What is the velocity and kinetic energy of a rolling disk on an inclined plane?

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Homework Help Overview

The discussion revolves around a solid disk rolling down an inclined plane, with specific parameters including mass, radius, and height. Participants are tasked with determining the linear velocity, number of revolutions, and total kinetic energy of the disk at the bottom of the incline.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between potential energy and kinetic energy, questioning how to account for both translational and rotational kinetic energy. Some express uncertainty about their calculations and the implications of the rolling condition.

Discussion Status

Several participants have attempted calculations and shared their results, but discrepancies with expected answers have led to further questioning and re-evaluation of their methods. Guidance has been offered regarding the inclusion of translational kinetic energy and the use of the moment of inertia.

Contextual Notes

Participants are working under the constraints of a homework assignment, which includes specific multiple-choice answers that they are trying to match. There is an ongoing discussion about the correct application of formulas and the assumptions made in their calculations.

cugirl
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Homework Statement



A solid disk of mass m = 2.2 kg and radius r = .20 m starts from rest at a height h = 3.2 m and rolls without slipping down an incline plane with an angle of 30 degrees above the horizontal.

a What is the linear velocity of the disk at the bottom of the incline?

b How many revolutions has the disk turned through when it reaches the bottom?

c What is the magnitude total Kinetic Energy of the sphere at the bottom of the incline?


Homework Equations


I=.5mr^2
Kinetic = .5*I*omega^2


The Attempt at a Solution


I started by solving for the hyp => sin 30 = 3.2 /x where x=6.4
That's where I am stuck
 
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cugirl said:

Homework Statement



A solid disk of mass m = 2.2 kg and radius r = .20 m starts from rest at a height h = 3.2 m and rolls without slipping down an incline plane with an angle of 30 degrees above the horizontal.

a What is the linear velocity of the disk at the bottom of the incline?

b How many revolutions has the disk turned through when it reaches the bottom?

c What is the magnitude total Kinetic Energy of the sphere at the bottom of the incline?

Homework Equations


I=.5mr^2
Kinetic = .5*I*omega^2

The Attempt at a Solution


I started by solving for the hyp => sin 30 = 3.2 /x where x=6.4
That's where I am stuck

It rolls without slipping ... an important condition. That means that the potential energy ... m*g*h ... is going to go into kinetic energy. But what kind of kinetic energy?

Won't it be rotational energy at 1/2*I*ω2 ?

Once you've found ω, it's all down hill from there, even though you are already at the bottom, since ω = v/r.
 
If 1/2*I*ω2 = mgh, I solve for the KE at about 69 and omega at 56. But, using ω = v/r, my v = 56*.2, or 11.2, which is not one of the multiple choice answers on the problem set.

I am also still uncertain about solving for the # revolutions the disk has turned through when it reaches the bottom.
 
Whoops, I did have a math error. I got omega = 39.6. But, again, using ω = v/r, my v = 39.6*.2, or 7.9, which is not one of the multiple choice answers on the problem set.
 
You must either include translational KE, or use the
moment of inertia about the point of contact.

Equating total KE to mgh is by far the simplest way of
finding v.

David
 
Last edited:
davieddy said:
You must either include translational KE, or use the
moment of inertia about the point of contact.

Equating total KE to mgh is by far the simplest way of
finding v.

David

Oops. Right you are. I left out the 1/2*m*v2 that is the translational energy. How silly.

Haste makes waste.

1/2*I*ω2 + 1/2*m*v2 = m*g*h

V = ω*r so ...

1/2*ω2*(I + r2) = g*h
 
LowlyPion said:
Oops. Right you are. I left out the 1/2*m*v2 that is the translational energy. How silly.

Haste makes waste.

1/2*I*ω2 + 1/2*m*v2 = m*g*h

V = ω*r so ...

1/2*ω2*(I + r2) = g*h

Sorry to nitpick but

1/2*ω2*(I + m*r2) = m*g*h
where I = (1/2)m*r^2

(Sort of confirms the parallel axis theorem)

David
 
Last edited:
davieddy said:
Sorry to nitpick but

1/2*ω2*(I + m*r2) = m*g*h
where I = (1/2)m*r^2

(Sort of confirms the parallel axis theorem)

David

Geez. No problem. You're absolutely right. Thus confirming that haste indeed makes waste in not converting I. Trying to do too many things is not good for concentration ... at least mine.
 
  • #10
Here's what I did but I am still not getting the answer:
mgh=0.5mv^2 (KE) + 0.5Iw^2

34.5 = .5Iomega^2
mgh*cos30 = 59.7

PLugging in:
59.7 = 1.1v^2 + 34.5
22.9 = v^2
4.8 = v

Not a possible answer on our sheet...
 
  • #11
davieddy said:
1/2*ω2*(I + m*r2) = m*g*h
where I = (1/2)m*r^2

David

You haven't even read how we got this, let alone tried it, have you?
 
  • #12
Yes, I did. I thought that was the theorem. My variable, v, is not in that equation you posted. Based on that equation, what would I be solving for? I went back and used LowlyPion's equation.
 
  • #13
cugirl said:
Yes, I did. I thought that was the theorem. My variable, v, is not in that equation you posted. Based on that equation, what would I be solving for? I went back and used LowlyPion's equation.

solve for w (omega)
v = wr

BTW LowlyPion was having a bad hair day, as you might have gathered
from our good-natured exchange.
Sorry if I sounded a bit cross in my last post.

David
 
  • #14
cugirl said:
Here's what I did but I am still not getting the answer:
mgh=0.5mv^2 (KE) + 0.5Iw^2

34.5 = .5Iomega^2
mgh*cos30 = 59.7

PLugging in:
59.7 = 1.1v^2 + 34.5
22.9 = v^2
4.8 = v

Not a possible answer on our sheet...

mgh=0.5mv^2 (KE) + 0.5Iw^2
is correct: PE lost = total KE gained.
What you did next was just crazy!

Now substitute w=v/r and solve for v.
 
  • #15
Good lord, I think I got it.
v=6.47
 
  • #16
cugirl said:
Good lord, I think I got it.
v=6.47

Whoopee:smile:
 

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