Moment of inertia problem involving a cylinder rolling down an incline

In summary, a can of tomato sauce is accelerated by an impulse on the kitchen counter and the initial energy is conserved.
  • #1
as2528
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Homework Statement
A metal can containing condensed mushroom soup has a mass of 215 g, a height of 10.8 cm, and a diameter of 6.38 cm. It is placed at rest on its side at the top of a 3.00-m-long incline that is at an angle of 25.0° to the horizontal and is then released to roll straight down. Assuming energy conservation, calculate the moment of inertia of the can if it takes 1.50 s to reach the bottom
of the incline. Which pieces of data, if any, are unnecessary for calculating the solution?
Relevant Equations
vf=vi+at
K=1/2*I*w^2+1/2*m*v^2
PE=mgh
v=rw
a=2/3*g*sin(B)
a=2/3*g*sin(25*(pi/180))=>a=2.8507 m/s^2
vf=vi+at=>vf=0+2.8507*1.50=>vf=4.2760 m/s

So the translational motion of the cylinder is 4.2760 m/s.

4.2760=R*w
w=134.04 rad/s
PE=mgh=>PE=215*9.8*.108=>PE=227.56 J
PE = KE at the end of the roll because of energy conservation.

227.56 = 1/2*I*w^2+1/2*m*v^2
227.56=0.5*I*(134.04)^2+0.5*215*(4.276)^2
I=-0.1931 kg*m^2

The answer is 1.21 *10^4 kg*m2

How is this solved and why is my approach wrong? I think I did all the calculations correctly, so I must have gone wrong in applying the physics.
 
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  • #2
as2528 said:
Relevant Equations::

PE=mgh
When calculating the PE, careful with the units. The mass is given in grams.

as2528 said:
a=2/3*g*sin(B)
You shouldn't assume that the can of soup is a uniform solid cylinder. So, this formula might not apply. You are given enough information to find the acceleration without this assumption.

The answer is 1.21 *10^4 kg*m2
Looks like you dropped the negative sign for the power of 10.
 
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  • #3
As an aside, I should add that the conservation of energy in this situation is temperature-dependent. I used to do a lecture demonstration in which two cans of the same soup (chunky beef) raced down an incline. One was at room temperature, the other had spent the night in the freezer and was still frozen. Guess which can won the race.
 
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  • #4
kuruman said:
As an aside, I should add that the conservation of energy in this situation is temperature-dependent. I used to do a lecture demonstration in which two cans of the same soup (chunky beef) raced down an incline. One was at room temperature, the other had spent the night in the freezer and was still frozen. Guess which can won the race.
Try a similar demonstration with a raw egg versus a hard-boiled egg.
 
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  • #5
as2528 said:
PE=215*9.8*.108
In addition to the mass units error, where does the .108 come from? It is not 3sin(25°).
 
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  • #6
kuruman said:
As an aside, I should add that the conservation of energy in this situation is temperature-dependent. I used to do a lecture demonstration in which two cans of the same soup (chunky beef) raced down an incline. One was at room temperature, the other had spent the night in the freezer and was still frozen. Guess which can won the race.
Is that to do with conservation of energy or just the fact that the contents are not rotating as fast?
 
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  • #7
haruspex said:
In addition to the mass units error, where does the .108 come from? It is not 3sin(25°).
The .108 was supposed to be the height of the incline.
 
  • #8
as2528 said:
The .108 was supposed to be the height of the incline.
3m at 25°, only 11cm? Looks like you used 2°.
 
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  • #9
haruspex said:
3m at 25°, only 11cm? Looks like you used 2°.
No I used it from the question. It said that the height was 10.8 cm. I converted that to meters and reasoned that since gravity is conservative I could use .108 as the height of the soup can.
 
  • #10
as2528 said:
It said that the height was 10.8 cm.
That is the height of the can as a cylinder, or length if that's clearer, not the height at which the can is placed.
 
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  • #11
haruspex said:
That is the height of the can as a cylinder, or length if that's clearer, not the height at which the can is placed.
Oh! I did not realize that. Thanks!
 
  • #12
haruspex said:
Is that to do with conservation of energy or just the fact that the contents are not rotating as fast?
The latter. I believe that the energy goes into viscous losses between the solid tidbits in the can and the broth plus the coupling between the fluid inside and the walls.

On edit:
Here is a homemade video of a can of tomato sauce given an impulse on my kitchen counter. What happens to the initial energy?

 
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FAQ: Moment of inertia problem involving a cylinder rolling down an incline

What is the moment of inertia of a solid cylinder?

The moment of inertia of a solid cylinder about its central axis is given by the formula \( I = \frac{1}{2} m r^2 \), where \( m \) is the mass of the cylinder and \( r \) is its radius.

How do you calculate the acceleration of a cylinder rolling down an incline?

The acceleration \( a \) of a solid cylinder rolling down an incline without slipping can be calculated using the formula \( a = \frac{2}{3} g \sin(\theta) \), where \( g \) is the acceleration due to gravity and \( \theta \) is the angle of the incline.

What forces act on a cylinder rolling down an incline?

The forces acting on a cylinder rolling down an incline include gravitational force, normal force, and frictional force. The gravitational force can be decomposed into components parallel and perpendicular to the incline. The frictional force is what prevents slipping and allows the cylinder to roll.

How does the angle of the incline affect the rolling motion of the cylinder?

The angle of the incline \( \theta \) affects the component of gravitational force acting parallel to the incline, which in turn affects the acceleration of the cylinder. A steeper incline (larger \( \theta \)) results in a greater component of gravitational force parallel to the incline, leading to greater acceleration.

What is the condition for rolling without slipping for a cylinder on an incline?

The condition for rolling without slipping is that the linear acceleration \( a \) of the center of mass must be equal to the angular acceleration \( \alpha \) times the radius \( r \) of the cylinder, i.e., \( a = r \alpha \). This ensures that the point of contact with the incline has zero relative velocity.

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