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•  CAPA due tomorrow night. •  Exam scores should be posted tonight. •  Today is circular motion and friction.

Solving force problems

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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 n mg

What happens if ? mgrmv <

2

Can 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 ≤

2

3

which gives

What about a bucket on a rope? For the bucket, the tension plays the same role as the normal force in the previous problem.

For the eraser or water in the bucket, the normal force from the bottom of the bucket provides the necessary force.

What is the minimum speed at the top so the rope stays taut, the water stays in the bucket, or the motorcycle stays in the loop?

This is where the normal (or tension) just reaches 0 so that gravity alone is providing the radial acceleration

mgrmv =

2So this critical speed is when rgvc =

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Clicker question 1 Set frequency to BA

A bucket containing a brick is swung in a circle at constant speed in a vertical plane as shown. The bucket is swung fast enough that the brick does not fall out. The net force on the brick as it is swung has maximum magnitude at which point?

R

B

T

A. Top B. Right C. Bottom D. The net force magnitude is the same at all positions

Acceleration magnitude is always the same so the

net force magnitude must be the same (Newton’s 2nd law).

arad =v2

r!

"#

$

%&

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Clicker question 2 Set frequency to BA A bucket containing a brick is swung in a circle at constant speed in a vertical plane. The bucket is swung fast enough that the brick does not fall out. The magnitude of the normal force of the bucket on the brick is a minimum at which point?

R

B

T

A. The normal force is the same at all positions B. Top C. Right D. Bottom

is the same everywhere rva2

rad =

At the top: Fnet,r = n+mg =marad so n =marad −mg

At the right: Fnet,r = n =marad so n =marad

Fnet,r = n−mg =marad so n =marad +mgAt the bottom:

smallest

largest

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Need to specify the coordinate system and breakdown forces into components

Cave man with a sling A cave man needs to throw his 1.6 kg rock at a speed of 45 mph (20 m/s) to win the fastest rock contest. The maximum force he can apply is 100 N. How long a rope does he need to get the rock spinning around at that speed?

. The tension in the rope is 100 N. Need to find the length for a given mass & speed of the rock.

T

Fg =mg

Have the free body diagram r

zOur coordinate system:

We add an angle (currently unknown)

θ

So: θcosTTr =θsinTTz =

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Using our free body diagram and force components we can apply Newton’s 2nd law in both dimensions.

Cave man with a sling A cave man needs to throw his 1.6 kg rock at a speed of 45 mph (20 m/s) to win the fastest rock contest. With a 100 N tension, how long must his rope be to get the needed speed?

T

Fg =mg

rz

θ

Fnet,r = T cosθ

Fnet,z = T sinθ −mg and there is no motion in

the z direction so 0=zma mgT =θsinand so

and the acceleration is radial so rmvmar

2=

rmvT

2cos =θSo we have two equations: and mgT =θsin

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Cave man with a sling A cave man needs to throw his 1.6 kg rock at a speed of 45 mph (20 m/s) to win the fastest rock contest. With a 100 N tension, how long must his rope be to get the needed speed?

T

Fg =mgθ

rmvT

2cos =θSo we have two equations: and mgT =θsin

Can solve the second equation for θ

θ = sin−1 mgT

"

#$

%

&'= sin−1 1.6 kg ⋅9.8 m/s2

100 N"

#$

%

&'= 9.0°

Solve the first equation for r.

r = mv2

T cosθ=

1.6 kg ⋅ (20 m/s)2

100 N ⋅cos(9.0°)= 6.5 m

Is this the length of the rope?

No. Just the radius of the circle L = r

cosθ= 6.6 m

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Friction Friction occurs when two materials slide past one another

On a microscopic level, molecules in one material form bonds with molecules in the other material

Friction acts parallel to the surface (perpendicular to the normal force).

Friction only acts to oppose motion Not enough friction!