:: force/math question ::

[3roomlab]

hi folks,

its a calculation question. if i have a solid metal cylinder of say 100kg mass, 200cm diameter. assuming friction free condition. and if i apply a force at the edge to turn it, if i need to (example) accelerate the cylinder from zero to 180degrees in 1 second (half a turn). how do i calculate the force required (kgm/s ?). an approximate math or an exact math tip if any input ... much appreciated.

must 1 make a calculation to assume approximated mass at the edge of block?  ... tyvm yall

[Mechatrommer]

your question need to revise back to the college book.
your question can be complex since involving rotational motion.
your question is contradicting itself. if there is no friction, the cylinder wont roll, it just goes straight linear motion if you mean "edge" as the side of the cylinder and pushing right toward the center of gravity.

maybe for a start... F = ma coupled with kinematics...
http://www.schoolphysics.co.uk/age16-19/Mechanics/Kinematics/text/Equations_of_motion_proof/index.html

[ejeffrey]

For rotational motion you need to talk about torque and moment of inertia.  Basically all the equations of linear motion apply to rotational with the following equivalences:  mass -> moment of inertia.  force -> torque, x -> theta, velocity -> rotation frequency, acceleration -> angular acceleration.

The moment of inertia of a solid cylinder is 0.5 * m * r^2.  So the moment of inertia of your cylinder is 50 kg * m^2.

You want a 180 degree rotation in 1 second, so 0.5 * a * t^2 = pi.  Set t = 1 s, and solve to get a = 2*pi radian/ s^2.

torque = moment of inertia * angular acceleration = 50*pi * kg * m^2 / (s^2) ~ 157 N*m.

If you apply a force of 150 N at a distance 1 m from the axis of rotation, that should do the trick.  Of course you need a bearing at the axis of rotation, or you need to apply half the force on opposite points of the cylinder to get it to rotate but not translate.  If you apply the force closer or farther from the axis of rotation you need more or less force, but the torque is the same: 157 N*m.

[westfw]

Do you want it to stop after 180 degrees as well?  If so, you need to apply force to decelerate it (opposite and equal to the force you applied to get it going.)

I'm pretty sure you get multiple solutions, depending on how much you want to "coast" between acceleration and deceleration.  Rather like spaceflight: all the SF spaceships accelerate at 1G for half the trip, then "flip" and decelerate at 1G for the 2nd half, conveniently providing "normal gravity" for the passengers.  But most of the real ships kick themselves in the right direction and then coast most of the distance.

[IanB]

hi folks,

its a calculation question. if i have a solid metal cylinder of say 100kg mass, 200cm diameter. assuming friction free condition. and if i apply a force at the edge to turn it, if i need to (example) accelerate the cylinder from zero to 180degrees in 1 second (half a turn). how do i calculate the force required (kgm/s ?). an approximate math or an exact math tip if any input ... much appreciated.

must 1 make a calculation to assume approximated mass at the edge of block?  ... tyvm yall

For this question to be reasonable, you must assume the cylinder to be pivoted on a frictionless axis running through the center. You must also assume the cylinder to be of uniform density. Given those assumptions, you can use moment of inertia as mentioned above.

[3roomlab]

 thanks for the inputs folks, yea i think my question is missing bits n pieces.

OK got it .. moment of inertia !

[IanB]

I think you are getting a bit confused about force, work, energy and equations of motion.

Let's see if you can understand motion in a straight line before looking at rotation.

Suppose you had a stationary mass of 100 kg. What force would you need to apply to move it over a distance of 1 metre in 1 second?

How much work would be done on the object and how much energy would be expended?

[IanB]

Hint: you would use the following two equations:

1. Equation of distance moved under acceleration:

    s = ut + ½at²

(where s = distance moved, u = initial velocity, a = acceleration and t = time)

2. Newton's second law of motion:

    F = ma

(where F = force applied, m = mass, a = acceleration)

You can later solve the rotational problem using analogs of these equations.

[IanB]

See my post above.

The equation is:

    s = ut + ½at²

Now the distance s = 1 m, the initial velocity u = 0 m/s, the time t = 1 s.

So:

    1 = ½a

therefore

    a = 2 m/s²

Next, use:

    F = ma

We have m = 100 kg and a = 2 m/s²

Therefore

    F = 100 x 2 = 200 N

Since the force is applied over a distance of 1 m, the work done is force x distance, or

    W = 200 x 1 = 200 J

The work done is the same as the energy expended since no energy is lost to friction or other causes.

[IanB]

hmmm ok so a = 2m/s/s
which makes F = 200N
and workdone 200J

yes ?

that 1/2at2 formula ... is so familiar ...

Yes, that's right.

[IanB]

Now the equations are going to work in a similar way for rotational motion.

Instead of

    s = ut + ½at²

we are going to have

    (theta) = (omega) t + ½ (alpha) t²

Where (theta) is angular position in radians, (omega) is angular velocity in rad/s, and (alpha) is angular acceleration in rad/s².

The equivalent of Newton's second law is going to be

    (tau) = I (alpha)

where (tau) is the torque in Nm, I is the moment of inertia in kg m² and (alpha) is the angular acceleration in rad/s²

(Unfortunately this forum can't do Greek letters)

Also the work done when a torque moves through a given angle is going to be from the similar equation

    W = (tau)(theta)

[3roomlab]

ahhhh so this is what i missed, i think i confused myself cos i totally forgot the base stuff

-- i will be back after playing with more newtons

[ejeffrey]

Also be careful with units.  Torque has units of newton-meter, but it is not a measure of energy.  Torque is the rotational equivalent of force.  With linear motion we have the formula work = force * distance.  For angular motion that is work = torque * angle.

[3roomlab]

okay, based on the rectangular rotating block

Moment of inertia (rectangular) I = 1/12 x m(X^2+Y^2)
Assume X=2, Y=3. m = 180kg. Therefore I = 1/12 x 100(13) = 150kg/m2

To move/accelerate the object 1 complete rotation in 1 sec on its CG pivot (assume lossless), from position of rest. Angular distance moved = 360 x pi / 180 = 2(pi)
Angular velocity = 2(pi)
Angular acceleration = A (unknown)
Solving for A, we get 2(pi) = 0.5 * A * 1
A = 4(pi) rad/s/s

Solving for “F=ma”
T = I * A = 600pi Nm

And so work done, energy expended
W = 600(pi) x 2(pi) = 1200(pi)^2 joules

so this 1200pi.pi joules being expended in 1 second, also expressable as 1200pi.pi watts? yes ? (=11.8kNm ... 11.8kJ)

[IanB]

Moment of inertia (rectangular) I = 1/12 x m(X^2+Y^2)
Assume X=2, Y=3. m = 180kg. Therefore I = 1/12 x 100(13) = 150kg/m2

Be careful with your arithmetic and with your units of measure.

You have

    I = 1/12 m (x² + y²)

where

    m = 180 kg, x = 2 m, y = 3 m.

So

    I = 1/12 x (180 kg) x [(2 m)² + (3 m)²]

therefore

    I = 1/12 x 180 kg x (4 m² + 9 m²) = 195 kg m²

Note: this is not "kg/m²", it is "kg x m²".

To move/accelerate the object 1 complete rotation in 1 sec on its CG pivot (assume lossless), from position of rest. Angular distance moved = 360 x pi / 180 = 2(pi)

It's best to forget the degrees, they are only confusing. Everything is radians in physics and there are 2 pi radians in a complete revolution. (If you wonder why there are two pies in a single turn, you may go look up some YouTube videos about why "pi is wrong"  )

[IanB]

yea. i was fascinated by the radians thingy for a abit, thats how i got into this mess of figuring out where radians meet newtons.

Radians happen because they are a natural way of expressing turning.

If you have a point on the edge of a disk 1 m in diameter and you turn the disk through 1 radian the point will move through a distance of 1 m.

So with radians you can convert radius to distance moved just by multiplying by the angle turned.

[Mechatrommer]

the fun part about pushing something in pure freefall vaccum not in the line of its CG is that it will also moves linearly as it rotates "angularly" and then you need to split your exerted force into linear force and torque at the CG in somesort of superposition method, but thats the easy part. then you have 2 branches of formulation for linear and angular, redefine what you want (the angular and linear motion), and backtrace it to get your required force. you want angular motion alone? dont exert a force, exert a pure "twist", then thats the pure torque.