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.

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uss

Study of a ball rolling down an incline without slipping. Energy point of view

PHY221 Ch16: Rotational Dynamics2.

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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.

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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