Uniform Circular Motion Centripetal Force Torque El Paso Independent School District

Uniform Circular Motion Uniform circular motion is motion along a circular path in which there is no change in speed, only a change in direction. Constant velocity V tangent to path. v Fc Constant force Fc toward center. Question: Is there an outward force on the ball?

Uniform Circular Motion (Cont.) The question of an outward force can be resolved by asking what happens when the string breaks! Ball moves tangent to path, NOT outward as might be expected. v When central force is removed, ball continues in straight line. Centripetal force is needed to change direction.

Examples of Centripetal Force You are sitting on the seat next to the outside door. What is the direction of the resultant force on you as you turn? Is it away from center or toward center of the turn? Fc Car going around a curve. Force ON you is toward the center.

Deriving Central Acceleration Consider initial velocity at A and final velocity at B: vf vf B R vo -vo Dv R vo s A PAP

Deriving Acceleration (Cont.) vf -vo R vo Dv d ac = Dv t Definition: = Dv vf Similar Triangles mass m = Dv t v2 R ac = Centripetal acceleration: PAP

v m m = 3 kg R R = 5 m; v = 8 m/s ac = 12.8 m/s2 F = (3 kg)(12.8 m/s2) Example 1: A 3-kg rock swings in a circle of radius 5 m. If its constant speed is 8 m/s, what is the centripetal acceleration? R v m m = 3 kg R = 5 m; v = 8 m/s ac = 12.8 m/s2 F = (3 kg)(12.8 m/s2) Fc = 38.4 N

Example 2: A skater moves with 15 m/s in a circle of radius 30 m Example 2: A skater moves with 15 m/s in a circle of radius 30 m. The ice exerts a central force of 450 N. What is the mass of the skater? Draw and label a sketch Solve for mass(m) F = mv2/R FR = mv2 FR/v2 = m, substitute m=(450N)(30m)/(15m/s)2 450 N 30 m v = 15 m/s R Fc m=? Speed skater m = 60.0 kg

Car Negotiating a Flat Turn The centripetal force Fc is that of static friction fs: R v m Fc n Fc = fs fs R mg The central force FC and the friction force fs are not two different forces that are equal. There is just one force on the car. The nature of this central force is static friction. PAP PAP

Finding the maximum speed for negotiating a turn without slipping. Fc = fs n mg fs R R v m Fc The car is on the verge of slipping when FC is equal to the maximum force of static friction fs. Fc = mv2 R fs = msmg Fc = fs

Maximum speed without slipping (Cont.) Fc = fs n mg fs R v m Fc mv2 R = msmg v = msgR Velocity v is maximum speed for no slipping.

v R m Fc ms = 0.7 Fc = mv2 R fs = msmg From which: v = msgR Example 4: A car negotiates a turn of radius 70 m when the coefficient of static friction is 0.7. What is the maximum speed to avoid slipping? R v m Fc ms = 0.7 Fc = mv2 R fs = msmg From which: v = msgR g = 9.8 m/s2; R = 70 m v = 21.9 m/s

Motion in a Vertical Circle Fc = mv2 R Resultant force toward center T mg Consider TOP of circle: mg + T = mv2 R AT TOP: + T mg T = - mg mv2 R

Vertical Circle; Mass at bottom Fc = mv2 R Resultant force toward center T mg Consider bottom of circle: T - mg = mv2 R AT Bottom: T mg + T = + mg mv2 R

Visual Aid: Assume that the centripetal force required to maintain circular motion is 20 N. Further assume that the weight is 5 N. R v Resultant central force FC at every point in path! FC = 20 N Weight vector W is downward at every point. W = 5 N, down FC = 20 N at top AND at bottom.

FC = 20 N at top AND at bottom. Visual Aid: The resultant force (20 N) is the vector sum of T and W at ANY point in path. R v FC = 20 N at top AND at bottom. Top: T + W = FC WT + T + 5 N = 20 N T = 20 N - 5 N = 15 N + T W Bottom: T - W = FC T - 5 N = 20 N T = 20 N + 5 N = 25 N

For Motion in Circle v AT TOP: mv2 R + T = - mg R mg T AT BOTTOM: mv2

mg + T = mv2 R At Top: R v T T = - mg mv2 R T = 5.40 N Example 6: A 2-kg rock swings in a vertical circle of radius 8 m. The speed of the rock as it passes its highest point is 10 m/s. What is tension T in rope? mg + T = mv2 R At Top: R v T mg T = - mg mv2 R T = 5.40 N T = 25 N - 19.6 N

T - mg = mv2 R At Bottom: R v T T = + mg mv2 R T = 44.6 N Example 6 (cont): A 2-kg rock swings in a vertical circle of radius 8 m. The speed of the rock as it passes its lowest point is 10 m/s. What is tension T in rope? T - mg = mv2 R At Bottom: R v T mg T = + mg mv2 R T = 44.6 N T = 25 N + 19.6 N

mg + T = mv2 R At Top: R v T vc occurs when T = 0 mg = mv2 R vc = gR Example 7: What is the critical speed vc at the top, if the 2-kg mass is to continue in a circle of radius 8 m? mg + T = mv2 R At Top: R v T mg vc occurs when T = 0 mg = mv2 R vc = gR vc = 8.85 m/s v = gR = (9.8 m/s2)(8 m)

Definition of Torque Torque is defined as the tendency to produce a change in rotational motion. Examples:

Units for Torque Torque is proportional to the magnitude of F and to the distance r from the axis. Thus, a tentative formula might be: t = Fr Units: Nm or lbft t = (40 N)(0.60 m) = 24.0 Nm, cw 6 cm 40 N t = 24.0 Nm, cw

Torque is determined by Three Factors: The magnitude of the applied force. The direction of the applied force. The location of the applied force. 20 N Location of force The forces nearer the end of the wrench have greater torques. 20 N Magnitude of force 40 N The 40-N force produces twice the torque as does the 20-N force. Each of the 20-N forces has a different torque due to the direction of force. 20 N Direction of Force q

Torque is a vector quantity that has direction as well as magnitude. Direction of Torque Torque is a vector quantity that has direction as well as magnitude. Turning the handle of a screwdriver clockwise and then counterclockwise will advance the screw first inward and then outward.

Sign Convention for Torque By convention, counterclockwise torques are positive and clockwise torques are negative. ccw Positive torque: Counter-clockwise, out of page cw Negative torque: clockwise, into page

Torque = force x moment arm Calculating Torque Read problem and draw a rough figure. Extend line of action of the force. Draw and label moment arm. Calculate the moment arm if necessary. Apply definition of torque: t = Fr Torque = force x moment arm

Extend line of action, draw, calculate r. Example 1: An 80-N force acts at the end of a 12-cm wrench as shown. Find the torque. Extend line of action, draw, calculate r. r = 12 cm sin 600 = 10.4 cm t = (80 N)(0.104 m) = 8.31 N m

Alternate: An 80-N force acts at the end of a 12-cm wrench as shown Alternate: An 80-N force acts at the end of a 12-cm wrench as shown. Find the torque. positive 12 cm Resolve 80-N force into components as shown. Note from figure: rx = 0 and ry = 12 cm t = (69.3 N)(0.12 m) t = 8.31 N m as before

Universal Law of Gravitation

This is Newton’s Universal Law of Gravitation So because of Newton’s 3rd law every body in the universe exerts a force of attraction on every other body. This is Newton’s Universal Law of Gravitation The force between two objects, due to their masses, is called the gravitational force (Fg)- in this case it’s not Earth specific and is not 9.81 m/s2).

Universal Gravitational Constant- 6.67 x10-11 Masses of the 2 objects Distance between the objects

Example 1:What is the gravitational force between the Earth and the Moon? mEarth = M = 6.0 x 1024 kg mMoon = m = 7.4 x 1022 kg r = 3.8 x 108 m G = 6.67 x 10-11

Example 2: What is the gravitational force between the Earth and Venus? mEarth = M = 6.0 x 1024 kg mVenus = m = 5.0 x 1024 kg r = 3.8 x 1010 m G = 6.67 x 10-11 Answer: F = 1.386x1018 N

There is also a way to determine the gravitational field around one object: This is now the gravitational Field Strength(GFS)

Example 3: What is the Gravitational Field Strength in Earth? Radius of the Earth – 6.37 x 106 m

Example 4: What is the Gravitational field strength on the moon? Radius of the Moon – 1.7 x 106 m

What if the distance between the Earth and the Moon was doubled? What if the distance between the Earth and the Moon was tripled? What if the distance between the Earth and the Moon was quadrupled?

Both the ULG and the GFS follow the Inverse square law: ULG-If the distance between two objects is doubled the gravitational attraction is (1/4) of the original. OR GFS-If we travel beyond the Earth by a distance that is double it’s radius than we will only feel a quarter of Earth’s gravitational pull (9.81 m/s2/4 = 2.45 m/s2).

re = 6.4 x 106 m 1x 2x 3x 4x ¼ re= 2.24 m/s2 1/9 re = 1.09 m/s2

Example 5: The gravitational attraction between the Earth and Mars is 8.7 x 1016 N. The distance between the two planets is 5.5 x 1010m. Earth has a mass of 6.0 x 1024 kg. What’s the mass of Mars?