THE WORLD SEEMS TO BE SPINNING OUT OF CONTROL

Rules of This Activity#1 All students must perform all of

the work of the activity in their science journals.

#2 When you are sure of your answer, call me over to show me your work.

#3 If your answer is correct, become a student teacher to assist others with understanding of the concept.

Stating the Problem

The bank teller has been spun around due to a tornado. Is the bank teller rotating or revolving? What is the difference?

Working the Problem

He is rotating about an axis within the body. Revolution is around an axis external to the body.

Working the Problem

The 1500 kg Batallac travels south at 25 m/s along a curved road. If it takes 20 seconds to change directions by 90 degrees, what is T?

What is T?

(20 s) (4) = T 80 s = T

Cycle/Period = 1/T Known Wantm = 1700 kg (includes occupants) T t = 20 s 90* = ¼ cycle

Stating the Problem

The 1500 kg Batallac travels south at 25 m/s along a curved road. If it takes 20 seconds to change directions by 90 degrees, what is the radius of the curve in the road?

What is the radius of the road?

F = mv2/r and F = m4π2r/T2

Known Wantm = 1700 kg (includes occupants) r =

318m v = 25.00 m/sT = 80 s mv2/r = m4π2r/T2

v2/r = 4π2r/T2

v2 = 4π2r2/T2

r = √T2v2/4π2 r = √(80)2(25)2/4π2

Working the Problem

If an F1 tornado was generated at 144.84 km/h (able to over turn a car!) and has a diameter of 10 meters at the center of its vortex, what is T?

What is T?V = 2πr/T

Known Want

V = 144.84 km/h T r = 5 mConvert km to m and h to s V = 2πr/T T = 2πr/v T = 31.42/40.23 T = .78s

Stating the Problem

What is the centripetal acceleration of the tornado that was generated?

Working the Problem

ac = 4π2r/T2

Known WantV = 40.23m/s

ac

r = 5 mt = .78 s

ac = 197.39/(.78s)2

ac = 324.44 m/s2

Stating the Problem

If this TV reporter weighs 490N, what is the centripetal force he is experiencing due to the tornado?

Working the Problem

Fc = m ac

Known Wantm = 50 kg Fc

ac = 324.44 m/s2

Fc = (50kg)(324.44 m/s2)

Fc = 16222 N

Stating the Problem

In terms of angular momentum, in which direction is the wind inside the tornado moving, right to left or left to right? How do you know?

Working the Problem

It must be moving from left to right. Using the right hand rule on wind movement from left to right, your right thumb would point upward. That is the direction that the taxi is being moved.

Stating the Problem

The Batallac is lifted 30 meters in the air by a hoist with a radius of 10 meters. If it the Batallac is traveling at 6 rpm, what is its tangential velocity?

Working the Problem

V = 2πr/T Known Want Rpm = 6 V r = 10 m(6 rotations/min)(min/60sec) = 6/60 = 1/10, T =

10s

V = 2πr/TV = 6.28 m/s

Stating the Problem

Batfink is being spun around at the surface of the Earth by a tornado. If his mass is 50 kg, his velocity is 25 m/s and the radius from the tip of his wing to the center of his body is 3 meters, what is his angular momentum?

Working the Problem

L = mvr Known Wantm = 50 kg L r = 3 mV= 25.00 m/s

L = (50)(25)(3)L = 3750 kgm2/s

Stating the Problem

The bad guy wants to spin Batfink into the ground, so he pins his wings against his body with rope. Is this a smart idea? Why?

Working the Problem

Yes it is smart, because it will cause Batfink to spin faster.

Stating the Problem

What is Batfink’s new velocity if the radius was decreased from 3 meters to 1 meter by the rope?

Working the Problem

L(before) = L(after)

L = mvr Known

WantL = 3750 kgm2/s V

3750 kgm2/s = mvr3750 kgm2/s = (50)(v)(1)

V = 75 m/sBatfink’s tangential velocity increased

by 3 times!

Stating the Problem

The tornado is causing a 8 kg dog to be swung around a 1.5 meter rope in a horizontal circle. The dog moves at a rate of 1.25 revolutions per second. What is the tangential velocity of the dog?

Working the Problem V = 2πr/T

Known Wantr = 1.5 m vm = 8 kgRPS = 1.25(1.25 revolutions/1s)/(1.00/1.25) = .8s

V = (6.28)(1.5)/.8V = 11.78 m/s

Stating the Problem

What is the dogs centripetal acceleration?

Working the Problem

ac = v2/r

Known Wantv = 11.78 m/s ac

r = 1.5 mac = v2/r

ac = 138.77/1.5

ac = 92.51 m/s2

Stating the Problem

Determine the tension that was in the rope as the dog was being spun around in the air.

Working the Problem

F = ma Known Wantm = 8 kg F ac = 92.51 m/s2

F = (8)(92.51)F = 740 N

Stating the Problem

Determine the dog’s angular momentum.

Working the Problem

L = mvr Known Wantm = 8 kg L

v = 11.78 m/s r = 1.5 m

L = (8)(11.78)(1.50)L = 141.36 kgm2/s

Stating the Problem

Where in the vortex would you expect the tangential velocity in the tornado to be the greatest? Why?

Working the Problem

You would probably expect it to be at the top because of the increased radius, however, the bottom of the vortex has the highest winds because barometric pressure increases. Winds near the top move in a circular pattern but winds near the bottom are spiraling downward. A tornado is not on a fixed platform! Doesn’t that just blow your mind?

Don’t forget to study about Universal Gravitation for the Exam!