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Debora [2.8K]
4 years ago
15

Commercially-available hybrid vehicles, such as the Toyota Prius, use electrical batteries to store energy for later use. Howeve

r, Formula One racecars have used a flywheel to store energy. When the car applies its brakes, translational kinetic energy of the car is converted to rotational kinetic energy of a large wheel; later, when the driver wants extra power, the wheel can be coupled to the drivetrain of the car, using that kinetic energy to supplement the power from the engine. The “Flybrid” hybrid system uses a 5 kg wheel with a radius of 12 cm. When this system is engaged, it can provide an extra 60 kW of power to the car for 7 seconds.(a) How much energy does this system store?(b) At what angular velocity must the wheel spin to store this much energy? Convert your answer to revolutions per minute. Does this result surprise you?(c) What is the centripetal acceleration of a point on the edge of this wheel? Can you think of any engineering or safety challenges in building these systems?
Physics
1 answer:
Bad White [126]4 years ago
7 0

(A) 4.2\cdot 10^5 J

The energy stored by the system is given by

E=Pt

where

P is the power provided

t is the time elapsed

In this case, we have

P = 60 kW = 60,000 W is the power

t = 7 is the time

Therefore, the energy stored by the system is

E=(60,000 W)(7 s)=4.2\cdot 10^5 J

(B) 4830 rad/s

The rotational energy of the wheel is given by

E=\frac{1}{2}I \omega^2 (1)

where

I is the moment of inertia

\omega is the angular velocity

The moment of inertia of the wheel is

I=\frac{1}{2}MR^2=\frac{1}{2}(5 kg)(0.12 m)^2=0.036 kg m^2

where M is the mass and R the radius of the wheel.

We also know that the energy provided is

E=4.2\cdot 10^5 J

So we can rearrange eq.(1) to find the angular velocity:

\omega=\sqrt{\frac{2E}{I}}=\sqrt{\frac{2(4.2\cdot 10^5 J)}{0.036 kg m^2}}=4830 rad/s

(C) 2.8\cdot 10^6 m/s^2

The centripetal acceleration of a point on the edge is given by

a=\omega^2 R

where

\omega=4830 rad/s is the angular velocity

R = 0.12 m is the radius of the wheel

Substituting, we find

a=(4830 rad/s)^2 (0.12 m)=2.8\cdot 10^6 m/s^2

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