Problems and solutions I

# This is the result we had in our table relating

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Unformatted text preview: to the tangential velocity, which is then related to the angular velocity. pi¶ = mi vit = mi ri w The angular momentum of the ith mass becomes Li = ri pi¶ = ri mi vit = mi ri2 w The total angular momentum is the sum over all these terms L = ⁄i Li . Using I = ⁄i mi ri2 we get the angular momentum of a rotating rigid body L = I w. This is the result we had in our table relating rotations about a fixed axis to one dimensional linear motion. The Second Law We can now, finally, derive the rotational equivalent of the second law tnet = I a. Start with the momentum form of the second law. ext tnet = „ „t Ltot Now take the component along the axis of rotation. When the system is the rigid body then the net external torque on the rigid body is just the torque on it. Similarly, the total angular momentum is just I w the angular momentum of the body. We get tnet = „ „t L. Using L = I w and a = „ w ê „ t we get our result. tnet = I a 6 | Chapter I - Rotational Dynamics and Equilibrium The Torque Due to Gravity We saw earlier that to calculate the potential energy due to gravity we treat the object as if all the mass is at the center of mass. The same is true for finding the torque due to gravity. ” tgrav = rcm ä M g It is straightforward to verify this. Write the torque as the sum over the torques on all the m...
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## This document was uploaded on 02/26/2014 for the course PHYS 2425 at Blinn College.

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