Angular kinematics-Moment of Inertia

•  CAPA Set #9 due next Tuesday + CAPA Set #10 due next Friday.

•  Exam MC scores and Exam solutions will available on D2L soon.

•  Covering Material in Chap. 10 – this week and part of next.

Web page: http://www.colorado.edu/physics/phys1110/phys1110_sp12/

Announcements:

Center of Mass Calculations A 60 cm length of uniform wire of 60 g mass, is bent into a right triangle. The x- and y- coordinates of the center of mass, in cm, are closest to: A) (8,3) B) (8,5) C) (9,4) D) (10,3) E) (10,5)

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xcm =(10g)(0) + (24g)(12cm) + (26g)(12cm)

60g=10cm

€

ycm =(10g)(5cm) + (24g)(0) + (26g)(5cm)

60g= 3cm

€

r cm =mi r i∑

mi∑

Work Problem An 8.0 Kg block is released from rest, v1=0 m/s, on a rough incline. The block moves a distance of 1.6 m down the incline, in a time interval of 0.80 sec, and acquires a velocity of v2=4.0 m/s. The average rate at which the friction force does work during the 0.80 sec time interval is closest to:

A)  Zero

B) -18W

C) -20 W

D) -22 W

E) -24 W €

Average work = Ef − E i

Δt

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Ei = mgh = (8kg)(9.8 ms2)(1.6msin(40)) = 80.8J

Ef =12mv 2 = .5(8kg)(4 m

s)2 = 64J

€

=64J − 80.6J

.8s=−16.6J.8s

= −20.8J

Your exam score •  Exam questions

– Exam questions can be addressed to TA’s or the professors (Dan Dessau or myself)

– Exams will be returned in recitation next Thursday – For A&S students – last day to drop the course

without petitioning the Dean is Mar. 23 – next Friday.

Angular kinematics The same equations which were derived for constant linear (or translational) acceleration apply for constant angular (or rotational) acceleration

Constant angular acceleration only!

Constant linear acceleration only!

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x = x0 + v0xt + 12 axt

2

Missing

Variables

Position

Final v

time

Relationships to linear acceleration What about acceleration? If speed is then

We know that centripetal (or radial) acceleration is . Using v = rω, this can be rewritten as .

Total linear acceleration is composed of both tangential and radial acceleration which are always perpendicular to each other.

Clicker question 1 Set frequency to BA

A.  4 cm/s B.  5 cm/s C.  10 cm/s D.  20 cm/s E.  None of the above

5 cm CD slows from angular velocity of 4 rad/s (dust has speed of 20 cm/s) to a stop in 4 seconds with constant angular acceleration. What is the dust speed halfway through deceleration?

Clicker question 1 Set frequency to BA

A.  4 cm/s B.  5 cm/s C.  10 cm/s D.  20 cm/s E.  None of the above

5 cm CD slows from angular velocity of 4 rad/s (dust has speed of 20 cm/s) to a stop in 4 seconds with constant angular acceleration. What is the dust speed halfway through deceleration?

At t=2 seconds the angular velocity is:

This gives a linear velocity of

Clicker question 2 Set frequency to BA

A.  20 cm/s2 B.  5 cm/s2 C.  21 cm/s2 D.  0.8 cm/s2 E.  None of the above

5 cm CD has an angular acceleration of -1 rad/s2 and an angular velocity of 2 rad/s. What is the magnitude of the acceleration of the dust particle?

Clicker question 2 Set frequency to BA

A.  20 cm/s2 B.  5 cm/s2 C.  21 cm/s2 D.  0.8 cm/s2 E.  None of the above

5 cm CD has an angular acceleration of -1 rad/s2 and an angular velocity of 2 rad/s. What is the magnitude of the acceleration of the dust particle?

Total acceleration is composed of radial and tangential acceleration.

Kinetic Energy of Rotation If we have a object spinning, then it must have some kinetic energy, but we can’t apply KE = mv2/2. The center of mass is not moving. So we have to develop an expression for the moving energy.

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K =12m1v1

2 +12m2v2

2 +12m3v3

2 + ......

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K =12mivi

2∑Note vi is not the same for all particles in the object, we solve this by writing vi=ωri.

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K =12mivi

2∑ =12∑ mi(ωri)

2 =12

miri2∑( )ω 2

Kinetic Energy of Rotation Cont.

Term in the parentheses tells us how the mass of a rotating body is distributed about its axis of rotation and we call it rotational inertia or Moment of Inertia. €

K =12mivi

2∑ =12

miri2∑( )ω 2

€

I = miri2 (moment of inertia)∑

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K =12Iω 2 Note this is due to pure rotation. If the

object is also translating, then we have to add

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12MvCoM

2 ; where M = mi∑

Moment of inertia Consider an object of mass m at the end of a massless rod of length R spinning around an axis with angular velocity ω. What is the kinetic energy of the mass?

R m

For a particle which is a distance R from the axis (or pivot) the moment of inertia is

As the particle moves further out (R increases), ω doesn’t change but velocity increases so kinetic energy increases. This comes from the moment of inertia increasing

Moment of inertia If you have a bunch of particles rotating about an axis, can find the total moment of inertia by adding up the moment of inertia of all the particles

If there is a smooth distribution of matter then the sum becomes an integral but the idea is the same.

Remember that r is the distance from the axis.

Therefore, the moment of inertia depends on the axis location.

Moment of inertia (ring & cylindrical shell) Rings and cylindrical shells (hollow cylinders) behave like a bunch of little particles all located at a distance of R from the axis.

This is only true for the axis shown.

Moment of inertia What about a solid sphere spinning about it’s axis?

Not all of the bits of matter are at a distance R from the axis.

End up with something less than MR2.

A hollow sphere of radius R still doesn’t have all bits of matter a distance of R from the axis. (hollow sphere)

Moments of inertia for other bodies can be found in Table 10.2

in Halliday and Resnick

(solid sphere)

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I = r2∫ dm

€

I = MR2

€

I =12MR2

€

I =25MR2

€

I =112

ML2

€

I =23MR2

€

I =12MR2

€

I =12M(R1

2 + R22)

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I =14MR2 +

112

ML2

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I =112

M(a2 + b2)

Moment of Inertia Demos Any freely sliding object will beat any rotating object on the same incline because none of its potential energy is given to rotational kinetic energy. Objects which contain a liquid which does not roll with the can, means that most of the liquid effectively slides down the incline inside the rolling can. The can of liquid therefore has very little rotational inertia compared to its mass. A solid object on the other hand, is made to rotate, giving the can more rotational inertia. It will rotate more slowly! We will find that for an object

rolling down an incline plane

€

a =g sinθ

(1+Icmmr2 )

Clicker question 3 Set frequency to BA

For a rotating object, kinetic energy is . If a ring, solid sphere, and a hollow sphere, each with mass M and radius R, are rolled down the same incline, which will have the largest kinetic energy at the bottom. Assume no energy is lost to friction. A.  ring B.  solid sphere C.  hollow sphere D.  All will have the same E.  Impossible to tell

Clicker question 3 Set frequency to BA

For a rotating object, kinetic energy is . If a ring, solid sphere, and a hollow sphere, each with mass M and radius R, are rolled down the same incline, which will have the largest kinetic energy at the bottom. Assume no energy is lost to friction. A.  ring B.  solid sphere C.  hollow sphere D.  All will have the same E.  Impossible to tell

By conservation of energy, they will all have the same kinetic energy. They all start with potential energy of Mgh and end with that amount of kinetic energy.

Since K is equal and I is not equal, this means ω is not equal.

Parallel axis theorem The moments of inertia for a ring, hoop, solid sphere, and hollow sphere were for an axis through the center of mass

What if the axis is not through the center of mass?

If the axis is parallel to the center of mass axis then can use the parallel axis theorem . = moment of inertia for an axis through the center of mass

= distance between the two parallel axes