How Do You Calculate the Linear Speed of a Rolling Disc?

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Homework Help Overview

The discussion revolves around calculating the linear speed of a rolling disc under the influence of a constant force. The problem involves concepts from dynamics and rotational motion, including mass, diameter, rotational inertia, and the relationship between linear and angular quantities.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between linear acceleration and angular acceleration, with attempts to apply Newton's second law and rotational dynamics. There are discussions about the role of friction and the equations governing motion.

Discussion Status

Some participants have provided insights into the relationships between linear and angular quantities, while others express confusion regarding their calculations. There is an ongoing exploration of the correct application of equations, and hints have been offered to guide the discussion without providing direct solutions.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is also a mention of the rotational inertia of the disc, which may influence the calculations but is not explicitly resolved in the discussion.

public_enemy720
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One more problem on a tough worksheet...I have tried it for a while, but can't find an equation(s) suitable for the problem...

A constant force of 12N is applied to the axle of a disc rolling along a flat plane. The disc has mass m=22kg, diameter D=.50m, and rotational inertia I=.688kgm^2. What is the linear speed of the center of the disc, V, after it has rolled for 12m?

I drew a free body diagram, and I know that the sum of the moments is equal to I times alpha, and I know the good ol' F=ma. But I can't seem to be able to find an acceleration with what I am given. A hint would be wonderful.
 
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Oh boy...I went somewhere, but I don't know if it was right. Here is what I did:

I used m*a=f-f(friction) and friction*radius = I*alpha. Alpha is equal to a*r, so I plugged that into make friction times radius = I*a*radius.

Plugging that into m*a=F-f(friction) i got:

m*a=F-((I*alpha*radius)/(radius))

This is rather confusing, but I think it is somewhat correct. Any suggestions or gross errors on my part?
 
public_enemy720 said:
Oh boy...I went somewhere, but I don't know if it was right. Here is what I did:

I used m*a=f-f(friction) and friction*radius = I*alpha. Alpha is equal to a*r, so I plugged that into make friction times radius = I*a*radius.

Plugging that into m*a=F-f(friction) i got:

m*a=F-((I*alpha*radius)/(radius))

This is rather confusing, but I think it is somewhat correct. Any suggestions or gross errors on my part?
Right idea, but alpha does not equal a*r. alpha and a are certainly related and you need that relationship, but that is not it.
 
it is...acceleration=radius x alpha
 
vijay123 said:
it is...acceleration=radius x alpha

[tex]a_{T}=\alpha \cdot r[/tex], i.e. tangential acceleration equals angular acceleration times radius.
 
I hope this can help you:

total K = (1/2)*m*v^2 + (1/2)*I*w^2

actually you don't have to use I = 0.688 which is given if you know the I of the disk is (1/2)*m*r^2.

plug all in, find out K = 3/4 * m * v^2

next,

the work does on the wheel also equals to the total K above. W = F*s = 12*12 = 144J

solve for v.

Good luck.

Minh T. Le
 

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