What causes UCM?

What causes UCM? UCM involves circular motion but at a constant speed.

What causes UCM? UCM involves circular motion but at a constant speed. What is the direction of the instantaneous acceleration at points A, B and D ? A B C D

What causes UCM? UCM involves circular motion but at a constant speed. The direction of the instantaneous acceleration at points A, B and D is “toward the centre”. A a B a C a D

What causes UCM? UCM involves circular motion but at a constant speed. The direction of the instantaneous acceleration at points A, B and D is “toward the centre”. A What is the direction of the net or unbalanced force at points A, B and D and why? a B a C a D

What causes UCM? UCM involves circular motion but at a constant speed. The direction of the instantaneous acceleration at points A, B and D is “toward the centre”. A Fnet The net force is also “towards the centre” because Newton's second law says that a and Fnet are always in the same direction. a=Fnet /m a Fnet B a C Fnet a D

What causes UCM? UCM involves circular motion but at a constant speed. The direction of the instantaneous acceleration at points A, B and D is “toward the centre”. A The net force is also “towards the centre” because Newton's second law says that a and Fnet are always in the same direction. a=Fnet /m Fnet a Fnet B a C Fnet The net force that keeps a body in UCM is always directed toward the centre and is therefore called a _____________ force. a D

What causes UCM? UCM involves circular motion but at a constant speed. The direction of the instantaneous acceleration at points A, B and D is “toward the centre”. A The net force is also “towards the centre” because Newton's second law says that a and Fnet are always in the same direction. a=Fnet /m Fnet a Fnet B a C Fnet The net force that keeps a body in UCM is always directed toward the centre and is therefore called a centripetal force. a D

Centripetal Force and Velocity

Centripetal Force and Velocity What is the direction of the instantaneous velocity at points A, B and D ? A Fnet a Fnet B a C up Fnet a D right

Centripetal Force and Velocity The directions of the instantaneous velocities at points A, B and D are given by the tangent-lines to the circle at those points. v A Fnet a Fnet B v a C up Fnet a D v right

Centripetal Force and Velocity The directions of the instantaneous velocities at points A, B and D are given by the tangent-lines to the circle at those points. v A The “net” or “centripetal” force for UCM is always 90° to the velocity. What effect does this “net” or “centripetal” force have on the object's velocity? Fnet a Fnet B v a C up Fnet a D v right

Centripetal Force and Velocity The directions of the instantaneous velocities at points A, B and D are given by the tangent-lines to the circle at those points. v A The “net” or “centripetal” force for UCM is always 90° to the velocity. This centripetal force only changes the direction of motion, but not its speed. Fnet a Fnet B v a C Fnet up a D v right

Is centripetal force a “Real” Force?

Is centripetal force a “Real” Force? Centripetal Force is not one of the four “real” forces of nature.

Is centripetal force a “Real” Force? Centripetal Force is not one of the four “real” forces of nature. It is just a special ______ force that keeps an object moving in uniform circular motion. (UCM)

Is centripetal force a “Real” Force? Centripetal Force is not one of the four “real” forces of nature. It is just a special net or unbalanced force that keeps an object moving in uniform circular motion. (UCM)

Is centripetal force a “Real” Force? Centripetal Force is not one of the four “real” forces of nature. It is just a special net or unbalanced force that keeps an object moving in uniform circular motion. (UCM) For UCM, the centripetal force must be supplied by one or more of the four real forces of nature.

What real force supplies the centripetal force for these cases? Case #1: An electron is revolving in UCM around a proton in a hydrogen atom. (simple planetary model) ? = Fc + Fnet or FC = ?

What real force supplies the centripetal force for these cases? Case #1: An electron is revolving in UCM around a proton in a hydrogen atom. (simple planetary model) Fel = Fc The electrical force of the proton supplies the net or centripetal force on the electron. + Fnet or FC = Fel

What real force supplies the centripetal force for these cases? Case #1: An electron is revolving in UCM around a proton in a hydrogen atom. (simple planetary model) Fel = Fc The electrical force of the proton supplies the net or centripetal force on the electron. + Fnet or FC = Fel This is a vector statement and a good starting point in centripetal force problems

What real force supplies the centripetal force for these cases? Case #2: The moon revolves around the earth in approximate UCM every 27.3 days. ? = Fc Moon Earth Fnet or FC = ?

What real force supplies the centripetal force for these cases? Case #2: The moon revolves around the earth in approximate UCM every 27.3 days. Fg = Fc The gravitational force of the earth supplies the net or centripetal force on the moon. Moon Earth Fnet or FC = Fg

What real force supplies the centripetal force for these cases? Case #2: The moon revolves around the earth in approximate UCM every 27.3 days. Fg = Fc The gravitational force of the earth supplies the net or centripetal force on the moon. Moon Earth Fnet or FC = Fg This is a vector statement and a good starting point in centripetal force problems

What real force supplies the centripetal force for these cases? Case #3: A car rounds a unbanked flat circular curve at a constant speed. (See side view) car road Centre of curve Fnet or FC = ? Flat road surface ? = Fc

What real force supplies the centripetal force for these cases? Case #3: A car rounds a unbanked flat circular curve at a constant speed. (See side view) road car Fnet or FC = fs Centre of curve Flat road surface fs = Fc The static friction force of the road supplies the net or centripetal force on the car.

Centripetal Force Formulas

Centripetal Force Formulas Review☺: The centripetal force is just a special _______ force to keep something in UCM.

Centripetal Force Formulas Review☺: The centripetal force is just a special net force to keep something in UCM.

Centripetal Force Formulas Review☺: The centripetal force is just a special net force to keep something in UCM. In terms of mass and acceleration, what is Fnet ?

Centripetal Force Formulas Review☺: The centripetal force is just a special net force to keep something in UCM. │Fnet │= m │ ac │

Centripetal Force Formulas Review☺: The centripetal force is just a special net force to keep something in UCM. │Fnet │= m │ ac │ Fc = m ( ? ) in terms of v and r ??

Centripetal Force Formulas Review☺: The centripetal force is just a special net force to keep something in UCM. │Fnet │= m │ ac │ Fc = m v2/r Equation #1

Centripetal Force Formulas Review☺: The centripetal force is just a special net force to keep something in UCM. │Fnet │= m │ ac │ Fc = m v2/r Equation #1 Fc = m ( ? ) in terms of T and r ??

Centripetal Force Formulas Review☺: The centripetal force is just a special net force to keep something in UCM. │Fnet │= m │ ac │ Fc = m v2/r Equation #1 Fc = m 4 π2 r / T2 Equation #2

Centripetal Force Formulas Review☺: The centripetal force is just a special net force to keep something in UCM. │Fnet │= m │ ac │ Fc = m v2/r Equation #1 Fc = m 4 π2 r / T2 Equation #2 Fc = m ( ? ) in terms of f and r ??

Centripetal Force Formulas Review☺: The centripetal force is just a special net force to keep something in UCM. │Fnet │= m │ ac │ Fc = m v2/r Equation #1 Fc = m 4 π2 r / T2 Equation #2 Fc = m 4 π2 r f2 Equation #3

Centripetal Force Formulas Review☺: The centripetal force is just a special net force to keep something in UCM. │Fnet │= m │ ac │ Fc = m v2/r Equation #1 Fc = m 4 π2 r / T2 Equation #2 Fc = m 4 π2 r f2 Equation #3 Memorize please!

Ex #1: A 1525 kg car is moving around a circular track of radius 300 Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car

Ex #1: A 1525 kg car is moving around a circular track of radius 300 Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: ?

Ex #1: A 1525 kg car is moving around a circular track of radius 300 Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: ? = 1525 kg

Ex #1: A 1525 kg car is moving around a circular track of radius 300 Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: m = 1525 kg

Ex #1: A 1525 kg car is moving around a circular track of radius 300 Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: m = 1525 kg ? = 72.00 km/h

Ex #1: A 1525 kg car is moving around a circular track of radius 300 Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: m = 1525 kg v = 72.00 km/h

Ex #1: A 1525 kg car is moving around a circular track of radius 300 Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: m = 1525 kg v = 72.00 km/h = ?

Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s

Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s ? = 300.0 m Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s ? = 300.0 m

Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m

Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Unknown: ? Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Unknown: ?

Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Unknown: Fc = ?

Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Unknown: Fc = ? Formula: ?

Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Unknown: Fc = ? Formula: Fc = ?

Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Unknown: Fc = ? Formula: Fc = mv2/r

Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Unknown: Fc = ? Formula: Fc = mv2/r Sub: ?

Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Unknown: Fc = ? Formula: Fc = mv2/r Sub: Fc = (1525)(20.0)2/(300.0) = ?

Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Unknown: Fc = ? Formula: Fc = mv2/r Sub: Fc = (1525)(20.0)2/(300.0) = 2033 N

Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Unknown: Fc = ? Formula: Fc = mv2/r Sub: Fc = (1525)(20.0)2/(300.0) = 2033 N b) ?

Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Ex #1: A 1525 kg car is moving around a circular track of radius 300.0 m at a constant speed of 72.00 km/h. Find: a) the magnitude of the centripetal force required b) the real force that supplies the centripetal force on the car Given: m = 1525 kg v = 72.00 km/h = 20.0 m/s r = 300.0 m Unknown: Fc = ? Formula: Fc = mv2/r Sub: Fc = (1525)(20.0)2/(300.0) = 2033 N b) The static friction of the road on the car supplies the centripetal force and is 2033 N.

Try. : The earth has a mass of 6. 0 x 1024 kg Try!: The earth has a mass of 6.0 x 1024 kg. It goes around the sun in approximate UCM with an orbital radius of 1.5 X 1011 m. The orbital period is 365.25 days. Find: a) the size of the centripetal force b) the real force that supplies Fc c) the mass of the sun ( G = 6.67 X 10-11 mks units )

Try. : The earth has a mass of 6. 0 x 1024 kg Try!: The earth has a mass of 6.0 x 1024 kg. It goes around the sun in approximate UCM with an orbital radius of 1.5 X 1011 m. The orbital period is 365.25 days. Find: a) the size of the centripetal force b) the real force that supplies Fc c) the mass of the sun ( G = 6.67 X 10-11 in mks units ) Given: m = 6.0 x 1024 kg r = 1.5 X 1011 m T = 365.25 d X 24 h/d X 3600 s/ h =3.2 X 107 seconds Unknown: Fc = ? Formula: Fc = m 4 π2 r / T2 Fc = (6.0 x 1024 kg) 4 π2 (1.5 X 1011 m) / (3.2 X 107 s)2 = 3.5 X 1022 N The sun exerts a gravitational force of 3.5 X 1022 N on the earth and supplies the centripetal force that keeps the earth in UCM. Fg = GMm/r2 so M = Fgr2/Gm = ( 3.5 X 1022 ) (1.5 X 1011 )2 / (6.0 x 1024 X 6.67 X 10-11) mks Fg = 2.0 X 1030 kg

What real forces supply the centripetal force to keep the mass in UCM? UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s What real forces supply the centripetal force to keep the mass in UCM? m = 5.00 kg String of r = 1.60 m v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s The tension force of the string and the earth's force of gravity provide the centripetal force to keep the mass in UCM. m = 5.00 kg String of r = 1.60 m v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s The tension force of the string and the earth's force of gravity provide the centripetal force to keep the mass in UCM. a) Find the magnitude of the centripetal force needed to keep the mass in UCM. m = 5.00 kg String of r = 1.60 m v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s The tension force of the string and the earth's force of gravity provide the centripetal force to keep the mass in UCM. a) Find the magnitude of the centripetal force needed to keep the mass in UCM. Fc = mv2/r = 5(6)2 /1.6 = 113 N m = 5.00 kg String of r = 1.60 m v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s b) Find the tension force of the string at the top of the circle. m = 5.00 kg String of r = 1.60 m v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s b) Find the tension force of the string at the top of the circle. Step 1: ? m = 5.00 kg String of r = 1.60 m v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s b) Find the tension force of the string at the top of the circle. Step 1: Draw an FBD at the top of the circle. m = 5.00 kg String of r = 1.60 m v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s b) Find the tension force of the string at the top of the circle. Step 1: Draw an FBD at the top of the circle. m = 5.00 kg String of r = 1.60 m T Fg a v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s b) Find the tension force of the string at the top of the circle. Step 1: Draw an FBD at the top of the circle. m = 5.00 kg String of r = 1.60 m T Fg = ? a v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s b) Find the tension force of the string at the top of the circle. Step 1: Draw an FBD at the top of the circle. m = 5.00 kg String of r = 1.60 m T Fg = 50.0 N a v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s b) Find the tension force of the string at the top of the circle. Step 1: Draw an FBD at the top of the circle. Step 2: ? m = 5.00 kg String of r = 1.60 m T Fg = 50 N a v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s b) Find the tension force of the string at the top of the circle. Step 1: Draw an FBD at the top of the circle. Step 2: Write a vector statement of the real forces supplying the centripetal force. m = 5.00 kg String of r = 1.60 m T Fg = 50 N a v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s b) Find the tension force of the string at the top of the circle. Step 1: Draw an FBD at the top of the circle. Step 2: Write a vector statement of the real forces supplying the centripetal force. T + Fg = Fc m = 5.00 kg String of r = 1.60 m T Fg = 50 N a v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s b) Find the tension force of the string at the top of the circle. Step 1: Draw an FBD at the top of the circle. Step 2: Write a vector statement of the real forces supplying the centripetal force. T + Fg = Fc ? Step 3 Scalar statement m = 5.00 kg String of r = 1.60 m T Fg = 50 N a v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s b) Find the tension force of the string at the top of the circle. Step 1: Draw an FBD at the top of the circle. Step 2: Write a vector statement of the real forces supplying the centripetal force. T + Fg = Fc T - 50 = - 113 Step 3 Scalar statement m = 5.00 kg String of r = 1.60 m T Fg = 50 N a v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s b) Find the tension force of the string at the top of the circle. Step 1: Draw an FBD at the top of the circle. Step 2: Write a vector statement of the real forces supplying the centripetal force. T + Fg = Fc T - 50 = - 113 Step 3 Scalar statement T = ? m = 5.00 kg String of r = 1.60 m T Fg = 50 N a v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s b) Find the tension force of the string at the top of the circle. Step 1: Draw an FBD at the top of the circle. Step 2: Write a vector statement of the real forces supplying the centripetal force. T + Fg = Fc T - 50 = - 113 Step 3 Scalar statement T = 50 – 113 = ? m = 5.00 kg String of r = 1.60 m T Fg = 50 N a v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s b) Find the tension force of the string at the top of the circle. Step 1: Draw an FBD at the top of the circle. Step 2: Write a vector statement of the real forces supplying the centripetal force. T + Fg = Fc T - 50 = - 113 Step 3 Scalar statement T = 50 – 113 = - 63.0 N m = 5.00 kg String of r = 1.60 m T Fg = 50 N a v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s b) Find the tension force of the string at the top of the circle. Step 1: Draw an FBD at the top of the circle. Step 2: Write a vector statement of the real forces supplying the centripetal force. T + Fg = Fc T - 50 = - 113 Step 3 Scalar statement T = 50 – 113 = - 63.0 N At the top of the circle, the tension force of the string on the mass is 63.0 N [down] m = 5.00 kg String of r = 1.60 m T Fg = 50 N a v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s c) Find the tension force of the string at the bottom of the circle. m = 5.00 kg String of r = 1.60 m v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s c) Find the tension force of the string at the bottom of the circle. m = 5.00 kg String of r = 1.60 m T Fg a v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s c) Find the tension force of the string at the bottom of the circle. m = 5.00 kg String of r = 1.60 m T Fg = ? a v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s c) Find the tension force of the string at the bottom of the circle. m = 5.00 kg String of r = 1.60 m T Fg = 50 N a v= 6.00 m/s Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s c) Find the tension force of the string at the bottom of the circle. m = 5.00 kg String of r = 1.60 m T Fg = 50 N a v= 6.00 m/s Vector Statement ? Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s c) Find the tension force of the string at the bottom of the circle. m = 5.00 kg String of r = 1.60 m T Fg = 50 N a v= 6.00 m/s Vector Statement T + Fg = Fc Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s c) Find the tension force of the string at the bottom of the circle. m = 5.00 kg String of r = 1.60 m T Fg = 50 N a v= 6.00 m/s Vector Statement T + Fg = Fc Scalar statement ? Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s c) Find the tension force of the string at the bottom of the circle. m = 5.00 kg String of r = 1.60 m T Fg = 50 N a v= 6.00 m/s Vector Statement T + Fg = Fc Scalar statement T – 50 = 113 Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s c) Find the tension force of the string at the bottom of the circle. m = 5.00 kg String of r = 1.60 m T Fg = 50 N a v= 6.00 m/s Vector Statement T + Fg = Fc Scalar statement T – 50 = 113 T = ? Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s c) Find the tension force of the string at the bottom of the circle. m = 5.00 kg String of r = 1.60 m T Fg = 50 N a v= 6.00 m/s Vector Statement T + Fg = Fc Scalar statement T – 50 = 113 T = 50 + 113 Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s c) Find the tension force of the string at the bottom of the circle. m = 5.00 kg String of r = 1.60 m T Fg = 50 N a v= 6.00 m/s Vector Statement T + Fg = Fc Scalar statement T – 50 = 113 T = 50 + 113 T = 163 N Y+ up X + right

UCM Dynamics in a Vertical Plane: Using a string, a mass of 5 UCM Dynamics in a Vertical Plane: Using a string, a mass of 5.00 kg is twirled in a circle of radius 1.60 m on a vertical plane with a constant speed of 6.00 m/s c) Find the tension force of the string at the bottom of the circle. m = 5.00 kg String of r = 1.60 m T Fg = 50 N a v= 6.00 m/s Vector Statement T + Fg = Fc Scalar statement T – 50 = 113 T = 50 + 113 T = 163 N At the bottom of the circle, the tension force of the string on the mass is 163 N [up] Y+ up X + right

Try this yourself: Using a string, a mass of 8 Try this yourself: Using a string, a mass of 8.00 kg is twirled in a circle of radius 0.900 m on a vertical plane with a constant speed of 4.00 m/s a) Find the magnitude of the centripetal force required for UCM b) Find the tension force of the string... i) at the bottom of the circle ii) at the top of the circle m = 8.00 kg String of r = 0.900 m v= 4.00 m/s Y+ up X + right

Try this yourself: Using a string, a mass of 8 Try this yourself: Using a string, a mass of 8.00 kg is twirled in a circle of radius 0.900 m on a vertical plane with a constant speed of 4.00 m/s a) Find the magnitude of the centripetal force required for UCM b) Find the tension force of the string... i) at the bottom of the circle ii) at the top of the circle m = 8.00 kg String of r = 0.900 m a) Fc = mv2/r = 8(4)2/0.9 = 142 N v= 4.00 m/s Y+ up X + right

Try this yourself: Using a string, a mass of 8 Try this yourself: Using a string, a mass of 8.00 kg is twirled in a circle of radius 0.900 m on a vertical plane with a constant speed of 4.00 m/s a) Find the magnitude of the centripetal force required for UCM b) Find the tension force of the string... i) at the bottom of the circle ii) at the top of the circle m = 8.00 kg String of r = 0.900 m a) Fc = mv2/r = 8(4)2/0.9 = 142 N T b) part i) T + Fg= Fc T – 80 = 142 T = 222 N [up] a v= 4.00 m/s Fg= 80 N Y+ up c) part ii) T + Fg= Fc T – 80 = -142 T = - 62 = 62 N [d] X + right T Fg= 80 N a

Minimum speed example: Using a string, an unknown mass is twirled in a circle of radius 0.900 m on a vertical plane with just enough speed to go around in UCM. Find the minimum speed required. m = ? kg String of r = 0.900 m v= ? Y+ up X + right

Minimum speed example: Using a string, an unknown mass is twirled in a circle of radius 0.900 m on a vertical plane with just enough speed to go around in UCM. Find the minimum speed required. Strategy: Focus at the top of the circle. It is at that point that tension has it minimum value. When moving at minimum speed to go around in UCM, the tension force will be zero at the top of the circle. m = ? kg String of r = 0.900 m v= ? Y+ up X + right

Minimum speed example: Using a string, an unknown mass is twirled in a circle of radius 0.900 m on a vertical plane with just enough speed to go around in UCM. Find the minimum speed required. Strategy: Focus at the top of the circle. It is at that point that tension has it minimum value. When moving at minimum speed to go around in UCM, the tension force will be zero at the top of the circle. m = ? kg String of r = 0.900 m FBD = ? v= ? Y+ up X + right

Minimum speed example: Using a string, an unknown mass is twirled in a circle of radius 0.900 m on a vertical plane with just enough speed to go around in UCM. Find the minimum speed required. Strategy: Focus at the top of the circle. It is at that point that tension has it minimum value. When moving at minimum speed to go around in UCM, the tension force will be zero at the top of the circle. m = ? kg String of r = 0.900 m v= ? a Y+ up Fg = ? X + right

Minimum speed example: Using a string, an unknown mass is twirled in a circle of radius 0.900 m on a vertical plane with just enough speed to go around in UCM. Find the minimum speed required. Strategy: Focus at the top of the circle. It is at that point that tension has it minimum value. When moving at minimum speed to go around in UCM, the tension force will be zero at the top of the circle. m = ? kg String of r = 0.900 m v= ? a Y+ up Fg = mg X + right

Minimum speed example: Using a string, an unknown mass is twirled in a circle of radius 0.900 m on a vertical plane with just enough speed to go around in UCM. Find the minimum speed required. Strategy: Focus at the top of the circle. It is at that point that tension has it minimum value. When moving at minimum speed to go around in UCM, the tension force will be zero at the top of the circle. m = ? kg String of r = 0.900 m Vector Statement ? v= ? a Y+ up Fg = mg X + right

Minimum speed example: Using a string, an unknown mass is twirled in a circle of radius 0.900 m on a vertical plane with just enough speed to go around in UCM. Find the minimum speed required. Strategy: Focus at the top of the circle. It is at that point that tension has it minimum value. When moving at minimum speed to go around in UCM, the tension force will be zero at the top of the circle. m = ? kg String of r = 0.900 m Fg = FC v= ? a Y+ up Fg = mg X + right

Minimum speed example: Using a string, an unknown mass is twirled in a circle of radius 0.900 m on a vertical plane with just enough speed to go around in UCM. Find the minimum speed required. Strategy: Focus at the top of the circle. It is at that point that tension has it minimum value. When moving at minimum speed to go around in UCM, the tension force will be zero at the top of the circle. m = ? kg String of r = 0.900 m Fg = FC Scalar statement = ? v= ? a Y+ up Fg = mg X + right

Minimum speed example: Using a string, an unknown mass is twirled in a circle of radius 0.900 m on a vertical plane with just enough speed to go around in UCM. Find the minimum speed required. Strategy: Focus at the top of the circle. It is at that point that tension has it minimum value. When moving at minimum speed to go around in UCM, the tension force will be zero at the top of the circle. m = ? kg String of r = 0.900 m Fg = FC -mg = -mv2/r v= ? a Y+ up Fg = mg X + right

Minimum speed example: Using a string, an unknown mass is twirled in a circle of radius 0.900 m on a vertical plane with just enough speed to go around in UCM. Find the minimum speed required. Strategy: Focus at the top of the circle. It is at that point that tension has it minimum value. When moving at minimum speed to go around in UCM, the tension force will be zero at the top of the circle. m = ? kg String of r = 0.900 m Fg = FC -mg = -mv2/r v = ? v= ? a Y+ up Fg = mg X + right

Minimum speed example: Using a string, an unknown mass is twirled in a circle of radius 0.900 m on a vertical plane with just enough speed to go around in UCM. Find the minimum speed required. Strategy: Focus at the top of the circle. It is at that point that tension has it minimum value. When moving at minimum speed to go around in UCM, the tension force will be zero at the top of the circle. m = ? kg String of r = 0.900 m Fg = FC -mg = -mv2/r v = ( r g )1/2 v= ? a Y+ up Fg = mg X + right

Minimum speed example: Using a string, an unknown mass is twirled in a circle of radius 0.900 m on a vertical plane with just enough speed to go around in UCM. Find the minimum speed required. Strategy: Focus at the top of the circle. It is at that point that tension has it minimum value. When moving at minimum speed to go around in UCM, the tension force will be zero at the top of the circle. m = ? kg String of r = 0.900 m Fg = FC -mg = -mv2/r v = ( r g )1/2 v = ( 0.900 X 10 )1/2 v= ? a Y+ up Fg = mg X + right

Minimum speed example: Using a string, an unknown mass is twirled in a circle of radius 0.900 m on a vertical plane with just enough speed to go around in UCM. Find the minimum speed required. Strategy: Focus at the top of the circle. It is at that point that tension has it minimum value. When moving at minimum speed to go around in UCM, the tension force will be zero at the top of the circle. m = ? kg String of r = 0.900 m Fg = FC -mg = -mv2/r v = ( r g )1/2 v = ( 0.900 X 10 )1/2 v = 3.00 m/s v= ? a Y+ up Fg = mg X + right

Minimum speed example: Using a string, an unknown mass is twirled in a circle of radius 0.900 m on a vertical plane with just enough speed to go around in UCM. Find the minimum speed required. Strategy: Focus at the top of the circle. It is at that point that tension has it minimum value. When moving at minimum speed to go around in UCM, the tension force will be zero at the top of the circle. m = ? kg String of r = 0.900 m Fg = FC -mg = -mv2/r v = ( r g )1/2 v = ( 0.900 X 10 )1/2 v = 3.00 m/s v= ? a Y+ up Fg = mg X + right The minimum speed for the mass to undergo UCM is 3.00 m/s. Note the formula only depends on the radius and not the mass.

Critical Thinking Problems Critical Thinking: will take-up in class An airplane is moving in “loop the loop” in a vertical circle with radius 500.0 m and a constant speed of 540.0 km/h. At the bottom of the loop, the 80.0 kg pilot has an apparent weight not equal to her actual weight of 800.0 N. What is her apparent weight? (Ans 4400 N or 5.5 times her actual weight) Using a string, a mass “ m “ kg is twirled in a circle of radius “ r “ m on a vertical plane with a constant speed of “ v ” m/s. As it twirls, the string makes an angle Θ with a vertical line from the center to the top of the circle. Derive or create a formula to find the tension force in terms of m, g , v and Θ. We will take up in class. The sun has a mass of 2.0 X 1030 kg. The earth orbits the sun in approximate UCM with an orbital radius of 1.5 X 1011 m. Find the orbital speed of the earth. G = 6.67 x 10-11 Ans 30 km/s More practice next slide.

Centripetal Force Practice Read over your notes on this lesson or review the power point presentation on Moodle. Do the centripetal force handouts. New textbook: p123 Q1,Q3,Q4 ans same page p124 Q1,Q2, Q4,Q6 ans p715 Old textbook: p 133 Q3,Q4, Q6, Q7, Q8 ans same page p 138 Q2-Q7 ans p782