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T = I*An
We also know that
T = R*f
where
Therefore,
T = I*An, or
An = T/I, or
An = R*f/I, or
f = An*I/R
where
As we discussed earlier, when an object is rolling with a given angular velocity, the translational velocity of the axis of rotation is proportional tothe radius of the object. The greater the radius, the greater will be the translational velocity. Thus, the translational velocity is proportional to theangular velocity with the radius being the proportionality constant. We can write
Vcm = W*R
If the object is not slipping, the rate of change of the translational velocity must also be proportional to the rate of change of the angular velocitythrough the same proportionality constant, which is the radius.
Therefore, we can write
Atr = R*An, or
An = Atr/R
The translational acceleration is also given by the net force divided by the mass. The net force is the component of the weight pointing down the inclineminus the force of friction pointing up the incline. Therefore, we can write
M*g*sin(U) - f = M*Atr
Substitution from above yields
M*g*sin(U) - (An*I/R) = M*Atr
Further substitution from above yields
M*g*sin(U) - ((Atr/R)*I/R) = M*Atr, or
M*g*sin(U) - (Atr*I/(R^2)) = M*Atr
Solving for Atr yields
M*Atr + (Atr*I/(R^2)) = M*g*sin(U), or
Atr * (M + I/(R^2)) = M*g*sin(U), or
Atr = (M*g*sin(U))/(M + I/(R^2)), or
Atr = (g*sin(U))/(1 + I/(M*R^2))
For a solid cylinder,
I = (1/2)*M*R^2
By substitution
Atr = (g*sin(U))/(1 + ((1/2)*M*R^2)/(M*R^2)), or
Atr = (g*sin(U))/(1 + (1/2)), or
Atr = (g*sin(U))/(3/2), or
Therefore, the translational acceleration of the rolling solid cylinder is given by
Atr = (2/3)*(g*sin(U))
If the cylinder were sliding in the total absence of friction, from what you learned in earlier modules, the acceleration would simply be
Atr = g*sin(U)
Therefore, if there is sufficient friction to cause the cylinder to roll without slipping, the translational acceleration will be only (2/3) of thesliding acceleration. Once again, this is the result of a portion of the potential energy being transformed into rotational kinetic energy, resulting inless translational velocity, and less translational acceleration.
I encourage you to work through the computations that I have presented in this lesson to confirm that you get the same results. Experiment withthe scenarios, making changes, and observing the results of your changes. Make certain that you can explain why your changes behave as they do.
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