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

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SUMMARY

The discussion focuses on calculating the linear speed of a rolling disc subjected to a constant force. A force of 12N is applied to a disc with a mass of 22kg, diameter of 0.50m, and rotational inertia of 0.688kgm². The relationship between linear acceleration and angular acceleration is clarified as a = r * α. The total kinetic energy equation is provided as K = (1/2) * m * v² + (1/2) * I * ω², leading to the conclusion that the work done on the wheel equals the total kinetic energy, allowing for the calculation of linear speed.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of rotational dynamics, specifically the relationship between torque and angular acceleration
  • Familiarity with kinetic energy equations for both translational and rotational motion
  • Basic principles of friction and its role in rolling motion
NEXT STEPS
  • Study the relationship between linear and angular motion in rolling objects
  • Learn about the conservation of energy in mechanical systems
  • Explore the effects of friction on rolling motion and its calculations
  • Investigate the derivation of rotational inertia for various shapes, including discs
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Physics students, mechanical engineers, and anyone interested in understanding the dynamics of rolling motion and energy conservation principles.

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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