#### Transcript R - IBPhysicsLund

Topic 2: Mechanics 2.4 Uniform circular motion 2.4.1 Draw a vector diagram to illustrate that the acceleration of a particle moving with constant speed in a circle is directed towards the center of the circle. 2.4.2 Apply the expressions for centripetal acceleration. 2.4.3 Identify the force producing circular motion in various situations. Examples include friction of tires on turn, gravity on planet/moon, cord tension. 2.4.4 Solve problems involving circular motion. Topic 2: Mechanics 2.4 Uniform circular motion On the next pass, however, Helen failed to clear the mountains. r m v What force must be applied to Helen to keep her moving in a circle? How does it depend on the Helen’s radius r? How does it depend on Helen’s velocity v? How does it depend on Helen’s mass m? Topic 2: Mechanics 2.4 Uniform circular motion Draw a vector diagram to illustrate that the acceleration of a particle moving with constant speed in a circle is directed towards the center of the circle. A particle is said to be in uniform circular motion if it travels in a circle (or arc) with constant speed v. v red Observe that the velocity vector is always r blue tangent to the circle. y Note that the magnitude of the v velocity vector is NOT changing. r Note that the direction of the velocity vector IS changing. x Thus, there is an acceleration, even though the speed is not changing! Topic 2: Mechanics 2.4 Uniform circular motion Draw a vector diagram to illustrate that the acceleration of a particle moving with constant speed in a circle is directed towards the center of the circle. In order to find the direction of the acceleration (a = ∆v/∆t ) we observe two nearby snapshots of the particle: v red The direction of the acceleration is gotten r blue from ∆v = v2 – v1 = v2 + (-v1): v y 2 v2 v1 -v1 v1 -v1 ∆v ∆v The direction of the acceleration is toward the center of the circleyou must be able to sketch this. x Topic 2: Mechanics 2.4 Uniform circular motion Apply the expressions for centripetal acceleration. Centripetal means center-seeking. How does centripetal acceleration ac depend on r and v? We define the centripetal force Fc: Fc = mac centripetal force Picture yourself as the passenger in a car that is rounding a left turn: The sharper the turn, the harder you and your door push against each other. (Small r = big Fc.) The faster the turn, the harder you and your door push against each other. (Big v = big Fc.) Fc Topic 2: Mechanics 2.4 Uniform circular motion Apply the expressions for centripetal acceleration. PRACTICE: For each experiment A and B, label the control, independent, and dependent variables. manipulated No change r r r ac manipulated ac v ac no change ac v v responding F c responding Fc r Fc CONTROL: r INDEPENDENT: v A DEPENDENT: Fc and ac Fc CONTROL: v INDEPENDENT: r B DEPENDENT: Fc and ac v Topic 2: Mechanics 2.4 Uniform circular motion Apply the expressions for centripetal acceleration. We know the following things about ac: If v increases, ac increases. If r increases, ac decreases. v ac = r first guess formula From dimensional analysis we have ac = v r m ? = s2 m/s m ? = 1 s What can we do to v or r to “fix” the units? ac = v2/r ac v2 = r centripetal acceleration m ? m2/s2 ? m = 2 = 2 s m s This is the correct one! Topic 2: Mechanics 2.4 Uniform circular motion Apply the expressions for centripetal acceleration. ac = v2/r centripetal acceleration Fc = mac centripetal force EXAMPLE: A 730-kg Smart Car negotiates a 30. m radius turn at 25. m s-1. What is its centripetal acceleration and force? What force is causing this acceleration? SOLUTION: ac = v2/r = 252/30 = 21 m s-2. Fc = mac = (730)(21) = 15000 n. The centripetal force is caused by the friction force between the tires and the pavement. Topic 2: Mechanics 2.4 Uniform circular motion Apply the expressions for centripetal acceleration. The period T is the time for one complete revolution. One revolution is one circumference C = 2r. Therefore v = distance / time = 2r/T. Thus v2 = 42r2/T2 so that ac = v2/r = 42r2/T2r = 42r/T2. ac = v2/r ac = 42r/T2 centripetal acceleration Topic 2: Mechanics 2.4 Uniform circular motion Apply the expressions for centripetal acceleration. ac = 42r/T2 ac = v2/r centripetal acceleration EXAMPLE: Albert the 2.50-kg physics cat is being swung around your head by a string harness having a radius of 3.00 meters. He takes 5.00 seconds to complete one fun revolution. What are ac and Fc? What is the tension in the string? SOLUTION: ac = 42r/T2 = 42(3)/(5)2 = 4.74 m s-2. Albert the Fc = mac = (2.5)(4.74) = 11.9 n. Physics The tension is causing the Cat centripetal force, so the tension is Fc = 11.9 n. Topic 2: Mechanics 2.4 Uniform circular motion Solve problems involving circular motion. EXAMPLE: Dobson is watching a 16-pound bowling ball being swung around at 50 m/s by Arnold. If the string is cut at the instant the ball is next to the ice cream, what will the ball do? (a)It will follow path A and strike Dobson's ice cream. (b)It will fly outward along curve path B. (c)It will fly tangent to the original circular path along C. A C B Topic 2: Mechanics 2.4 Uniform circular motion Solve problems involving circular motion. EXAMPLE: A 3.0-kg mass is tied to a string having a length of 1.5 m, and placed in uniform circular motion as shown. The string traces out a cone with a base angle of 60°, with the mass traveling the base of the cone. (a) Sketch in the forces acting on the mass. SOLUTION: The ONLY two forces acting on the mass are its weight W and the tension T in the string T. Don’t make the mistake of drawing Fc into the diagram. Fc is the resultant of T and W, as the next questions will illustrate. W Topic 2: Mechanics 2.4 Uniform circular motion Solve problems involving circular motion. EXAMPLE: A 3.0-kg mass is tied to a string having a length of 1.5 m, and placed in uniform circular motion as shown. The string traces out a cone with a base angle of 60°, with the mass traveling the base of the cone. (b) Draw a FBD in the space provided. Then break down the tension force in terms of the unknown tension T. Ty T T SOLUTION: Tx = T cos = T cos 60° = 0.5T. Tx Ty = T sin = T sin 60° = 0.87T. W W Topic 2: Mechanics 2.4 Uniform circular motion Solve problems involving circular motion. EXAMPLE: A 3.0-kg mass is tied to a string having a length of 1.5 m, and placed in uniform circular motion as shown. The string traces out a cone with a base angle of 60°, with the mass traveling the base of the cone. (c) Find the value of the components of the tension Tx and Ty. SOLUTION: Ty Note that Ty = mg = 4(10) = 40 n. T But Ty = 0.87T. Thus 40 = 0.87T so that T = 46 n. Tx = 0.5T = 0.5(46) = 23 n. Tx W Topic 2: Mechanics 2.4 Uniform circular motion Solve problems involving circular motion. EXAMPLE: A 3.0-kg mass is tied to a string having a length of 1.5 m, and placed in uniform circular motion as shown. The string traces out a cone with a base angle of 60°, with the mass traveling the base of the cone. (d) Find the speed of the mass as it travels in its circular orbit. SOLUTION: The center of the UCM is here: Since r / 1.5 = cos 60°,r = 0.75 m. Fc = Tx = 23 n. Thus Fc = mv2/r 23 = 3v2/0.75 r v = 2.4 m s-1. Topic 2: Mechanics 2.4 Uniform circular motion Solve problems involving circular motion. EXAMPLE: Suppose a 0.500-kg baseball is placed in a circular orbit around the earth at slightly higher that the tallest point, Mount Everest (8850 m). Given that the earth has a radius of RE = 6400000 m, find the speed of the ball. SOLUTION: The ball is traveling in a circle of radius r = 6408850 m. Fc is caused by the weight of the ball so that Fc = mg = (0.5)(10) = 5 n. Since Fc = mv2/r we have 5 = (0.5)v2/6408850 v = 8000 m s-1! Topic 2: Mechanics 2.4 Uniform circular motion Solve problems involving circular motion. EXAMPLE: Suppose a 0.500-kg baseball is placed in a circular orbit around the earth at slightly higher that the tallest point, Mount Everest (8850 m). How long will it take the ball to return to Everest? SOLUTION: We want to find the period T. We know that v = 8000 m s-1. We also know that r = 6408850 m. Since v = 2r/T we have T = 2r/v T = 2(6408850)/8000 T = (5030 s)(1 h / 3600 s) = 1.40 h. Topic 2: Mechanics 2.4 Uniform circular motion Solve problems involving circular motion. EXAMPLE: Explain how an object can remain in orbit yet always be falling. SOLUTION: Throw the ball at progressively larger speeds. In all instances the force of gravity will draw the ball toward the center of the earth. When the ball is finally thrown at a great enough speed, the curvature of the ball’s path will match the curvature of the earth’s surface. The ball is effectively falling around the earth! Topic 2: Mechanics 2.4 Uniform circular motion Preparation for future topics. UCM will be revisited in Topic 4 Oscillations and waves. We now extend UCM a bit to get ready. In uniform circular motion the x and y coordinates of the particle at any instant can be determined using the following relationships: x = r cos θ relationship between rotational and linear variables y = r sin θ We call the angular position of the particle. Take note that varies with time. y r θ x y x Topic 2: Mechanics 2.4 Uniform circular motion Preparation for future topics. x = r cos θ relationship between rotational and linear variables y = r sin θ EXAMPLE: Find the x- and y-components of a particle whose path follows a circle of radius 3.00 m at the precise instant its angular position is θ = 135°. Then sketch it into the grid. (-2.12 m, 2.12 m ) y SOLUTION: r = 3 m and θ = 135°. x = r cos θ = 3 cos 135° = -2.12 m. y = r sin θ = 3 sin 135° = +2.12 m. x Topic 2: Mechanics 2.4 Uniform circular motion Preparation for future topics. x = r cos θ relationship between rotational and linear variables y = r sin θ EXAMPLE: A particle in uniform circular motion at a radius of 3.00 m has an x-coordinate of -2.00 m and a y-coordinate that is negative. Sketch the particle. Then find the values of y and θ. SOLUTION: r2 = x2 + y2 so that y 32 = 22 + y2 or y = 2.24 m. x = 2 (adj) and r = 3 (hyp) so cos θ = adj/hyp = 2/3 θ = cos-1(2/3) = 48.2°. r From +x dir, 180° + 48.2° = 228°. x Topic 2: Mechanics 2.4 Uniform circular motion Preparation for future topics. Just as we defined displacement in linear motion as ∆x = x2 – x1 we can define angular displacement in circular motion to be ∆ = 2 – 1. And just as we defined a velocity as v = ∆x/∆t, we define angular velocity = ∆/∆t. ∆ = 2 – 1 ang. displacement angular displacement and angular velocity = ∆/∆t angular velocity y At first glance we would expect t2 that the units for = ∆/∆t are ∆ (deg/s) but we would be wrong! t1 θ2 The next slides will explain. θ 1 x Topic 2: Mechanics 2.4 Uniform circular motion Preparation for future topics. Some of you may already know the formula for arc length s. s = r∆θ ∆θ in radians arc length s Radians is a more natural way to measure angular displacement than degrees. Here is how they are related. rad = 180° = 1/2 rev radian-degree-revolution 2 rad = 360° = 1 rev conversions Arc length s is measured in meters (m) and tells us how far we have traveled in an arc. Angular velocity (or angular speed) is measured in radians per second (rad/s). Topic 2: Mechanics 2.4 Uniform circular motion Preparation for future topics. s = r∆θ ∆θ in radians arc length s rad = 180° = 1/2 rev radian-degree-revolution 2 rad = 360° = 1 rev conversions EXAMPLE: How far have you traveled if you are on the outer perimeter of a merry-go-round having a diameter of 7.00 m and it has turned 135°? Sketch your arc. SOLUTION: 90° 135° We are looking for arc length s. 45° ∆θ = 135°. But s needs radians. 135°( rad/180°) = 2.356 rad. 180° 0° r = diameter/2 = 7/2 = 3.5 m. s = r∆θ = (3.5)(2.356) = 8.25 m. 225° 315° The red arc is 8.25 m long. 270° Topic 2: Mechanics 2.4 Uniform circular motion Preparation for future topics. s = r∆θ ∆θ in radians rad = 180° = 1/2 rev 2 rad = 360° = 1 rev arc length s radian-degree-revolution conversions PRACTICE: How far have you traveled if you are 2.75 m from the center of a merry-go-round having a diameter of 7.00 m and it has turned 135°? Sketch your arc. We are looking for arc length s. 90° ∆θ = 135°. But s needs radians. 135° 45° 135°( rad/180°) = 2.356 rad. But r = 2.75 m so that 0° s = r∆θ = (2.75)(2.356) = 6.48 m. 180° FYI Arc length depends on arc radius. 225° 315° 270° Topic 2: Mechanics 2.4 Uniform circular motion Preparation for future topics. s = r∆θ ∆θ in radians arc length s Recall that velocity v is just the displacement s divided by the time. Thus v = s/∆t = r∆θ/∆t = r(∆θ/∆t) = r Angular speed is therefore related to linear speed v like this: v = r in rad/s relation between v and 12 Topic 2: Mechanics 2.4 Uniform circular motion 9 3 6 Preparation for future topics. v = r in rad/s relation between v and EXAMPLE: The second hand of a clock has a length of 12.0 cm. What are its angular speed (in rad/s) and the linear speed, of its tip, and its base? SOLUTION: The second hand makes one revolution each 60 s. One revolution is 2 rad. Therefore = 2 rad / 60 s = 0.105 rad s-1. Angular speed is the same for the base and tip. v = r = (0.12)(0.105) = 0.013 m s-1 for the tip. v = r = (0)(0.105) = 0 m s-1 for the base. Topic 2: Mechanics 2.4 Uniform circular motion Preparation for future topics. We define the frequency f of uniform circular motion as how many revolutions or cycles the object in UCM executes per unit time. Since the period T is how many seconds per cycle, the relationship between T and f is simply f = 1/T T = 1/f relation between T and f Frequency f is measured in (cycles / second) which are known as (Hertz). Since v = 2r/T and v = r, we can derive 2r/T = r 2/T = 2f = = 2/T = 2f relation between , T and f Topic 2: Mechanics 2.4 Uniform circular motion Preparation for future topics. f = 1/T T = 1/f relation between T and f = 2/T = 2f relation between , T and f EXAMPLE: The wheel of a car rotates once each 0.500 s. Find its period, frequency, and angular speed. SOLUTION: The period T is just 0.500 s. The frequency f = 1/T = 1/0.5 = 2.00 Hz. The angular speed can be found in two ways. METHOD 1: = 2f = 2(2) = 12.6 rad s-1. METHOD 2: = ∆/∆t = (2 rad)/(0.5 s) = 12.6 rad s-1.