0.35 Phy1350: angular momentum -- the physics of rolling objects  (Page 5/5)

T = I*An

We also know that

T = R*f

where

• T is the torque
• R is the radius
• f is the unknown frictional force

Therefore,

T = I*An, or

An = T/I, or

An = R*f/I, or

f = An*I/R

where

• T represents the torque
• I represents the moment of inertia
• R represents the radius
• f represents the unknown frictional force

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.

Work through the computations

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.

Resources

I will publish a module containing consolidated links to resources on my Connexions web page and will update and add to the list as additional modulesin this collection are published.

Miscellaneous

This section contains a variety of miscellaneous information.

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