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•  CAPA due Tuesday night. •  Exam solutions posted on D2L •  Exam scores will be posted

either tonight or Monday night. •  Today will be multiple body

problems and the dynamics of circular motion.

Solving force problems

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Pulleys Until we get to Chapter 10, all of our pulleys will be ideal pulleys which are massless and frictionless.

An ideal pulley simply changes the direction of the tension force in the rope which goes around the pulley.

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Statics problem

30 kg

A massless rope is looped over a pulley with one end connected to a 30 kg box and the other end held by a person. What force must the person exert to keep the box stationary?

Fnet,y = T −mg = 0

Free body diagram for the hand handF

TFree body diagram for the box mgFG =

T

mgT =

Fnet,y = T −Fhand = 0mgTF ==hand

.

In a massless rope the tension is the same everywhere!

2hand m/s 9.8kg 30 ⋅=F

N 290=

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from the last problem we know

Statics problem

30 kg

What is the magnitude of the force exerted by the rod holding the pulley to the ceiling?

Free body diagram for the pulley mgT =

Fnet,y = Frod − 2T = 0

N 590m/s 9.8kg 30222 2rod =⋅⋅=== mgTF

.

This is the force on the ceiling

T

rodF

T

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30 kg

The person holding the rope steps back a few feet. What is the magnitude of the force exerted by the rod on the ceiling now (note the angle)?

.

Clicker question 1 Set frequency to BA

A.  B.  C.  D.  E.  Impossible to tell

mgF 2rod =mgFmg 2rod <≤

mgF <rod

mgF 2rod >

TrodF

TPulley free body diagram

Frod < 2T and T=mg. Given the angles, we could calculate Frod.

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An Atwood's machine is a pulley with two masses connected by a string as shown. The mass of object A, mA, is twice the mass of object B, mB. The tension T in the string is...

Clicker question 2 Set frequency to BA

A. B. C. Neither of these gmT A= gmT B= A B

Fnet,yA = T −mAg =mAay

A

gmF AG =

ATA Fnet,y

B = T −mBg =mBayB

gmF BG =

BTB

Since the weights are unequal, it should be clear that there will be some non-zero acceleration. Thus, since and are not zero, neither A nor B can be correct.

AyAam

ByBam

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A B

AyAA amgmT =−

gmF AG =

ATA B

yBB amgmT =−

gmF BG =

BTB

We defined up as positive for both masses which means By

Ay aa −=

ByAA amgmT −= B

yBB amgmT +=& give ByBB

ByAA amgmamgm +=−

T = mBg+mBayB = mA +mB

mA +mB

mBg+mA −mB

mA +mB

mBg = 2mAmB

mA +mB

g

Full solution to Atwood’s machine

gmmmmaBA

BABy +

−=Solving for acceleration: so the tension is

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Remember circular motion

1v!

2v!

a!tanarada

What is the acceleration for circular motion with varying speed?

Can divide acceleration vector into two parts

Radial acceleration is perpendicular to the velocity vector and points to the inside of the curve with magnitude r

va2

rad =

Tangential acceleration is related to change in speed and is parallel to the velocity vector with magnitude dt

vda

!=tan

Since they are perpendicular: 2rad

2tan aaa +=

!

We can consider the coordinate system as r,t rather than the more usual x,y.

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What is the force?

A car going around a corner on a level surface

For each of these scenarios, what is the force which causes the radial acceleration?

Friction between tires and pavement

A car going around a banked curve Friction between tires and pavement and a component of the normal force

Satellite orbiting the Earth Earth’s gravity

We know we need forces to cause acceleration.

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Vertical circular motion Consider a motorcycle doing a vertical loop.

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Vertical loops What does the free body diagram look like at the bottom of the loop? n

mgLet’s apply Newton’s 2nd law

Fnet,r = n−mgAdding the forces:

We will have radial acceleration rva2

rad = so rmvmar

2=

Therefore rmvmgn

2=− Can also write

rmvmgn

2+=

making it clear that mgn >

The normal force is what you feel as weight which is why you feel heavier at the bottom of the roller coaster.

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What is the free body diagram at the top?

Let’s apply Newton’s 2nd law: Fnet,r = n+mg =maradTherefore

rmvmgn

2=+ Can also write mgr

mvn −=2

A. B. C. D. E.

Clicker question 3 n mg

What happens if ? mgrmv <

2Can the normal force be negative?

No. In this case, n=0 and there is more than radial acceleration; it is in free fall and follows a parabola. It falls off the loop.

For , n = 0 so you feel weightless. mgrmv ≤

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