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A Steamroller on a Slope (Moment of Inertia)

  • Thread starter srecko97
  • Start date
  • #1
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Homework Statement


There is a steamroller with total mass 12000 kg on a 4° slope. It has 2 wheels (full homogeneous cylinders) with mass 3000 kg (each one 3000 kg) and diameter 1,2 m (radius 0,6 m). How big would the acceleration of the steamroller be if brakes got broken.

Result: 0.56 m/s^2

Homework Equations


J=1/2 mr^2

M=J*α

(J-moment of inertia, M-torque, α-angular acceleration)

The Attempt at a Solution


i assumed that the gravity forces of steamroller on the axis of wheels is equal (half-half). I tried to write torques on one of the wheels. 12000 kg (whole steamroller) - 3000 kg (wheel1) - 3000 kg(wheel2) = 6000 kg (driver, cabin, engine ....) --> 3000 kg on each wheel[/B]

image.jpg
 

Answers and Replies

  • #2
686
172
I had a hard time following your work. Maybe I just didn't take enough time to look at it closely.

But if I had a simple mass on a 4 degree frictionless slope, the acceleration of that mass would be much less than the answer you came up with for the steamroller. So there is definitely something wrong with your solution, since the acceleration of the steamroller should be less than a mass sliding on a frictionless surface.

If you haven't already, I recommend that you draw a free body diagram for the steam roller and sum the forces based on that FBD. Then you can sum torques on one of the rollers. This should result in 2 equations and 2 unknowns.
 
  • #3
haruspex
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Homework Statement


There is a steamroller with total mass 12000 kg on a 4° slope. It has 2 wheels (full homogeneous cylinders) with mass 3000 kg (each one 3000 kg) and diameter 1,2 m (radius 0,6 m). How big would the acceleration of the steamroller be if brakes got broken.

Result: 0.56 m/s^2

Homework Equations


J=1/2 mr^2

M=J*α

(J-moment of inertia, M-torque, α-angular acceleration)

The Attempt at a Solution


i assumed that the gravity forces of steamroller on the axis of wheels is equal (half-half). I tried to write torques on one of the wheels. 12000 kg (whole steamroller) - 3000 kg (wheel1) - 3000 kg(wheel2) = 6000 kg (driver, cabin, engine ....) --> 3000 kg on each wheel[/B]

image.jpg
Two errors.
  1. You have neglected the acceleration of the body of the steamroller.
  2. The instantaneous centre of rotation of the wheel is the point of contact with the road, not the wheel's centre. There are two ways to fix that. You can either use the parallel axis theorem or include the mass of the wheel in the linear acceleration of the body of the steamroller.
 
  • #4
2,169
572
The given answer seems a little high. I get the acceleration = 0.54728 m/s^2.
The difference could lie in the value used for g. I used g = 9.807 m/s^2.
 

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