Final velocity involving a can of soup and an inclined plane

In summary, the problem involves a can rolling down a 3.00 meter board at a 25 degree angle with a given acceleration of 4.1417 m/s^2. The original answer of 4.9850 m/s for final velocity is incorrect because it assumes sliding without friction instead of rolling without slipping. With the additional information of 1.50 seconds for the time it takes to roll down the board, the correct answer can be calculated using the average velocity and starting velocity. This highlights the difference between sliding and rolling, where in the rolling scenario, some of the gravitational energy is converted into rotational energy, resulting in a slower descent down the board.
  • #1
as2528
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Homework Statement
A can of soup travels down an inclined plane which is 3.00 m long and makes an angle of 25 degrees with the horizontal. It travels this distance in 1.50 seconds. What is the final velocity, assuming the can of soup started at rest, a perfect roll is occurring, and friction is negligible?
Relevant Equations
(vf+vi)/2=vavg
vf^2=vi^2+2*a*s
a=gsin(theta)
a = 9.8*sin(25*pi/180)=>a=4.1417 m/s^2

vf^2=vi^2+2*a*s=>vf=sqrt(0^2+2*4.1417*3)=>vf=4.9850 m/s

Meanwhile the correct answer is:
(vf+vi)/2=>(vf+0)/2=2=>vf=4 m/s

Why is my answer wrong? It seems that the acceleration is what is wrong, but I don't understand why.
 
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  • #2
This looks a lot like a recent thread that I can no longer locate. However, in that thread the problem specified that it took 1.50 seconds from top of ramp to bottom. This additional fact would explain the 4 m/s final velocity.

Could you please quote the problem as given?
 
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  • #3
as2528 said:
the acceleration is what is wrong, but I don't understand why.
They don’t mean friction is negligible. It rolls without sliding. It's the rolling resistance that’s negligible.
 
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  • #5
jbriggs444 said:
This looks a lot like a recent thread that I can no longer locate. However, in that thread the problem specified that it took 1.50 seconds from top of ramp to bottom. This additional fact would explain the 4 m/s final velocity.

Could you please quote the problem as given?
Yes the 1.50 seconds was also part of the problem. I accidentally omitted the sentence. I have updated the problem.
 
  • #6
as2528 said:
Yes the 1.50 seconds was also part of the problem. I accidentally omitted the sentence. I have updated the problem.
With the 1.50 second figure in hand, we know the average velocity. 3 meters in 1.50 seconds is 2 meters per second. With the average velocity and the starting velocity in hand, we can solve for the final velocity.

Your original post already shows that calculation: "(vf+vi)/2=>(vf+0)/2=2=>vf=4 m/s"

Your competing calculation started by calculating the acceleration for a puck sliding without friction down a 25 degree slope: "a = 9.8*sin(25*pi/180)=>a=4.1417 m/s^2"

That part matches what I get for such a puck. But, as @haruspex points out, the problem is assuming rolling without slipping. The calculation you used assumes sliding without friction.

Under the sliding interpretation, the board length of 3.00 meters, the board angle of 25 degrees and the resulting acceleration of 4.1417 m/s^2 mean that the elapsed time cannot be 1.50 seconds. That is how we can know that the sliding interpretation is wrong unintended. [Well, that and the fact that the problem statement says "perfect roll is occurring"].

Under the rolling interpretation, the moment of inertia of the can is an unknown parameter. We have enough information to solve for it.
 
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  • #7
jbriggs444 said:
With the 1.50 second figure in hand, we know the average velocity. 3 meters in 1.50 seconds is 2 meters per second. With the average velocity and the starting velocity in hand, we can solve for the final velocity.

Your original post already shows that calculation: "(vf+vi)/2=>(vf+0)/2=2=>vf=4 m/s"

Your competing calculation started by calculating the acceleration for a puck sliding without friction down a 25 degree slope: "a = 9.8*sin(25*pi/180)=>a=4.1417 m/s^2"

That part matches what I get for such a puck. But, as @haruspex points out, the problem is assuming rolling without slipping. The calculation you used assumes sliding without friction.

Under the sliding interpretation, the board length of 3.00 meters, the board angle of 25 degrees and the resulting acceleration of 4.1417 m/s^2 mean that the elapsed time cannot be 1.50 seconds. That is how we can know that the sliding interpretation is wrong unintended. [Well, that and the fact that the problem statement says "perfect roll is occurring"].

Under the rolling interpretation, the moment of inertia of the can is an unknown parameter. We have enough information to solve for it.
Got it! Thanks I didn't realize the formula I was using was assuming slipping. Guess I learned a new thing about it!
 
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  • #8
as2528 said:
Got it! Thanks I didn't realize the formula I was using was assuming slipping. Guess I learned a new thing about it!
To be totally clear, in the "slipping" scenario, all of the gravitational potential energy is going into increasing the kinetic energy and hence the velocity of the slipping object. The object slides down at the expected speed.

By contrast, in the "rolling" scenario, some of the gravitational energy goes into rotational energy. There is less available for plain old kinetic energy and hence, linear velocity. The object rolls down more slowly than it would have slid.

This can happen because of the force of static friction between the rolling object and the slope upon which it is rolling. The slope exerts a retarding force on the rolling object, slowing its descent. At the same time, that retarding force acts as a torque around the object's center of mass, increasing its rotation rate.
 
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  • #9
jbriggs444 said:
To be totally clear, in the "slipping" scenario, all of the gravitation potential energy is going into increasing the kinetic energy and hence the velocity of the slipping object. The object slides down at the expected speed.

By contrast, in the "rolling" scenario, some of the gravitational energy goes into rotational energy. There is less available for plain old kinetic energy and hence, linear velocity. The object rolls down more slowly than it would have slid.

This can happen because of the force of static friction between the rolling object and the slope upon which it is rolling. The slope exerts a retarding force on the rolling object, slowing its descent. At the same time, that retarding force acts as a torque aroujnd the object's rotational axis, increasing its rotation rate.
Oh, I see! Thanks for clearing that up! While I knew about static friction, I didn't know about rotational energy leaching gravitational energy. That perfectly explains why my acceleration was higher using the sliding formula. Thanks!
 
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  • #10
This is all assuming that the liquid in the can follows the motion of the can as a rigid body. If viscosity is neglected, the fluid will not rotate with the can. The only moment of inertia will be given by the can's walls. In reality, some (or all ) fluid will rotate but not necessary with all parts having the same angular velocity. But if there is viscosity, conservation of energy is not really justified. The author of the problem should have picked a rigid object as a subject of the problem.
 
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  • #11
nasu said:
This is all assuming that the liquid in the can follows the motion of the can as a rigid body. If viscosity is neglected, the fluid will not rotate with the can. The only moment of inertia will be given by the can's walls. In reality, some (or all ) fluid will rotate but not necessary with all parts having the same angular velocity. But if there is viscosity, conservation of energy is not really justified. The author of the problem should have picked a rigid object as a subject of the problem.
No, I believe it is intentional that the contents do not necessarily rotate as a solid. You only have to assume constant acceleration to solve the problem. Whether that would be roughly true, even for a Newtonian fluid, I am unsure.
 
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  • #12
Then what's the point of mentioning "perfect roll"? You are right, assuming constant acceleration of the CofM produces their result. This does not make it a meaningful problem
 
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  • #13
The other problem states its a can of condensed cream of mushroom soup. It has the consistency of paste. Flow is probably negligible.
 
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  • #14
nasu said:
Then what's the point of mentioning "perfect roll"?
To my mind "perfect roll" means nothing more than "rolling without slipping". Or, perhaps, "rolling without slipping and without rolling resistance".

Based on the level at which the problem appeared to be aimed and the lack of details on whether we were dealing with vegetable beef (liquid) or condensed cream of mushroom soup (effectively solid) the best guess at the intention seems to be cream of mushroom. [As @erobz confirmed from the other post]

If @haruspex is unsure of the behavior of a can with semi-liquid contents then a hapless first year student stands no chance whatsoever!
 
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  • #15
Apparently there is some interesting behaviors that arise depending on what is in the can, and how much. Fast forwarding to 24 minutes in the video we can see what happens with rice!

 
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Related to Final velocity involving a can of soup and an inclined plane

What is the final velocity of a can of soup rolling down an inclined plane?

The final velocity of a can of soup rolling down an inclined plane depends on factors such as the angle of the incline, the height from which it starts, and the friction between the can and the plane. Using principles of energy conservation or kinematics, you can calculate the final velocity. For a frictionless incline, it can be determined using the equation \( v_f = \sqrt{2gh} \), where \( g \) is the acceleration due to gravity and \( h \) is the height of the incline.

How does the angle of the inclined plane affect the final velocity of the can of soup?

The angle of the inclined plane influences the component of gravitational force acting along the plane. A steeper angle results in a greater component of gravitational force in the direction of motion, leading to a higher acceleration and, consequently, a higher final velocity. The relationship can be described by the equation \( v_f = \sqrt{2gL\sin(\theta)} \), where \( L \) is the length of the incline and \( \theta \) is the angle.

Does friction play a significant role in determining the final velocity of the can of soup?

Yes, friction plays a significant role. If there is significant friction between the can and the inclined plane, it will oppose the motion, reducing the acceleration and therefore the final velocity of the can. The final velocity can be calculated by accounting for the work done against friction, which modifies the energy conservation equation to \( v_f = \sqrt{2gh - 2\mu gL\cos(\theta)} \), where \( \mu \) is the coefficient of friction.

Can we use kinematic equations to find the final velocity of the can of soup?

Yes, kinematic equations can be used to find the final velocity, especially when the motion is uniformly accelerated. For an inclined plane, the acceleration \( a \) can be expressed as \( g\sin(\theta) \). Using the kinematic equation \( v_f^2 = v_i^2 + 2aL \), where \( v_i \) is the initial velocity (which is zero if starting from rest), \( a \) is the acceleration, and \( L \) is the length of the incline, you can solve for the final velocity \( v_f \).

How would the final velocity change if the can of soup was replaced with a different object?

The final velocity would change if the object has different properties, such as mass distribution or shape, which affect rolling resistance and friction. For objects with different moments of inertia, the distribution of mass affects how they accelerate down the incline. For example, a solid cylinder and a

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