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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? Explicitly show how you follow the steps in the Problem-Solving Strategy for Rotational Energy.

Sagot :

Answer:

Explanation:

moment of inertia of flywheel = 1/2 m R²

= .5  x 1500 x .6²

= 270 kg m²

If required angular velocity be ω

rotational kinetic energy = 1/2 I ω²

= .5 x 270 x ω² = 135 ω²

kinetic energy of bus when its velocity is 20 m/s

= 1/2 x 10000 x 20²

= 2000000 J

Given 90 % of rotational kinetic energy is converted into bus's kinetic energy

135 ω² x 0.9 = 2000000 J

ω²=16461

ω = 128.3 radian /s

b )

Let the height required be h .

Total energy of bus at the top of hill = mgh + 1/2 m v²

m ( gh + .5 v²)

= 10000 ( 9.8h + .5 x 3²)

From conservation of mechanical energy theorem

10000 ( 9.8h + .5 x 3²) = 2000000

9.8h + .5 x 3² = 200

9.8h  = 195.5

h = 19.95 m .

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