hi,need help please.thank.
A car is designed to get its energy from a rotating flywheel (solid disk) with a radius of 2.00 m and a mass of 525 kg. Before a trip, the flywheel is attached to an electric motor, which brings the flywheel's rotational speed up to 4000 rev/min.
(a) Find the kinetic energy stored in the flywheel.
1 J
(b) If the flywheel is to supply energy to the car as would a 20.0-hp motor, find the length of time the car could run before the flywheel would have to be brought back up to speed.
2 h
physics - drwls, Sunday, November 22, 2009 at 6:58pm
Are those your answers? 1 J and 2 h? How dd you obtain them?
a) Use the formula E = (1/2) I w^2, where w is the angular speed in rad/s and I is the moment of inertia.
(b) Time = (Full speed energy)/Power
You will have to convert horsepower to Watts when using the formula.
I'm sorry I did try to pluck in the number but, it not right could you help me please in part a and b. Greatly appriciate.
I already did. Note spelling of the subject
Sure, I can help you with parts (a) and (b) of the problem.
(a) To find the kinetic energy stored in the flywheel, we need to use the formula:
E = (1/2) * I * w^2
where E is the kinetic energy, I is the moment of inertia, and w is the angular speed in radians per second.
Given that the radius of the flywheel is 2.00 m, we can find the moment of inertia using the formula:
I = (1/2) * m * r^2
where m is the mass of the flywheel.
Substituting the given values, we have:
I = (1/2) * 525 kg * (2.00 m)^2
I = 1050 kg * m^2
Now, we need to convert the rotational speed from revolutions per minute (rev/min) to radians per second (rad/s). We know that 1 revolution is equal to 2π radians. So, we can calculate the angular speed as follows:
w = (4000 rev/min) * (2π rad/rev) * (1 min/60 s)
w = 418.88 rad/s
Finally, substituting the values into the formula for kinetic energy, we get:
E = (1/2) * (1050 kg * m^2) * (418.88 rad/s)^2
E = 115,957,001.6 J
Therefore, the kinetic energy stored in the flywheel is approximately 115,957,001.6 Joules.
(b) To find the length of time the car could run before the flywheel would have to be brought back up to speed, we can use the formula:
Time = (Full speed energy) / Power
Given that the car needs energy equivalent to a 20.0-horsepower (hp) motor, we need to convert horsepower to watts. 1 horsepower is equal to 746 watts.
So, the power required for the car is:
Power = 20.0 hp * 746 W/hp
Power = 14,920 W
Now, substituting the values into the formula for time, we have:
Time = (115,957,001.6 J) / (14,920 W)
Time = 7,766.90 s
Since there are 3600 seconds in an hour, the time in hours is:
Time (hours) = 7,766.90 s / 3600 s/h
Time (hours) = 2.16 h
Therefore, the car could run for approximately 2.16 hours before the flywheel would have to be brought back up to speed.
I hope this helps! Let me know if you have any further questions.