How Do You Calculate the Vertical Acceleration of a Unrolling Disk?

Click For Summary

Homework Help Overview

The problem involves a solid uniform disk unrolling from a string, with the goal of finding its vertical acceleration in terms of gravity. The discussion centers around the forces acting on the disk, specifically tension and weight, and how they relate to both translational and rotational motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between tension, torque, and acceleration, with some attempting to derive equations involving these variables. Questions arise regarding the distinction between translational and rotational acceleration, as well as how to express tension in terms of the disk's acceleration.

Discussion Status

The discussion is active, with participants sharing various equations and attempting to clarify the relationships between the forces and accelerations involved. Some guidance has been offered regarding the use of kinematical constraints and the need to eliminate angular acceleration from the equations to find the desired vertical acceleration.

Contextual Notes

Participants note the importance of considering both force dynamics and rotational motion, and there is an ongoing exploration of how to manipulate the equations without substituting values prematurely. The discussion reflects a mix of understanding and confusion regarding the application of concepts.

veronicak5678
Messages
143
Reaction score
0

Homework Statement


A solid uniform disk unrolls from a string which is wrapped around it. If tension and weight are the only important forces, find the vertical acceleration of the disk in terms of gravity.


Homework Equations





The Attempt at a Solution



mg - T = ma

a = mg - T /m

I know it's not this simple, but I don't know what else to do.
 
Physics news on Phys.org
HINT: The tension creates a torque on the disc.
 
So far I have α = 2 * Tension force / mass*radius. I don't understand if the answer is supposed to be in translational or rotational, and I don't know how to get the tension force if the disk is accelerating at an unknown rate.
 
veronicak5678 said:
So far I have α = 2 * Tension force / mass*radius. I don't understand if the answer is supposed to be in translational or rotational, and I don't know how to get the tension force if the disk is accelerating at an unknown rate.

the answer is supposed to be translational, since it does not say angular acc.
also, if you consider the axis of rotation through the centre of mass, and equate torque to
I*[angular acc], using a kinematical constraint, you can gat another equation containing tension. Hope i helped...
 
idid equate torque to I (angular)

torque tension = I / alpha

rearranged to find alpha

alpha = tension * 2 / m*r

I don't know what you mean by a kinematical constraint. Could you please tell me how I can find the tension to solve this equation?
 
well, a kinematical constraint is simply a relation between two kinematical variable, which remains true irrespective of their value. in this particular case,
[linear acc] = [radius of disk]* [angular acceleration]
 
Oh! I've never seen that. But after plugging 2f/mr into that, I end up with acc = 2*f / m. How can I use that to answer the question " vertical acceleration in terms of g?"
 
since you have solved the problem, i guess i will give the solution:

mg - T = ma...( i )

T*r = [0.5*mr^2][alpha]...( ii )

using the kinematical constraint,

a = r*[alpha]...( iii )

so you have three equations three unknowns, just don't substitute the value of g here, to get the answer in terms of g :)
 
I still don't understand...

I just keep manipulating this in circles. I have a = 2 (g -a). Why did I just find angular velocity?
 
  • #10
angular velocity, energy conservation need not be applied here. you are asked to find out the acceleration, so i say stick to force dynamics...:wink:
 
  • #11
OK. I guess I haven't understood what you've been saying. Starting over, with just force dynamics, I found
mg - T = ma
a = mg - T /m
This was my first post.
 
  • #12
Just because we're using force and dynamics doesn't mean that we can ignore rotational motion. Let's return to these three equations:
aniketp said:
mg - T = ma...( i )

T*r = [0.5*mr^2][alpha]...( ii )

a = r*[alpha]...( iii )
Since we don't want the solution in terms of the angular acceleration, the first thing to do is eliminate [alpha]. Can you do this?

Use the second and third equations
 
  • #13
T*r = 1/2 m*r^2 (a/r)

T = 1/2 m a
a = 2T/m
 
  • #14
veronicak5678 said:
T*r = 1/2 m*r^2 (a/r)

T = 1/2 m a
Good :approve:.

So now can you use this to eliminate T from equation (i) and then solve for a?
 
  • #15
m*g - 1/2 ma = ma
g - a/2 = 2a
2g = a
 

Similar threads

Replies
23
Views
7K
Replies
4
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
18
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
12K
Replies
4
Views
2K