia I.

Newton’s 2nd Law for Rotation

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Newton’s 2nd Law for Rotation

R

4 kg

wF wo = 50 rad/s

R = 0.20 m

F = 40 N

How many revolutions are required to stop?

How many revolutions are required to stop?

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A disk of mass and radius is used to draw water from a well (Fig. 1). A bucket of mass is attached to a cord that is wrapped around the disk.a. What is the tension T in the cord and acceleration a of the bucket?b. If the bucket starts from rest at the top of the well and falls for 3.00 s before hitting the water, how far does it fall? c. About how many revolutions are made by the disk in the situation stated in (b)?

Note: Fig. 1

Seatwork: ½ sheet of pad paper (crosswise)

Time allotted: 40 minutes

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Work and Power for Rotation

∆ 𝜃F

F

∆ 𝑠since

since

𝑃𝑟𝑜𝑡=𝜏 ∆ 𝜃∆ 𝑡

∆ 𝑠=𝑅 ∆𝜃

𝑅

𝑊=𝐹 ∆𝑠 ∆ 𝑠=𝑅 ∆𝜃

𝑊=𝐹𝑅∆𝜃 𝐹𝑅=𝜏

𝑊 𝑟𝑜𝑡=𝜏 ∆𝜃

since

❑❑

∆ 𝜃∆ 𝑡

=𝜔

𝑃𝑟𝑜𝑡=𝑊 𝑟𝑜𝑡

∆ 𝑡

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Example:

The rotating disk has a radius of 40 cm and a mass of 6 kg. Find the work and power needed in lifting the 2-kg mass 20 m above the ground in 4 s in uniform motion.

qF

F=W

s

s = 20 m

2 kg6 kg

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Rotational Kinetic Energy

m2

m3

m4

m

m1

axis

w

v = wR

Object rotating at constant .w

Consider tiny mass m:

since 𝑣=𝜔𝑅

To find the total kinetic energy:

(½w2 same for all m )

𝐾=12𝑚𝑣2

𝐾=12𝑚 (𝜔𝑅 )2

𝐾=12

(∑𝑚𝑅2 )𝜔2

∑𝑚𝑅2=𝐼since

𝑲 𝒓𝒐𝒕=𝟏𝟐𝑰 𝝎𝟐

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The Work-Energy Theorem

Recall for linear motion that the work done is equal to the change in linear kinetic energy:

Using angular analogies, we find the rotational work is equal to the change in rotational kinetic energy:

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Example:

Applying the Work-Energy Theorem:

R

4 kg

wF wo = 60 rad/s

R = 0.30 m

F = 40 N

What work is needed to stop wheel rotating in the figure below?