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blsea [12.9K]
3 years ago
10

Find the equation of the line below.

Physics
1 answer:
Musya8 [376]3 years ago
7 0

Answers all in picture below

:

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A cylinder with rotational inertia I1=2.0kg·m2 rotates clockwise about a vertical axis through its center with angular speed ω1=
Mrac [35]

Answer:

<em>a) 0.67 rad/sec in the clockwise direction.</em>

<em>b) 98.8% of the kinetic energy is lost.</em>

Explanation:

Let us take clockwise angular speed as +ve

For first cylinder

rotational inertia I = 2.0 kg-m^2

angular speed ω = +5.0 rad/s

For second cylinder

rotational inertia I = 1.0 kg-m^2

angular speed = -8.0 rad/s

The rotational momentum of a rotating body is given as = Iω

where I is the rotational inertia

ω is the angular speed

The rotational momenta of the cylinders are:

for first cylinder = Iω = 2.0 x 5.0 = 10 kg-m^2 rad/s

for second cylinder = Iω = 1.0 x (-8.0) = -8 kg-m^2 rad/s

The total initial angular momentum of this system cylinders before they were coupled together = 10 + (-8) = <em>2 kg-m^2 rad/s</em>

When they are coupled coupled together, their total rotational inertia I_{t} = 1.0 + 2.0 = 3 kg-m^2

Their final angular rotational momentum after coupling = I_{t}w_{f}

where I_{t} is their total rotational inertia

w_{f} = their final angular speed together

Final angular momentum = 3 x w_{f} = 3w_{f}

According to the conservation of angular momentum, the initial rotational momentum must be equal to the final rotational momentum

this means that

2 =  3w_{f}

w_{f} = final total angular speed of the coupled cylinders = 2/3 = <em>0.67 rad/s</em>

From the first statement, <em>the direction is clockwise</em>

b) Rotational kinetic energy = \frac{1}{2} Iw^{2}

where I is the rotational inertia

w is the angular speed

The kinetic energy of the cylinders are:

for first cylinder = \frac{1}{2} Iw^{2} = \frac{1}{2}*2*5^{2} = 25 J

for second cylinder = \frac{1}{2}*1*8^{2} = 32 J

Total initial energy of the system = 25 + 32 = 57 J

The final kinetic energy of the cylinders after coupling = \frac{1}{2}I_{t}w^{2} _{f}

where

where I_{t} is the total rotational inertia of the cylinders

w_{f} is final total angular speed of the coupled cylinders

Final kinetic energy =  \frac{1}{2}*3*0.67^{2} = 0.67 J

kinetic energy lost = 57 - 0.67 = 56.33 J

percentage = 56.33/57 x 100% = <em>98.8%</em>

6 0
3 years ago
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