Solid disk on inclined plane

In summary: 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
  • #1
17
0

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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  • #2
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.
 
  • #3
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.
 
  • #5
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.
 
  • #6
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
 
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  • #7
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
 
  • #8
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:
  • #9
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:
 

1. What is a solid disk on an inclined plane?

A solid disk on an inclined plane is a simple physical system where a circular disk is placed on a flat surface that is tilted at an angle. This system is commonly used in physics experiments to study the effects of gravity and friction on objects.

2. How does the angle of inclination affect the motion of a solid disk on an inclined plane?

The angle of inclination plays a crucial role in determining the motion of a solid disk on an inclined plane. The steeper the incline, the faster the disk will accelerate down the plane due to the force of gravity. As the angle decreases, the acceleration of the disk also decreases.

3. What forces are acting on a solid disk on an inclined plane?

There are two main forces acting on a solid disk on an inclined plane: the force of gravity and the normal force. The force of gravity pulls the disk downward, while the normal force acts perpendicular to the surface of the plane to prevent the disk from sinking into the surface.

4. What is the relationship between the mass of the disk and its acceleration on an inclined plane?

The mass of the disk does not directly affect its acceleration on an inclined plane. The acceleration is primarily determined by the angle of inclination and the force of gravity. However, a heavier disk may experience more friction on the surface of the plane, which can slow down its acceleration.

5. How does friction affect the motion of a solid disk on an inclined plane?

Friction plays a significant role in the motion of a solid disk on an inclined plane. It opposes the motion of the disk and can slow down or even stop the disk from sliding down the plane. The amount of friction depends on the surface of the plane and the weight of the disk.

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