PHY205 Ch16: Rotational Dynamics 1.Combination Translational and Rotational motion and Atwood machine 2.Discuss Ball rolling down incline from 3 different points of view: Point of Contact (acceleration) Center of Mass (acceleration) Energy (velocity) showing consistency with previous 2 answers

PHY221 Ch16: Rotational Dynamics 1. Main Points Combination Translational and Rotational motion

PHY221 Ch16: Rotational Dynamics 2. Discuss Study of a ball rolling down an incline without slipping. CM point of view. (see slides 6-7 for the proof that: even if CM accelerated

PHY205 Ch16: Rotational Dynamics 2. Discuss Study of a ball rolling down an incline without slipping. Point of Contact point of view.

PHY221 Ch16: Rotational Dynamics 2. Discuss Study of a ball rolling down an incline without slipping. Energy point of view

PHY221 Ch16: Rotational Dynamics 2. Discuss Proof that the torque around the CM is given by:even if CM is accelerated!!! We start from the definition of the torque around CM (1) (*** See note at the end) Since we want the CM to possibly be accelerated, it is not an inertial frame and thus we CANNOT write: since it would assume that the cm is inertial to have NII valid. Thus the only option is to write the force on particle mi as: and proceed from there. Since O is assumed inertial we have: We insert this result in the definition of torque (1) and get: and thus: The 1 st term is zero because it gives the vector to the CM from the CM (because of the * in r). We need to work on the 2 nd term: Now we need to be a little careful: can be decomposed into two components, a centripetal one, call it a* r which is anti // to r* i and a tangential one whose value is as we’ve seen in CH15: r* i  i. where  i is the angular acceleration around z of the mass m i. Since a* r is anti// to r* i, its cross product with it is zero.

PHY221 Ch16: Rotational Dynamics 2. Discuss Since it is a rigid body, all the particle m i have the same angular acceleration  and thus the previous equation gives: (notice that we got rid of the subscript i on  ! Inserting our result in the expression of the torque around CM on the previous slide we get our final result: (*** note): Technically the only relevant components of both r* I and F i are because we are assuming rotation around only a z-axis through CM! Thus even if our rigid body is extended in the z direction the z component of the torque only comes from the x-y components of the r* I and F i because of the properties of the cross product On the other hand since the tangential acceleration is perpendicular to r* i its cross product will be along z, and of magnitude: Where we just wrote the z component, the others being zero because, again, we assume a rotation around z only. Notice that I wrote  to emphasize that the angular acceleration in this case is measured around the CM. So the center of mass behave just AS IF it was inertial. This is a CRUCIAL RESULT Note: If the torque around the CM is zero then  =0 and thus  is constant which implies that is also constant