Torsion Involving an Off Axis Thrust

In summary: Yes, in order to calculate the total mass including the counterweight, you need to know the force applied by the car. However, in order to calculate the mass including the counterweight, you need to know the force applied by the counterweight.
  • #36
Devin-M said:
Correct me if I’m wrong
Look carefully at the dimensions in the FBD. The portion of the vertical rod that is above the COM is not shown because it has no mass, and therefore no effect on the solution. The wheels are massless, so the FBD has exactly two masses, both of which are shown as point masses:
1) The COM of the vertical rod.
and
2) The added mass out in front.

Since this problem is defined as translation only, with zero rotation, all masses are shown at their COM. If you were working on a different problem involving angular acceleration, you would need to show distributed masses as distributed masses. The vertical rod would be one such mass.

The secret to solving these types of problems is to put most of your effort into getting the FBD correct, and without any extraneous detail. The best FBD has only the information needed to solve the problem.

If you want to better understand FBD's, solve this one several different ways:
1) By sum of moments about point P, the point of application of the force from the car.
and
2) By sum of moments about the COM of the vertical rod.
and
3) By sum of moments about the mass M.
and
4) By sum of moments about the center of the wheels.

Devin-M said:
According to this page,
Looking for an equation online will, all too often, lead you to a wrong answer because there is no guarantee that the online equation is correct for your FBD. If the FBD is correct, then the correct equation is easily found.
 
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  • #37
jrmichler said:
The portion of the vertical rod that is above the COM is not shown because it has no mass, and therefore no effect on the solution.
That portion weighs 50kg-- the 2m uniform rod is 100kg, so the half above that point is 50kg.
 
  • #38
Only the horizontal rod has negligible mass.
 
  • #39
OOPS, my bad. Good catch. But, because this problem is translation only, the vertical rod is shown as a 100 kg point mass.
 
  • #40
jrmichler said:
OOPS, my bad. Good catch. But, because this problem is translation only, the vertical rod is shown as a 100 kg point mass.
The rod in the problem is not a point mass, so depending where the force is applied (below or above the center of mass), there is either a backwards tilting or forwards tilting torque from the thrust from the car (in this case a backwards tilting torque from the thrust from the car, because the thrust is applied 0.5m below the center of mass of the vertical rod).
 
  • #41
Devin-M said:
The rod in the problem is not a point mass, so depending where the force is applied (below or above the center of mass), there is either a backwards tilting or forwards tilting torque from the thrust from the car (in this case a backwards tilting torque from the thrust from the car, because the thrust is applied 0.5m below the center of mass of the vertical rod).
I haven't been following your thread lately, but my impression all along was that you want to do the calculation for the situation where the vertical rod does not rotate, right? So there are no MOI considerations...
 
  • #42
berkeman said:
I haven't been following your thread lately, but my impression all along was that you want to do the calculation for the situation where the vertical rod does not rotate, right? So there are no MOI considerations...
The problem is that without the counterweight, the main rod would tend to tilt backwards (top towards car) as it is accelerated from the thrust from the car below the rod's center of mass (horizontal thrust applied below center of mass of main 100kg uniform rod). The counterweight is added to counter this torque resulting in net 0 torque composed of 2 equal torques in opposite directions, and an acceleration from the force from the car in the horizontal direction.
 
  • #43
I keep thinking that it will all end in a tangled mess. As the car accelerates steadily things are OK, but when the car becomes torque limited at high RPM, the counterweight will fall forwards, and the car will run over the contraption, hence, the tangled mess.
 
  • #44
equation.jpg


I believe I was able to solve counterweight mass KG for a desired acceleration...

A = Acceleration m/s^2 = 1
B = Counterweight Mass kg =
C = Car Force Distance From COM meters = 0.5
D = Main Rod Mass kg = 100
G = Acceleration of Gravity m/s^2 = 9.8

B = (D*C^2*A+D^2*C*A)/(D*G-C^2*A-D*C*A)

B = (100*0.5^2*1+100^2*0.5*1)/(100*9.8-0.5^2*1-100*0.5*1)

5.4kg = (100*0.5^2*1+100^2*0.5*1)/(100*9.8-0.5^2*1-100*0.5*1)

^5.4kg counterweight mass is needed for 1m/s^2 acceleration
 
  • #45
What happens when you set the required acceleration to 10.0 ?
 
  • #46
Please disregard the previous formula I posted-- it's in error.

I went back to specifying the counterweight mass (without knowing acceleration) and discovered there is a certain point when you keep increasing the counterweight mass the acceleration starts going back down... for example 300kg counterweight has lower acceleration than 100kg counterweight.
 
  • #47
90kg Counterweight:
Center of Mass
((90*0)+(100*1))/(90+100) = 0.5263m
^Center of mass is 0.5263m from Counterweight
Torque from counterweight:
90kg*9.8m/s^2*0.5263m = 464.19Nm torque from counterweight
Force from car:
464.19Nm / 0.5m = 928.38 N force from car
Acceleration from force & mass:
928.38N / (90kg + 100kg) = 4.8862m/s^2 acceleration

100kg Counterweight:
Center of Mass
((100*0)+(100*1))/(100+100) = 0.5m
^Center of mass is 0.5m from Counterweight
Torque from counterweight:
100kg*9.8m/s^2*0.5m = 490Nm torque from counterweight
Force from car:
490Nm / 0.5m = 980 N force from car
Acceleration from force & mass:
980N / (100kg + 100kg) = 4.9m/s^2 acceleration

110kg Counterweight:
Center of Mass
((110*0)+(100*1))/(110+100) = 0.476m
^Center of mass is 0.476m from Counterweight
Torque from counterweight:
110kg*9.8m/s^2*0.476m = 513.128Nm torque from counterweight
Force from car:
513.128Nm / 0.5m = 1026.25 N force from car
Acceleration from force & mass:
1026.25N / (110kg + 100kg) = 4.88m/s^2 acceleration

300kg Counterweight:
Center of Mass
((300*0)+(100*1))/(300+100) = 0.25m
^Center of mass is 0.25m from Counterweight
Torque from counterweight:
300kg*9.8m/s^2*0.25m = 735Nm torque from counterweight
Force from car:
735Nm / 0.5m = 1470 N force from car
Acceleration from force & mass:
1470N / (300kg + 100kg) = 3.67m/s^2 acceleration
 
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  • #48
It looks like acceleration peaks at 1/2 of Earth's gravitational acceleration, when the counterweight mass equals the main rod mass, then acceleration decreases with additional counterweight mass... which implies you can't calculate the counterweight mass from the desired acceleration because there's more than 1 solution, and the peak horizontal acceleration is 1/2 of Earth's 9.8m/s^2 vertical gravitational acceleration.
 
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  • #49
If there were no counterweight and stable acceleration were achieved by setting an initial forward lean angle, would there be any further acceleration to be gained by using a counterweight in addition to the forward lean angle?
IMG_0192.jpeg
 
  • #50
Devin-M said:
If there were no counterweight and stable acceleration were achieved by setting an initial forward lean angle, would there be any further acceleration to be gained by using a counterweight in addition to the forward lean angle?
The reason I ask is some electric skateboard riders showed me they are getting around 1g acceleration from leaning alone…

IMG_0195.jpeg

IMG_0196.png

IMG_0197.png

IMG_0204.png
 
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  • #51
Devin-M said:
The reason I ask is some electric skateboard riders showed me they are getting around 1g acceleration from leaning alone…
You can get arbitrarily high acceleration by leaning alone. Take the arc tangent of the horizontal acceleration in g's. The acceleration in g's is also the required coefficient of friction.

A drag racer can pull 5 g's. The arc tangent of 5 is around 79 degrees. If you measured the angle from the contact patch under the rear tires to the center of gravity, it would need to be under 12 degrees from the horizontal. [Ignoring aerodynamic effects]

You say that a skater can pull over 1 g. The arc tangent of 1 is 45 degrees. So the lean angle will be over 45 degrees from the vertical. You measure that angle from the contact patch under the skateboard to the position of the center of gravity of the skater plus counterweight.

If a motorcycle can corner at 2 g's then the lean angle will be 64 degrees from the vertical. You measure that angle from the contact patch to the center of gravity of rider plus bike. [Ignoring gyroscopic effects]
 
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  • #52
Thank you that is a very informative answer.
 
  • #53
I suggested they hold the battery forward— no weight penalty.
IMG_0317.jpeg

(AI generated)
 
  • #54
Devin-M said:
electric skateboard riders showed me they are getting around 1g acceleration from leaning alone…
I assume the forward lean (weight destitution) controls the electric motor thrust? Your phrasing is really weird. It's like saying: Fighter pilots get 9g acceleration from pulling on a stick alone.
 

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