A Steamroller on a Slope (Moment of Inertia)

In summary, a steamroller with a total mass of 12000 kg on a 4° slope with 2 wheels, each with a mass of 3000 kg and a diameter of 1.2 m, would have an acceleration of 0.56 m/s^2 if its brakes were broken. This calculation takes into account the moment of inertia, torque, and angular acceleration of the steamroller, as well as the gravity forces and mass distribution on the wheels. However, the given answer of 0.56 m/s^2 may be slightly high due to differences in the value used for acceleration due to gravity.
  • #1
srecko97
82
13

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
 
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  • #2
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
srecko97 said:

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
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.
 

FAQ: A Steamroller on a Slope (Moment of Inertia)

1. What is the moment of inertia?

The moment of inertia is a physical property of a rigid body that represents its resistance to rotational motion. It is a measure of how spread out the mass of an object is from its axis of rotation.

2. How is the moment of inertia calculated?

The moment of inertia is calculated by multiplying the mass of an object by the square of its distance from the axis of rotation. The formula is I = mr², where I is the moment of inertia, m is the mass, and r is the distance from the axis of rotation.

3. What is the significance of a steamroller on a slope in relation to moment of inertia?

The steamroller on a slope is a classic example used to explain the concept of moment of inertia. As the steamroller moves down the slope, its moment of inertia changes due to the changing distance of its mass from the axis of rotation.

4. How does the shape of an object affect its moment of inertia?

The shape of an object has a significant impact on its moment of inertia. Objects with more mass concentrated near the axis of rotation have a lower moment of inertia compared to objects with the same mass spread out farther from the axis of rotation.

5. How is the moment of inertia related to rotational motion?

The moment of inertia is directly related to rotational motion. The greater the moment of inertia, the more force is required to cause an object to rotate. This is why objects with a larger moment of inertia, such as a spinning top, are more stable and harder to change their rotational motion.

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