103 Phys Ch10 Part3

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Dr. Abdallah M. Azzeer 1

Two forms of

Newton’s second law

TORQUE AND ANGULAR ACCELERATIONTORQUE AND ANGULAR ACCELERATION

Σ ⇒ α

K =1

2mv

2

K R =1

2 I ω2

Σ F ⇒ a Σ F = ma

Σ = I α

∆s ⇔ ∆θ

v ⇔ ω

a ⇔ αt ⇔ t

m ⇔ I


1010--8.8. Work ,Power and Energy in Rotational MotionWork ,Power and Energy in Rotational Motion

Let’s consider a motion of a rigid body with a single external force F exerting

on the point P, moving the object by ds.

The work done by the force F as the

object rotates through the

infinitesimal distance ds=rdθ is

dW sd F ⋅= ( ) θ φ rd F sin=

dW θ τ d =( ) θ φ d rF sin=

Pdt

dW =

dt

d θ τ = τω =

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∑τ α I = ⎟ ⎠ ⎞⎜

⎝ ⎛ =dt

d I ω ⎟ ⎠ ⎞⎜

⎝ ⎛ ⎟

⎠ ⎞⎜

⎝ ⎛ =

dt

d

d

d I θ

θ

ω ⎟ ⎠ ⎞⎜

⎝ ⎛ =

θ ω ω

d

d I

dW θ τ d ∑= ω ω d I =

W ∫ ∑= f

d θ

θ ι

θ τ ∫= f

d I ω

ω ι

ω ω 22

2

1

2

1i f

I I ω ω −=

The rotational work done by an external force equals the change in rotational

energy.


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example 10.14

READ EXAMPLE 10.14 & try t o sol ve


Sample Problem

Figure 11-17a shows a uniform disk, with mass

M = 2.5 kg and radiusR = 20 cm, mounted on a

fixed horizontal axle. A block with massm = 1.2

kg hangs from a massless cord that is wrapped

around the rim of the disk. Find the acceleration

of the falling block, the angularacceleration of

the disk, and the tension in the cord. The cord

does not slip, and there is no frictionat the axle.

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αIR T

=

R

IT

α=

SOLUTION:

Both clockwise

R a αTherefore,

2

21

2

2

21

2 aM

R

a)R M(

R

aIT = Positive a is down.

2

a)Mm2(aMamTamgm

amTgm

21

= Positive direction is down.


)kg2.1()2(kg5.2

)kg2.1()2()s/m8.9(

m2M

m2ga

2

=

=

2s/m8.4 • Positive direction of a is down.

)s/m8.4()kg.2(MaT 221

21

=

N0.6

22

s/rad 24m20.0

s/m8.4

R

a=α

• The angular acceleration is clockwise.

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Let the disk in the previous Sample Problem start

from rest at time t = 0. What is its rotational kineticenergyK at t = 2.5 s?

tt0t0 αααωω =

2

4122

21

212

21 )tR (M)t()MR (IK ααω =

22

41 )]s5.2()s/rad 24)(m20.0[()kg5.2(

J90

Note that alpha is clockwise.

Sample Problem


)(R T)(W if if θθθθτ

2

212

212

21

iif tt0tt αααωθθ =

2

212

21

if tR T)t(R T)(R TWK ααθθ =22

21 )s5.2()s/rad 24()m20.0() N0.6(

J90

WW0WK K i =

An alternative way :

Note that the torque is clockwise

Clockwise

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Two masses m1 (5 kg) and m2 (10 kg) arehanging from a pulley of mass M (3 kg) and

radius R (0.1 m), as shown. There is no slip

between the rope and the pulleys.

(a) What will happen when the masses are

released?

(b) Find the velocity of the masses after they

have fallen a distance of 0.5 m.

(c) What is the angular velocity of the pulley at

that moment?

example 10.15

READ EXAMPLE 10.15 & try t o sol ve


Sample Problem

To throw an 80 kg opponent with abasic judo hip throw, you intend topull his uniform with a force F and amoment arm d 1 = 0.30 m from a pivotpoint (rotation axis) on your right hip(see Fig.). You wish to rotate himabout the pivot point w ith an angular

acceleration of - 6.0 rad/s2

- that is,with an angular acceleration that isclockwise in the figure. Assume thathis rotational inertia I relative to thepivot point is 15 kg·m2.

(a) What must the magnitude of F be if, before you throw him, you bend your

opponent forward to bring his center of mass to your hip?

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SOLUTION:SOLUTION:

m30.0

)s/rad 0.6()mkg15(

d

IF

22

1

⋅

=

α N300

The angular acceleration and torque are both clockwise.

(b) What must the magnitude of F be if your opponent remains upright before

you throw him, so that F g has a moment arm d 2 = 0.12 m from the pivot

point?

αImgd Fd 21

m30.0

)s/m8.9()kg80()m12.0( N300

d

mgd

d

IF

2

1

2

1

α N610 N6.613 ≈

The clockwise torque is negative.

Note that alpha is + 6.0 rad/s 2.

To minimize the required force, you should bend your opponent to bring his

center of mass to your hip.


Problem: Two bicycles roll down a hill that is 20m high. Both bicycleshave a total mass of 12kg and 700 mm diameter wheels (r =0.350 m). Thefirst bicycle has wheels that weigh 0.6 kg each, and the second bicycle haswheels that weigh 0.3 kg each. Neglecting air resistance, which bicyclehas the faster speed at the bottom of the hill? (Consider the wheels to bethin hoops).

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The only friction is static friction, so there are no non-conservative forces. (Static

friction involves no motion and since work is defined as W = Fd , when there is no

distance involved, there is no work/energy used).Mechanical Energy is Conserved, so:

ΣEi = ΣEf

KEi +PEi = KEf +PEf

(1/2)mvi2 + (1/2)Iωi

2 + mghi = (1/2)mvf 2 + 2*(1/2)Iωf

2 + mghf

mghi = (1/2)mvf 2 + 2(1/2)Iωf

2 (The two for two wheels).

Now w is given by ω=v / r where v is the velocity of the rim of the wheel and is the

same as the velocity of the bike. I is given by Mr2 where M is the mass of the tire.

mgh = (1/2)mvf 2 + Mr2vf

2 /r2= (1/2)mvf 2 + Mvf

2

So for the first bike vf = 19.3 m/s

For the second bike, the wheels have a mass of 0.6 kg, and we get

vf = 18.9 m/s

Why is this true? More of the potential energy goes into rotational kinetic energy

and less into translational kinetic energy when the wheel has more mass.


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103 Phys Ch10 Part3