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

AI Thread Summary
To calculate the linear speed of a rolling disc under a constant force, apply the relationship between linear and angular acceleration, where tangential acceleration equals angular acceleration times the radius. The total kinetic energy of the system can be expressed as the sum of translational and rotational kinetic energy. Given the force applied and the distance rolled, the work done on the disc can be calculated. By equating the work done to the total kinetic energy, the linear speed can be derived. This approach simplifies the problem and leads to the correct calculation of the disc's linear speed.
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

a_{T}=\alpha \cdot r, 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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