Accelerating and object in a straight line vs in a circle

[d_n_t]

More of a physics question I guess....

I'm trying to verify the difference in force that is required to move the same mass rotationally vs in just a straight line.

A real world example would be adding 40lbs to a car vs adding 10lbs to each car wheel (distributed evenly).

In the case where you just add the weight, you just have the force required to move that additional weight, but in the case of adding it to the wheel, you can the force needed to accelerate not only the additional weight, but the force required to spin the wheel to move the weight (or so it seems to me).

What I have so far looks like this...

Formula for the amount of torque it would take to rotate a given mass:

ΣT = Iα

The summation though for my purposes can go to just T as I don't need to worry about friction acting against the object (say a car wheel with the tread causing friction against the air as it spins or the friction of the axle itself, etc.). Given that, the formula goes to:

T = Iα

Now, for the formula for accelerating the mass in a straight line (like if a 40lb weight was added to the trunk) I have:
F=ma

Here is what I'm coming up with in a real world example...is this correct? If not, where I am going wrong please?

Diameter of tire and wheel: 28.7" -> .729 meters
Weight of tire and wheel: 49.8lbs -> 22.589 kg
Starting revolutions per min: 0
Ending revolutions per min: 704
Time to get to ending RPM: 4 seconds

Looking up the moment of inertia, I decided it best to go with a solid disc. I know that's not exact, but the best approximation I think of a tire on a wheel. That formula then is: 1/2MR².
So, in this case that goes to -> .5*22.589*(.729/2 * .729/2) = 1.501

To calculate the work force I calculate the starting and ending velocity in radians per second:
The ending velocity of 704RPM * 2π rads per rev / 60 seconds = 23.467π rads/second
The starting velocity of 0RPM = 0

To calculate α in π rads/s², I then take (23.467π rads/second - 0π rads/second) / 4 seconds = 5.867π rads/s²

Then I take the moment of inertia and α and multiply that by π to get the force in newton meters.
So, 1.501 * 5.867 * 3.14159265 = 27.655

Is that correct?

Assuming my calculations are correct (I know I rounded BTW) then I should be able to compare that force (27.655NM) to the force it would normally take to move that mass in a straight line correct?

So, F=ma....

22.589kg * acceleration.

To find acceleration I need to find the ending RPM (704) - the starting RPM (0) * the distance each revolution makes and divide by 60 seconds and then by 4 seconds (the total time to accelerate)

So, (704 - 0) * 2πr (.729 meters/2) / 60 = 6.718 m/s²

Does that look right so far?

I then plug that in and get that the force in NM is....

22.589 * 6.718 = 151.748NM

So to move the object takes 151.748NM, and it takes an additional 27.655NM to spin it (as in the case of a wheel).

Does that sound right?

Thanks!

 

[Deleted member 4993]

More of a physics question I guess....

I'm trying to verify the difference in force that is required to move the same mass rotationally vs in just a straight line.

A real world example would be adding 40lbs to a car vs adding 10lbs to each car wheel (distributed evenly).

In the case where you just add the weight, you just have the force required to move that additional weight, but in the case of adding it to the wheel, you can the force needed to accelerate not only the additional weight, but the force required to spin the wheel to move the weight (or so it seems to me).

What I have so far looks like this...

Formula for the amount of torque it would take to rotate a given mass:

ΣT = Iα

The summation though for my purposes can go to just T as I don't need to worry about friction acting against the object (say a car wheel with the tread causing friction against the air as it spins or the friction of the axle itself, etc.). Given that, the formula goes to:

T = Iα

Now, for the formula for accelerating the mass in a straight line (like if a 40lb weight was added to the trunk) I have:
F=ma

Here is what I'm coming up with in a real world example...is this correct? If not, where I am going wrong please?

Diameter of tire and wheel: 28.7" -> .729 meters
Weight of tire and wheel: 49.8lbs -> 22.589 kg
Starting revolutions per min: 0
Ending revolutions per min: 704
Time to get to ending RPM: 4 seconds

Looking up the moment of inertia, I decided it best to go with a solid disc. I know that's not exact, but the best approximation I think of a tire on a wheel. That formula then is: 1/2MR².
So, in this case that goes to -> .5*22.589*(.729/2 * .729/2) = 1.501

To calculate the work force I calculate the starting and ending velocity in radians per second:
The ending velocity of 704RPM * 2π rads per rev / 60 seconds = 23.467π rads/second
The starting velocity of 0RPM = 0

To calculate α in π rads/s², I then take (23.467π rads/second - 0π rads/second) / 4 seconds = 5.867π rads/s²

Then I take the moment of inertia and α and multiply that by π to get the force in newton meters.
So, 1.501 * 5.867 * 3.14159265 = 27.655

Is that correct?

Assuming my calculations are correct (I know I rounded BTW) then I should be able to compare that force (27.655NM) to the force it would normally take to move that mass in a straight line correct?

So, F=ma....

22.589kg * acceleration.

To find acceleration I need to find the ending RPM (704) - the starting RPM (0) * the distance each revolution makes and divide by 60 seconds and then by 4 seconds (the total time to accelerate)

So, (704 - 0) * 2πr (.729 meters/2) / 60 = 6.718 m/s²

Does that look right so far?

I then plug that in and get that the force in NM is....

22.589 * 6.718 = 151.748NM

So to move the object takes 151.748NM, and it takes an additional 27.655NM to spin it (as in the case of a wheel).

Does that sound right?

Thanks!

You have many problems here.

In your problem, the unit of "Force" is N (= kg*m/sec^2) and unit Torque is T (= N*m). Although the unit of work and unit of Torque are same - those (work and torque) are different animals.
You say -
"To find acceleration I need to find the ending RPM (704) - the starting RPM (0) * the distance each revolution makes and divide by 60 seconds and then by 4 seconds (the total time to accelerate)"

I think you you are trying to calculate the angular acceleration and you need to be careful about units again.

What are you trying to calculate - finally? Please post the exact problem - verbatim.

 

[topsquark]

More of a physics question I guess....

I'm trying to verify the difference in force that is required to move the same mass rotationally vs in just a straight line.

A real world example would be adding 40lbs to a car vs adding 10lbs to each car wheel (distributed evenly).

In the case where you just add the weight, you just have the force required to move that additional weight, but in the case of adding it to the wheel, you can the force needed to accelerate not only the additional weight, but the force required to spin the wheel to move the weight (or so it seems to me).

What I have so far looks like this...

Formula for the amount of torque it would take to rotate a given mass:

ΣT = Iα

The summation though for my purposes can go to just T as I don't need to worry about friction acting against the object (say a car wheel with the tread causing friction against the air as it spins or the friction of the axle itself, etc.). Given that, the formula goes to:

T = Iα

Now, for the formula for accelerating the mass in a straight line (like if a 40lb weight was added to the trunk) I have:
F=ma

Here is what I'm coming up with in a real world example...is this correct? If not, where I am going wrong please?

Diameter of tire and wheel: 28.7" -> .729 meters
Weight of tire and wheel: 49.8lbs -> 22.589 kg
Starting revolutions per min: 0
Ending revolutions per min: 704
Time to get to ending RPM: 4 seconds

Looking up the moment of inertia, I decided it best to go with a solid disc. I know that's not exact, but the best approximation I think of a tire on a wheel. That formula then is: 1/2MR².
So, in this case that goes to -> .5*22.589*(.729/2 * .729/2) = 1.501

To calculate the work force I calculate the starting and ending velocity in radians per second:
The ending velocity of 704RPM * 2π rads per rev / 60 seconds = 23.467π rads/second
The starting velocity of 0RPM = 0

To calculate α in π rads/s², I then take (23.467π rads/second - 0π rads/second) / 4 seconds = 5.867π rads/s²

Then I take the moment of inertia and α and multiply that by π to get the force in newton meters.
So, 1.501 * 5.867 * 3.14159265 = 27.655

Is that correct?

Assuming my calculations are correct (I know I rounded BTW) then I should be able to compare that force (27.655NM) to the force it would normally take to move that mass in a straight line correct?

So, F=ma....

22.589kg * acceleration.

To find acceleration I need to find the ending RPM (704) - the starting RPM (0) * the distance each revolution makes and divide by 60 seconds and then by 4 seconds (the total time to accelerate)

So, (704 - 0) * 2πr (.729 meters/2) / 60 = 6.718 m/s²

Does that look right so far?

I then plug that in and get that the force in NM is....

22.589 * 6.718 = 151.748NM

So to move the object takes 151.748NM, and it takes an additional 27.655NM to spin it (as in the case of a wheel).

Does that sound right?

Thanks!

Torque is just going to confuse you. Look up "centripetal force."

-Dan

 

[d_n_t]

You have many problems here.

In your problem, the unit of "Force" is N (= kg*m/sec^2) and unit Torque is T (= N*m). Although the unit of work and unit of Torque are same - those (work and torque) are different animals.
You say -
"To find acceleration I need to find the ending RPM (704) - the starting RPM (0) * the distance each revolution makes and divide by 60 seconds and then by 4 seconds (the total time to accelerate)"

I think you you are trying to calculate the angular acceleration and you need to be careful about units again.

What are you trying to calculate - finally? Please post the exact problem - verbatim.

So, to accelerate a disc (a car wheel for example) from point A to point B, you have more than centripetal force to overcome correct?

You have the force required to spin the disc (torque) and the force needed to overcome the inertia.

I THINK I calculated each force required correctly.

I suppose the question then is, how are these two combined to come up with total force needed to be applied to spin the disc so that it accelerates to 704RPM in 4 seconds with the given mass and diameter as the disc moves along a plane from point A to point B?

Again, it's not just one force or the other, but the two combined (ignoring friction, etc.) correct?

To put it into a real world example, how much force needs to be applied to a wheel that is .729 meters in diameter, weighs 22.589kg, to accelerate it from 0 to 704RPM and do it in four seconds.

Note that the wheel will start at one point on the road, and end up some distance away as, while torque is being applied to the wheel, it is rolling on the road.