phy205 ch16: rotational dynamics 1.combination translational and rotational motion and atwood...
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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 acceleratedTRANSCRIPT
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 Dynamics1.
Mai
n Po
ints
Combination Translational and Rotational motion
PHY221 Ch16: Rotational Dynamics2.
Disc
uss
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/ /CM CMI
PHY205 Ch16: Rotational Dynamics2.
Disc
uss
Study of a ball rolling down an incline without slipping. Point of Contact point of view.
PHY221 Ch16: Rotational Dynamics2.
Disc
uss
Study of a ball rolling down an incline without slipping. Energy point of view
PHY221 Ch16: Rotational Dynamics2.
Disc
uss
Proof that the torque around the CM is given by: even if CM is accelerated!!!
*/
1
n
CM i ii
r F
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:2 *
2di
iid rF m
t
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: 2
2diiid
tF
rm
2 2 2 *2
2 2 2
*
2d d d d( )i CM CM
i i ii i
i ir rd r d r d r d
m m mF mt t t t
We insert this result in the definition of torque (1) and get: and thus: *
*2 2
/1
2 2d d
ni
CCM
iii
iMd r d
m mt
rt
r
** *
/1
** *
/1 1
* */
1
2 2
2 2
2 2
2 2
2 2
2
d d
d d
d
ni
CM i ii
n ni
CM i ii i
n
CM i
CMi i
CMi i
CMi i i
i
d r dm mt t
d r dm
r
mt t
r r
rr r
rr rd r dm mt
*
12d
ni
i t
The 1st 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 2nd term:
2
2
**
/1 d
ni
Ci
iM irm r d
t
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 mi. Since a*r is anti// to r*i , its cross product with it is zero.
*2
2dird
t
/CM CM CMI
PHY221 Ch16: Rotational Dynamics2.
Disc
uss
*2 *2/
1 1
n n
CM i ii CM CMi i
im mr r I
Since it is a rigid body, all the particle mi 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 Fi 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 Fi 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: 2
2
** * * * *2
di
i i Tangential i i i i ir
r r a rd
tr r
** *
22
2
di
i ir
tr
dr
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
2/
12CM CMK I