Finding Height (KE vs. Work by Gravity)

[johnhuntsman]

A thin solid disk of radius R = 0.5 m and mass M = 2.0 kg is rolling without slipping on a horizontal surface with a linear speed v = 5.0 m/s. The disk now rolls without slipping up an inclined plane that is at an angle of 60 degrees to the vertical. Calculate the maximum height that the disk rolls up the incline.

The disk will stop when kinetic energy K equals the work done by gravity W_g.

R = 0.5m
M = 2.0kg
v = 5.0m / s
θ = 60°
I = (1 / 2)MR²
ω = v / R
K = (1 / 2)Iω² + (1 / 2)Mv²
W_g = mgh = mgdcosθ

The fact the θ to the vertical is given suggests that I'm meant to find h using it, but it seems to me that I wouldn't actually need it. However solving for h that way doesn't get me the right answer which makes me think I need to use it somehow. Can someone please help?

P.S. The correct answer is 2.6m.

 

[PhanthomJay]

A thin solid disk of radius R = 0.5 m and mass M = 2.0 kg is rolling without slipping on a horizontal surface with a linear speed v = 5.0 m/s. The disk now rolls without slipping up an inclined plane that is at an angle of 60 degrees to the vertical. Calculate the maximum height that the disk rolls up the incline.

The disk will stop when kinetic energy K equals the work done by gravity W_g.

R = 0.5m
M = 2.0kg
v = 5.0m / s
θ = 60°
I = (1 / 2)MR²
ω = v / R
K = (1 / 2)Iω² + (1 / 2)Mv²
W_g = mgh = mgdcosθ

The fact the θ to the vertical is given suggests that I'm meant to find h using it, but it seems to me that I wouldn't actually need it. However solving for h that way doesn't get me the right answer which makes me think I need to use it somehow. Can someone please help?

P.S. The correct answer is 2.6m.

You are asked to solve for h. Is the work done by gravity positive or negative?? Write out your work-energy equation and please show your work.

 

[johnhuntsman]

It should be negative.

(1 / 2)Iω² + (1 / 2)Mv² - Mgh = 0J

or

(1 / 2)Iω² + (1 / 2)Mv² = Mgh

or

(1 / 2)Iω² + (1 / 2)Mv² = Mgdcosθ

[Edit]
With stuff substituted in:
(1 / 2)(1 / 2)MR²(v / R)² + (1 / 2)Mv² = Mgdcosθ

Simplifying:
(1 / 4)MR²(v / R)² + (1 / 2)Mv² = Mgdcosθ

(1 / 4)R²(v / R)² + (1 / 2)v² = gdcosθ

(1 / 4)v² + (1 / 2)v² = gdcosθ

(3 / 4)v² = gdcosθ

[(3 / 4)v²] / g = dcosθ

What then?
[Edit]

 

[PhanthomJay]

That will give you the correct answer for h, and since you are not asked to solve for d, you don't need the last equation. But, you are nonetheless not using your energy equation correctly. In the absence of work done by non conservative forces, what equation are you using? Work is a function of the change in energy.

 

[johnhuntsman]

That will give you the correct answer for h, and since you are not asked to solve for d, you don't need the last equation. But, you are nonetheless not using your energy equation correctly. In the absence of work done by non conservative forces, what equation are you using? Work is a function of the change in energy.

I'm afraid I don't understand. Kinetic energy will have changed into gravitational protential energy because of the work gravity has done on it between the bottom of the incline and the point on the incline where the disk stops. That is a change in energy isn't it? At least in kinetic energy right?

[Edit] The equation gives me h = 1.9m as opposed to h = 2.6m which is the correct answer. [Edit]

 

[PhanthomJay]

When gravity is the only force acting that does work, which is your case here, then total mechanical energy is conserved, and you can use eiither of the following 2 equations:

1. $W_g = \Delta KE$ or
2. $\Delta KE + \Delta PE =0$

Neither equation applies if work is being done by other forces like friction or applied forces. I am not sure if you have studied this yet.

But if you examine either equation above, they are not quite the same as the equation you gave in your solution. It's a signage thing related to the change, which becomes extremely important for the more general class of problems.

 

[johnhuntsman]

When gravity is the only force acting that does work, which is your case here, then total mechanical energy is conserved, and you can use eiither of the following 2 equations:

1. $W_g = \Delta KE$ or
2. $\Delta KE + \Delta PE =0$

Neither equation applies if work is being done by other forces like friction or applied forces. I am not sure if you have studied this yet.

But if you examine either equation above, they are not quite the same as the equation you gave in your solution. It's a signage thing related to the change, which becomes extremely important for the more general class of problems.

So does that change the equation I used in anyway, or should I simply know this because it will be germane later. Also, is the h I arrived at correct or not? The correct answer section has been wrong several times before and I just want to hear it from someone educated that it probably is wrong.

 

[PhanthomJay]

So does that change the equation I used in anyway, or should I simply know this because it will be germane later.

Using the work = change in KE approach, -mgh = (0 - ((1/2)mv^2 + (1/2)Iω^2)), which is the same equation you have.

Also, is the h I arrived at correct or not? The correct answer section has been wrong several times before and I just want to hear it from someone educated that it probably is wrong.

The center of the disc rises 1.9 m from its start point before it stops. I am unsure where the 2.6 m comes from.

 

[johnhuntsman]

Alright then. Thanks.