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harkovskaia [24]
3 years ago
10

A solid disk of mass m1 = 9.3 kg and radius R = 0.22 m is rotating with a constant angular velocity of ω = 31 rad/s. A thin rect

angular rod with mass m2 = 3.9 kg and length L = 2R = 0.44 m begins at rest above the disk and is dropped on the disk where it begins to spin with the disk.
1) What is the initial angular momentum of the rod and disk system?
6.97 kg-m2/s
2) What is the initial rotational energy of the rod and disk system?
108.14 J

3) What is the final angular velocity of the disk?

4) What is the final angular momentum of the rod and disk system?

5) What is the final rotational energy of the rod and disk system?

6) The rod took t = 6.5 s to accelerate to its final angular speed with the disk. What average torque was exerted on the rod by the disk?
Physics
1 answer:
mel-nik [20]3 years ago
5 0
We are given with
<span>m1 = 9.3 kg
R = 0.22 m
ω = 31 rad/s
m2 = 3.9 kg
L = 2R = 0.44 m

The angular momentum is determined using the formula
L  = I w
where I is the moment of inertia

For a solid disk
I = (1/4) MR^2
For a rectangular rod
I = (1/12) ML^2

The rotational energy has the formula
E = (1/2) I w^2

Substitute the given values for getting the answers
</span>
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The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.

<h3>What is Potential and Kinetic energy?</h3>

Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.

mass of rod, mab = 2.4kg

mass of rod, mbc = 4kg

conservation of energy

T_{1}  + V_{1} = T_{2}  + V_{2}

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kinetic energy T1 at position 1 is zero

potential energy at position 2 is zero

K.E at position 2,

T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} +  \frac{1}{2} m_{bc} v^{2}_{G} +  \frac{1}{2} lw^{2}_{bc}

l_{ab} =\frac{m_{ab} l^{2}_{ab}  }{3}

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l =\frac{m_{bc} l^{2}_{bc}  }{12}

= 1/12 *4 * (0.6)²

=0.12kg m²

on putting the values in above equation we get,

T₂ = 1.0667vb²

0 + 11.30112 = 1.0667vb² + 0

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