Summary Lecture 12Rotational MotionRotational Motion

10.810.8 TorqueTorque

10.910.9 Newton 2 for rotationNewton 2 for rotation

10.10 Work and Power10.10 Work and Power

11.211.2 Rolling motionRolling motion

Rotational MotionRotational Motion

10.810.8 TorqueTorque

10.910.9 Newton 2 for rotationNewton 2 for rotation

10.10 Work and Power10.10 Work and Power

11.211.2 Rolling motionRolling motion

Problems:Chap. 10:

21, 28 , 33, 39, 49, 54, 67

Problems:Chap. 10:

21, 28 , 33, 39, 49, 54, 67

Today

20-minute teston material in

lectures 1-7

at end of lecture

Today

20-minute teston material in

lectures 1-7

at end of lecture

Some Rotational Inertia

PLUS

Axis of Rotation

h

Parallel-axis TheoremParallel-axis Theorem

CM

The rotational inertia of a body about any parallel axis, is equal to its The rotational inertia of a body about any parallel axis, is equal to its

R.I. about an axis through its CM,R.I. about an axis through its CM,

R.I. of its CM about a parallel axis through the point of rotationR.I. of its CM about a parallel axis through the point of rotation

I = ICM + Mh2

One rotation about yellow axis involves one rotation of CM about this axis plus one rotation of body about CM.

I = Icm + Mh2

Proof of Parallel-axis Theorem

h

RI of CM about suspension point, distance

R away is

MR2.

So total RI is

2MR2

Example What is it about here?

R

RI of ring of mass M about CM is

MR2

The Story so far...

Rotational Variables, ,

relation to linear variables

vector nature

Rotational kinematics with const.

Rotation and Kinetic Energy

Analogue equations to linear motion

Rotational Inertia

Torque…is the “turning ability” of a force

Where would you put the door knob?

here? or here?

FF

rr

The magnitude of the torque is Fr,

and this is greater here!

If F and r are perpendicular

= rF (Unit: N m)

Torque

Axis r

F

The same F at larger r has bigger turning effect.

F Axis r

Torque is a vector

= r F sin

= r x perp component of F

Direction of :Perpendicular to r and F

Sense:Right-hand screw rule (out of screen)

(Hint: Which way would it accelerate the body?

Sense is same as change in ang vel. .)

= r x F

In GeneralIn General

Fsin

Newton 2 for Rotational motion

For Translational motion we had:

maFext

For Rotational motion we expect:

I

ext

What is I for rotating object?

Example

= I (Newton 2)

I = /

I = 6400/1.2

I = 5334 kg m2

Observe ang accel of 1.2 rad s-2

Apply a force of 3200 N

Fr

Distance r = 2 m from axis

= r x F = 6400 N m

R

Work done by a torque which rotates a body through an angle

w =

d.w ds.Fwcf

If is constant w =

Power

dt

τ.dθ

dt

dwp P =

Work

cf P = F.v

Example

A car engine has a power of 100 HP at 1800 RPM

What is the torque provided by the engine?

1 HP = 746 W

P = 74600 W

P = .

= P/

1

1

srad188

srad60

2x1800

RPM1800

= 74600/ 188

= 397 N m

vcm t = 2RBut in turning one revolution (2 radian) in time t, = 2/tSo that t = 2/

2R

vcm 2/= 2R vcm = R

Rolling Motion

R

Rolling Motion

total energy (KE) = ½ mv2 + ½ v2mR2/R2

Object of radius R

The kinetic energy is shared between translational and rotational motion.

= ½ mv2(1 + )

Cons. Energy says

PEinitial = KEfinal

mgh = Ktrans + Krot

= ½ mvcm2 + ½ I2

h

remember vcm= R or = vcm/R

so mgh = ½ mv2 + ½ Iv2/R2

is the coefficient in the expression for the Rotational Inertia I

I = mR2

is the coefficient in the expression for the Rotational Inertia I

I = mR2

= ½ mv2 + ½ mv2

h

The kinetic energy is shared between translational and rotational motion.

mgh = Ktotal = Ktrans + KRot

= ½ mvcm2 + ½ mvcm

2

The fraction of KE that is translational is

11

)1(mv

mv2

cm21

2cm2

1

β)(1

2ghv 2

cm

The larger , the more of the available energy goes into rotational energy, and the smaller the centre of mass velocity

h = ½ mvcm

2(1 + )