Demo spin stopper

• Is it accelerating?

• Does it have Fnet?

• Direction of Fnet?

• What causes Fnet?

• Direction of Fnet?

• Is tension doing work?

Newton’s Newton’s Laws of MotionLaws of Motion::

A net external force is necessary A net external force is necessary

to make an object accelerate follow a circular path.to make an object accelerate follow a circular path.

This force is called aThis force is called a

CENTRIPETALCENTRIPETAL ( (“center seeking“center seeking”) ”) FORCEFORCE..

It is the FIt is the Fnetnet for an object with curved motion. for an object with curved motion.

Circular or Curved Motion Equations

• vc = d/t

• = 2r/T

• ac = v2/r

• Fc = mac = mv2/r.

The term “Centripetal Force” decribes the direction of Fnet & acceleration. It is

NOT itself a force.

• Any force or forces can cause curved motion.

Centripetal force and acceleration Centripetal force and acceleration may be caused by:may be caused by:

• gravitygravity - -

• frictionfriction – –

• Tension in a rope or cordTension in a rope or cord--

• EM EM ––

• R (Fn) -R (Fn) -

1. Write an equation for centripetal acceleration in terms of r and T.

Substituting 2r for v in

T

accl eq. a = v2

r

ac = 42r

T2.

2. The distance from moon’s center to earth’s center = 3.8 x 108 m. Moon’s ac = 2.8 x 10-3 m/s2. What is moon’s period?

• 2.3 x 106 seconds

• ~ 27 days.

You feel acceleration!!That is because of your

inertia.

The car is turning due to the inward force, you feel as though you are being forced leftward or outward. The car is beginning its turning motion (to the right) while you continue in a straight line path.

DirectionVelocity Tangent to circular path.

Accl & force toward center of circular path.

Mud sticks to tire.

Sometimes a measured in g’s.Multiples of Earth’s a of gravity.

1g = 9.81 m/s2.2g = 19.6 m/s2.3g = 29.4 m/s2.

.

.

.etc.

3. An 80-kg astronaut experience a force of 2890-N when orbiting Earth. How many g’s does he feel?

• ac = F/m

• 2890 N / 80-kg = 36 m/s2.

• 36 m/s2 / 9.81 m/s2

• = 3.7 g.

Go to vertical circles.

Vertical Circles

Swing keys in vertical circle. Gravity Fg acts vertically.

Speed is less at top, greater at bottom.

Fc must be less at top, greater at bottom.

This is also true for rides which make vertical loops.

Fc is any force pointing into circle!!

At the top, Fc is combo of weight & Fn.

WFn

Taking center as +, Fc = mv2/r = mg + Fn.

Minimum v occurs just as Fn = 0, when you lose contact with your seat, you begin to fall.

Fc = mv2/r = mg + Fn When Fn = 0,

mv2/r = mg. So mass is irrelevant.

Rearrange, solve for vmin,

v = (gr)1/2. v depends on radius only.

R (FN) may also be called "apparent weight" this is the force of the seat on the rider and also describes what the rider "feels" .

If R is less than the rider's weight, rider will fall out. R might become negative, a safety restraint system -- seat belts, lap bars, shoulder restraints, are needed.

Tension in RopesSwinging Keys

To find the required tension at the top of the arc: Write a free body Fnet expression:

Fc = T + mg.Fc – mg = Tmv2/r – mg = T.

The tension is equivalent to R in the ride.

At the bottom the vel is max & Fc = T – W.

Fc = T – mg. mv2/r = T – mg.The necessary tension to swing in a circle is: mv2/r + mg = T.

W.

T.

4: A 10 kg package is swinging in a vertical circle of radius 5.0 m at the end of a rope. If the ac is 5.0 m/s2,

a. Sketch the free body diagram at the bottom of the loop.

b. what is the tension in the rope when the package is at the bottom of the loop?

• Fc = T – mg.

• mv2/r + mg = T.

• m (v2/r + g) = T.

• m(ac + g) = T.

• 10kg (5.0 m/s2 + 10 m/s2.)

• 150 N.

mg.

T.

5. A motorcyclist rides in a vertical circle of 20.0 m radius. What is the minimum speed he must have at the top to complete the loop?

14 m/s.

• Hwk Packet.

Banked Curves

Banked Curves.

On a flat road only friction acts as the Fc to pull car into curves.

Banked curves add a Fn component toward the

center of the circle.

If the curve is banked, there is a critical speed where friction =0, the car can travel the curve without slipping out of its circular path. 

Free Body Diagram Banked Curve

Normal Force Componentssin = Fc

ntan = Fc

mg

Fc

This is NOT the same vector sketch as for Fll.

Fc = mac points horizontally toward the center of the curve.

This component of the normal is supplying the Fc to keep the car moving through the banked curve.

At the critical speed, there is no need for any friction between the car and the road's surface. 

If the speed of the car were to exceed vcritical then the car would drift up the incline.

If the speed of the car is less than vcritical then the car would slip down the incline.

If friction is present both the horizontal component of N and the Friction force contribute to Fc.

FN tan θ + Ff = Fc.

IB Problem Packet Banked Curve

• http://www.youtube.com/watch?v=QmAdfLlhfzw