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A bus contains a 1500 kg flywheel (a disk that has a 0.600 m radius) and has a total mass of 10,000 kg. (a) Calculate the angular velocity the flywheel must have to contain enough energy to take the bus from rest to a speed of 20.0 m/s, assuming 90.0% of the rotational kinetic energy can be transformed into translational energy. (b) How high a hill can the bus climb with this stored energy and still have a speed of 3.00 m/s at the top of the hill?

Respuesta :

Answer:

[tex]\omega = 128.300 rad/sec[/tex]

H = 19.92 m

Explanation:

Given data:

weight of flywheel 1500 kg

disk radius is 0.6 m

total mass of bus is 10,000 kg

Kinetci energy of wheel is given as

[tex]K.E = \frac{1}{2} I\omega^2[/tex]

where

I denotes moment of inertia [tex]= \frac{Mr^2}{2}[/tex]

[tex]\omega[/tex] is angular velocity

from information given in question 90% of K.E( flywheel) = K.E of bus

K.E of bus [tex]= \frac{1}{2} m v^2[/tex]

[tex]0.9 \frac{1}{2} I\omega^2 = \frac{1}{2} m v^2[/tex]

substituing all value in above equation

[tex]I = = \frac{Mr^2}{2}[/tex]

HERE   M = 1500 kg and r = 0 , so

[tex]0.9\frac{1}{2} \frac{1500\times 0.6^2}{2} \times \omega^2 = \frac{1}{2} 10,000 \times 20^2[/tex]

solving for angular velocity

[tex]\omega = 128.300 rad/sec[/tex]

part b

by energy conservation theorem

[tex]E_i =E_f[/tex]

0.90 Kr = Ke +  Ue

[tex]0.9( \frac{1}{2} I\omega^2) = \frac{1}{2} m v^2[ + MgH[/tex]

[tex]0.9\frac{1}{2} \frac{1500\times 0.6^2}{2} \times 128.3^2 = \frac{1}{2} 10,000 \times 3^2 + 10000 \times 9.81\times H[/tex]

solving for H

H = 19.92 m

Lanuel

The angular velocity that the flywheel must have to contain enough energy to take the bus from rest is equal to 128.30 rad/s.

Given the following data:

Weight of flywheel = 1500 kg.

Radius of disk = 0.6 m.

Total mass of bus = 10,000 kg.

Speed = 20.0 m/s.

Percent of KE = 90.0% = 0.9.

How to calculate the angular velocity.

Mathematically, the rotational kinetic energy is given by this formula:

[tex]K.E_{rot}=\frac{1}{2} I \omega^2\\\\K.E_{rot}=\frac{1}{2} (\frac{mr^2}{2} ) \omega^2\\\\\frac{0.9mr^2}{4} \omega^2=\frac{1}{2} Mv^2\\\\\omega =\sqrt{\frac{4Mv^2}{1.8mr^2}} \\\\\omega =\sqrt{\frac{4 \times 10000 \times 20^2}{1.8 \times 1500 \times 0.6^2}} \\\\\omega =\sqrt{\frac{16000000}{972}}\\\\\omega =128.30\;rad/s[/tex]

How to calculate the height.

In order to determine the height of the hill, we would apply the law of conservation of energy:

[tex]E_i=E_f\\\\\frac{0.9mr^2}{4} \omega^2=\frac{1}{2} Mv^2+Mgh\\\\0.9mr^2 \omega^2=2Mv^2+Mgh\\\\Mgh=0.9mr^2 \omega^2-2Mv^2\\\\h=\frac{0.9mr^2 \omega^2-2Mv^2}{Mg} \\\\h=\frac{0.9\times 1500 \times 0.6^2 \times 128.30^2-2\times 10000 \times 20.0^2}{10000 \times 9.8}[/tex]

Height, h = 19.92 meters.

Read more on kinetic energy here: brainly.com/question/1242059

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