Hi, I want to asked a question about how to calculate the energy needed to powered a rotary motor to move a car. (assume that this motor will rotate the wheel of the car)

What I do is I use the formula of kinetic energy which is [ 1/2 x Mass x Velocity(square) ]to calculate it.

Let say a car with mass 2000kg travelling at a velocity of 50m/s . After calculation,
[1/2 x 2000 x 50(square)] the answer appear to be 25000j /s

I do not know whether it is correct to just linked the kinetic energy fomula to this if I am wrong please provide me with the correct method to calculate the energy needed to powered the motor to move car in this instant.

What you want is a unit of power. The unit of power that's appropriate for automobiles is horsepower.

1 J/s = 1 watt, There are 746 W/hp

So, you would need an approximate 33 HP motor without frictional losses. This should be the amount of energy to keep it moving.

Power conversion online.

If you look at this link 2007 Toyota Prius Hybrid Review - Edmunds.com

it gives an idea of the electric motor and gasoline motor that the Toyota Prius develops.

Typical 4 cylinder gasoline engines might develop about 115 HP.

Your term Energy is like referring to gallons of gasoline. It depends on your driving habits how much energy it takes to get from point a to point b. Watts is like fuel consumption like in gal/s.

To accelerate to a given speed in a certain amount of time will require a greater peak HP.

Quote:

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Originally Posted by KeepItSimpleStupid

What you want is a unit of power. The unit of power that's appropriate for automobiles is horsepower.

1 J/s = 1 watt, There are 746 W/hp

So, you would need an approximate 33 HP motor without frictional losses. This should be the amount of energy to keep it moving.

Typical 4 cylinder gasoline engines might develop about 115 HP.

Your term Energy is like referring to gallons of gasoline. It depends on your driving habits how much energy it takes to get from point a to point b. Watts is like fuel consumption like in gal/s.

To accelerate to a given speed in a certain amount of time will require a greater peak HP.

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So you are saying that the way I calculate the power for the rotary motor is correct using kinetic energy fomula?

(btw just to inform u that I am saying motor not engine I know car is power by engine just imagine it's a tamiya car motor)

Except that there are lots of loses. Somethat I can think of are drag, coeficient of rolling friction and wind speed and direction. Drag is going to take a major bite. Drag might be a 20-25% loss. The wind blowing against you at 15 mph or behind you influences the amount of HP required as well. You may also need to check the HP required to climb a hilll and at what incline.

Torque is going to influence how fast you can accelerate. So with units of ft-lbs/sec, you can convert that to HP too. Say you wanted to go 0-60 mph in 10 sec. You can find the distance traveled in 10 s. You also know the weight of the car and the time.

When I bought my last car, I wanted it to be about the same lb/HP of automobile as a judge of performance. The ratio was a little higher than my old car. Acceleration does depend on gear ratio and with overdrive off, I think it's comperable. Stick shifts give better performace. The aautomobile design combined with tires determines how far you can push it. I think the number was about 14 lb/hp.

Quote:

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Originally Posted by KeepItSimpleStupid

Except that there are lots of loses. Somethat I can think of are drag, coeficient of rolling friction and wind speed and direction. Drag is going to take a major bite. Drag might be a 20-25% loss. The wind blowing against you at 15 mph or behind you influences the amount of HP required as well. You may also need to check the HP required to climb a hilll and at what incline.

Torque is going to influence how fast you can accelerate. So with units of ft-lbs/sec, you can convert that to HP too. Say you wanted to go 0-60 mph in 10 sec. You can find the distance traveled in 10 s. You also know the weight of the car and the time.

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Ok so basically u are saying 1/2 x M x v^2 is correct except that I must still involve some calculation that will affect the speed right.

However in the first post u mentioned that 1 J/s = 1 watt and there are 746 W/hp but when I go to the volvo car specification website :Volvo V50 Specifications
to convert and calculate the maximum speed that the car(T5 AWD R-Design) can travel using kinetic formula and I found that the car can only travel maximum at speed of approximately 52 km/hr. and how can it be possible?

The car has 227 hp, mass 3570 lbs which is approximately 1620kg.

So 227 hp x 734.7 = 169273.9 watt =169273.9 j/s

And 169273.9 j/s = 1/2 x 1620(M) x Velocity^2
=810 x v^2

So v^2 = 169273.9/810 = approximately 208.98
Then v will be equals to root of 208.98 which is 14.456

14.456 is in meter/s so to convert into kilometer/hr we multiply it by 3.6
so the answer is approximately 52km/hr

How can a volvo car only travel up to 52 km/hr max?

I see what you mean. I guess it doesn't make sense. KE, I guess is used primarily for energy in collisions. You can convert to any unit. You can convert to Watts if that's a unit you relate too or you can convert to callories, but a colliding car is worth consuming x number of Pizza's may not make any sense.

Top speed may have more to do with engine limitations and gear ratios than anything else.

The energy involved in moving a car is going to be derived from gasoline and the efficiency of the engine and power train. At some point mechanical limitations rear their ugly head.

Moving an object may be dependent on torque. Think of a screw tightened to a given torque. The bolt may not even move until you exceed the tightened torque.

You raise an interesting point as to what would be the HP to move an object and what HP is required to maintain a particular speed. Also what factors determine how fast an object can be accelerated.

I definitely messed up.

Try posting the same question in the physics forum here https://www.askmehelpdesk.com/physics/ and see if you can get better results.