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Linear and angular acceleration.

  1. Sep 14, 2007 #1
    Hey.

    I have been working on a relative simple 2D physics engine for a game I am making, and as I was researching the subject (rigid body mechanics) I found this article on physics in games. This pretty much answers everything. Though this is one little thing that I cannot figure out. The article says to calculate linear and angular acceleration totally independent of each other but based on the same forces. So the linear acceleration of a body would be all the forces affecting the body summed and divided by the mass of the body. If this is true then the same forces (identical direction and magnitude) applied on the same body but at different points would always cause the same linear acceleration but not necessary the same angular acceleration. How is this possible? I would think that the more angular acceleration the less linear acceleration and vice versa. How else is the total energy level preserved?
    I am pretty sure it is just me who has misunderstood something; and an explanation would be very appreciated.
     
  2. jcsd
  3. Sep 14, 2007 #2

    Doc Al

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    The article is correct. The linear acceleration of the center of mass just depends on the net force on the object, not on where the force is applied. The angular acceleration about the center of mass depends on where the force is applied. (Both statements are just consequences of Newton's 2nd law.)

    Realize that the work you do on an object is force times the distance that the contact point moves. When you push the object with an off-center force the contact point moves more (compared to an equal on-center force), thus it takes more work to maintain the force--that extra work goes into the rotational energy.
     
  4. Sep 14, 2007 #3
    Thanks, you are totally right. I just needed to have my understanding of forces refreshed.
     
  5. Sep 15, 2007 #4
    After thinking a bit more about it, I must admit that there is something I still have not understood.
    If a rocket engine burning fuel at a constant rate is attached to a spaceship, does it not apply the same force on the spaceship regardless of its position?
     
    Last edited: Sep 15, 2007
  6. Sep 15, 2007 #5

    rcgldr

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    Assuming the space ship is in space (no drag), the force is the same regardless of it's velocity. What happens is burnt fuel is ejected from the rocket ship at very high velocity, accelerating the rocket ship and the remaining unburnt fuel. The force is equal to the average mass and it's average rate of acceleration at any instant. The power relative to the spaceship, is equal to the rate of increase of kinetic energy (increase in KE / time) of the burnt fuel.

    However if the frame of reference has the same velocity as the space ships initial velocity, then the force (thrust) remains constant while the rocket's speed increases, so the power is also increasing. One way to accept this is to realize that the burnt fuel increases the kinetic energy of the space ship and the remaining unburnt fuel, effectively increasing the kinetic energy and therefore the total energy of the remaining fuel, relative to the initial velocity based frame of reference. Also as fuel is burnt, the remaining mass of the rocket and unburnt fuel decreases over time. If the ratio of fuel to mass of rocket is high enough, the rocket can go faster than the velocity of the burnt fuel.
     
  7. Sep 15, 2007 #6
    I am sorry but it is still not clear to me.

    If a spaceship has to identical rocket engines, attached at different angles to the center of mass. So firing one engine will produce more angular acceleration than firing the other.
    Then would the engine producing less angular acceleration not produce more linear acceleration?

    But if the linear acceleration is found from the net force, then the engine producing more angular acceleration should apply less force.
    But then that means that the two identical engines does not apply the same force on the spaceship. Which does not make much sense to me.

    So what am I getting wrong?
     
  8. Sep 15, 2007 #7

    rcgldr

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    Say you have a small sphere that weighs 1 kg and is 2 meters in diameter.

    First case: The sphere is on a treadmill that begins to accelerate at the rate required to generate 1 Newton of force at the edge of the sphere, but the sphere is glued to the treadmill and can't roll. The sphere accelerates at the rate of 1 m / s^2, as well as the tread mill, in order to maintain 1 Newton of force on the sphere.

    Second case: The sphere is on a treadmill that begins to accelerate at the rate required to generate 1 Newton of force at the edge of the sphere, the sphere is free to roll, but there is no slippage. Obviously the sphere will offer less "resistance" to the treadmill and therefore the treadmill will have to accelerate faster than the first case in order to maintain 1 Newton of force at the edge of the sphere. Still as posted before, the linear inertial reaction force of the sphere will equal the 1 Newton of force applied by the treadmill. This means the sphere will be accelerating linearly at 1 m / s^2, however the treadmill will be accelerating at a faster rate, since the sphere is also free to roll.

    The sphere has a linear and a rotational kinetic energy:

    Angular moment of sphere = 2/5 m r^2.
    Linear moment of sphere = m.
    Let w = rate of rotation of sphere = angular velocity:

    linear KE = KEl = 1/2 m v^2

    angular KE = KEa = (1/2 x 2/5 m r^2 x w^2)
    linear speed at edge of sphere, v = w x r => w = v/r

    Let c = ratio between acceleration of sphere / acceleration of plane
    Since time is the same for both, c is also the ratio of velocity of sphere vs plane,
    and distance traveled by sphere vs plane.

    as = c x ap => ap = as/c
    vs = c x vp => vp = vs/c
    ds = c x dp => dp = ds/c
    w = (vp-vs)/r = (vs/c-vs)/r = (1/r)((1 - c)/c)vs

    Let fs = force on sphere:

    fs = m x as = m x dvs/dt (linear)

    Torque on sphere is then:

    fs x r = 2/5 m r^2 x o = 2/5 m r^2 (1/r)((1 - c)/c)dvs/dt

    divide torque equation by r:

    fs = (1/r)(2/5 m r^2 (1/r)((1 - c)/c)dvs/dt) = 2/5 m ((1-c)/c)dvs/dt

    Include linear equation for fs:

    fs = m dvs/dt = 2/5 m ((1-c)/c)dvs/dt

    Divide both sides by m dvs/dt:

    1 = (2/5) ((1-c)/c)
    c = (2/5) (1-c)
    c = 2/5 - (2/5) c
    (1+2/5) c = 2/5
    7/5 c = 2/5
    c = 5/7 x 2/5 = 2/7

    So the linear acceleration of the sphere is only 2/7th's the acceleration of the treadmill, and the rotational acceleration is equal to the (acceleration of the treadmill - acceleration of the sphere) / radius of the sphere.

    The treadmill has to accelerate at 7/2 m/s^2 in order to generate 1 Newton of force at the bottom edge of a rolling 1kg sphere (note radius doesn't matter), 3.5 times as fast as compared to the first case with the glued sphere.

    Using the change in energy to check this result:

    What is the kinetic energy after the treadmill has accelerated for 1 second?
    Distance = 1/2 (7/2 m/s^2) s^2 = 7/4 m.
    Work done = 7/4 N m = 7/4 joules.
    Speed of treadmill after 1 second = 1s x (7/2 m/s^2) = 7/2 m/s

    Linear speed of sphere = 2/7 speed of tread mill = 2/7 x (7/2 m/s^2) x 1 s = 1 m/s
    Linear speed of sphere related to rotation = speed of tread mill - speed of sphere = (7/2 - 1)m/s = 5/2 m/s

    Kinetic energy of sphere:

    KE = 1/2 kg (1 m/s)^2 + 1/2 kg 2/5 x (5/2 m/s)^2 = (1/2 + 5/4) kg (m/s)^2
    KE = 7/4 kg (m/s)^2
    N = kg m/(s^2) => kg = N s^2/m
    KE = 7/4 N (s^2/m) (m/s)^2 = 7/4 N m = 7/4 joules

    Since force is constant and the velocity of the treadmill increases linearly, then power (= force x speed) also increases linearly. In the first case, power = t watts, in the second case, 7/2 t watts where t is elapsed time/seconds. Average power over time t is 1/2 the power at t. After 1 second, average power in the first case would be 1/2 watt, corresponding to work done and an increase in KE of the glue sphere of 1/2 joules, and in the second case, it's 7/4 watt, correspoding to and increase of KE of the sphere of 7/4 joules. After 2 seconds, work done and increase in KE in the first case 2 joules, in the second case, 7 joules.
     
    Last edited: Sep 15, 2007
  9. Sep 15, 2007 #8

    rcgldr

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    If the engine is attached to the rocket, it's a different case, the direction of force from the engine will change if the rocket rotates. If the direction of force is not through the rockets center of mass, the rocket will follow a spiral path.
     
  10. Sep 16, 2007 #9
    Thanks for the long explanation, I'm still going through to make sure I get everything.
    But let me try asking in another way.

    If a rocket engine is burning fuel at a constant rate, will it always produce the same force? - eg. 200 newton.

    If the above is true then how can the below be true:

    Case 1:
    A rocket engine is placed on a spaceship with the direction of its force going through the center of the spaceships mass.

    Case 2:
    Same as case 1 one, but the rocket engine is moved to the force is parallel with the force in case 1, but not through the spaceships center of mass.

    In case 1 and 2 the netforces are identical, but in case 2 tourqe is also produced by the rocket engine.
    So how is the total energy preserved?

    I am sorry if you already answered this and I just cannot see it.
     
  11. Nov 30, 2008 #10

    rcgldr

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    I just linked to this thread in another post and realised I never answered the last post here:

    yes
    In both cases, the work done equals force times linear distance plus torque times rotation. In the second case, because the rocket rotates, the same force is applied over a longer distance, since the distance involved includes a linear and angular component, so more work is done and the rate of total energy increase of the rocket is higher in the second case.

    It may seem that the same rate of fuel burn is creating more power in the second case, but that is because the energy of the spent fuel ejected from the rocket engine was being ignored. In space, a rocket and it's fuel are a closed system. Both linear and angular momentum are preserved. As the rocket's linear and angular momentum changes, the spent fuel from the rocket engine experiences an equal an opposite momentum change. The chemical potential energy of the fuel is converted into kinetic energy of rocket and spent fuel, both linear and angular (plus heat). The total energy, potential energy, kinetic energy (and losses due to heat) remain constant.
     
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