Week 11 - Rotational Motion PDF

ROTATIONAL MOTION  ROTATIONAL MOTION Translational motion – straight line motion (horizontal, free fall, projectile motion) Rotational Motion - motion of a body that spins about an axis. - objects in rotational motion undergo circular motion FERRIS WHEEL Any point on a Fe...

ROTATIONAL MOTION  ROTATIONAL MOTION Translational motion – straight line motion (horizontal, free fall, projectile motion) Rotational Motion - motion of a body that spins about an axis. - objects in rotational motion undergo circular motion FERRIS WHEEL Any point on a Ferris wheel that spins about a fixed axis undergoes circular motion. To analyze the motion, it is convenient to set up a fixed reference line. And assume that at time t=0, a cart is on the reference line. Reference Line  After time interval t2 the cart advances to a new position. In this time interval, the line from the center to the cart moved through angle, θ with respect to the reference line. Likewise, the cart moved a distance s, measured from the circumference of the circle; s is arc length. cart (at time 𝒕𝟐 ) s 𝜃 Reference Line  Angles can be measured in RADIANS Radian - an angle whose arc length is equal to the radius, which is approximately equal to 57.3 degrees. Full circle has an angle of 2 radians. Thus, one radian is 360°/2 =  Any angle θ measured in radians is defined by: s  = r Radian is a pure number, with NO dimensions. Units cancel, and we use the abbreviation rad to identify unit.  is measured in degrees or radians (SI unit: radian) Conversion Factors Radian Degrees 2 360°  180°  90° 1 57.3°  Rotational Kinematics Concepts in translational kinematics will be applied in rotational kinematics but using different variable, symbols and units. Consider the rotation of rigid bodies. A rigid body is an extended object (not a point particle) with a definite shape. This means that the body’s particles stay in fixed positions relative to each other. Rotational Motion Linear motion Angular Displacement Distance/displacement  =  f −  i rad/radians 𝑑 = 𝑑𝑓 − 𝑑𝑖 meters, m Average angular speed Speed/velocity 𝜃𝑓 − 𝜃𝑖 Δ𝜃 𝜃 𝑑𝑓 − 𝑑𝑖 𝑑 𝜔lj = 𝑡𝑓 − 𝑡𝑖 = = Δ𝑡 𝑡 rad/s 𝑣= = m/s 𝑡𝑓 − 𝑡𝑖 𝑡 Average angular acceleration Acceleration  f − i  𝑣𝑓 − 𝑣𝑖 = = rad/s2 𝑎= m/s2 t f − ti t 𝑡 A screenshot of a computer Description automatically generated  𝑣 𝑣 The same is true with angular velocity. Every point of a rotating rigid body also has linear acceleration aside from its angular quantities. Because each particle is moving in a circle, the direction of this linear acceleration is tangent to the circular path. Thus, this linear acceleration, called tangential acceleration, is mathematically stated as, 𝑎𝑟𝑎𝑑 𝑎𝑡𝑎𝑛  Let us try this! A carousel is moving slowly at an angular speed of 0.480 rad/s. Leo and Judy are seated with distances of 1.50 m and 2.50 m from the center of the carousel, respectively. What are the linear speeds of Leo and Judy? A carousel is moving slowly at an angular speed of 0.480 rad/s. Leo and Judy are seated with distances of 1.50 m and 2.50 m from the center of the carousel, respectively. What are the linear speeds of Leo and Judy? 𝑟𝑎𝑑 Given: 𝜔 = 0.480 Asked: 𝑣 𝑠 𝑟𝐴 = 1.50 𝑚 Formula: 𝑣 = 𝑟𝜔 𝑟𝐵 = 2.50 𝑚 You can see here that Judy, who is farther away from the center of the carousel, has a higher linear speed than the one closer to the center, who is Leo.  Let us try this! In another carousel that is initially at rest, Ron is seating 3.0 m from the center. At an initial t = 0, the carousel is given a constant angular acceleration of a = 0.065 rad/s2. At t = 10.0 s, find the following quantities: (a) angular velocity of the carousel, (b) linear velocity of Ron, (c) his tangential acceleration, (d) his centripetal acceleration (e) his total linear acceleration In another carousel that is initially at rest, Ron is seating 3.0 m from the center. At an initial t = 0, the carousel is given a constant angular acceleration of a = 0.065 rad/s2. At t = 10.0 s, find the following quantities: (a) angular velocity of the carousel, (b) linear velocity of Ron, (c) his tangential acceleration, (d) his centripetal acceleration (e) his total linear acceleration Given: 𝑟 = 3.0 𝑚 Asked: a. 𝜔 𝑡 = 10.0 𝑠 b. 𝑣 𝑡𝑜 = 0 c. 𝑎𝑡𝑎𝑛 𝛼 = 0.065 𝑟𝑎𝑑/𝑠 2 d. 𝑎𝑟𝑎𝑑 e. 𝑎𝑎𝑐𝑡𝑢𝑎𝑙 Solution: In another carousel that is initially at rest, Ron is seating 3.0 m from the center. At an initial t = 0, the carousel is given a constant angular acceleration of a = 0.065 rad/s2. At t = 10.0 s, find the following quantities: (a) angular velocity of the carousel, (b) linear velocity of Ron, (c) his tangential acceleration, (d) his centripetal acceleration (e) his total linear acceleration In another carousel that is initially at rest, Ron is seating 3.0 m from the center. At an initial t = 0, the carousel is given a constant angular acceleration of a = 0.065 rad/s2. At t = 10.0 s, find the following quantities: (a) angular velocity of the carousel, (b) linear velocity of Ron, (c) his tangential acceleration, (d) his centripetal acceleration (e) his total linear acceleration In another carousel that is initially at rest, Ron is seating 3.0 m from the center. At an initial t = 0, the carousel is given a constant angular acceleration of a = 0.065 rad/s2. At t = 10.0 s, find the following quantities: (a) angular velocity of the carousel, (b) linear velocity of Ron, (c) his tangential acceleration, (d) his centripetal acceleration (e) his total linear acceleration In another carousel that is initially at rest, Ron is seating 3.0 m from the center. At an initial t = 0, the carousel is given a constant angular acceleration of a = 0.065 rad/s2. At t = 10.0 s, find the following quantities: (a) angular velocity of the carousel, (b) linear velocity of Ron, (c) his tangential acceleration, (d) his centripetal acceleration (e) his total linear acceleration In another carousel that is initially at rest, Ron is seating 3.0 m from the center. At an initial t = 0, the carousel is given a constant angular acceleration of a = 0.065 rad/s2. At t = 10.0 s, find the following quantities: (a) angular velocity of the carousel, (b) linear velocity of Ron, (c) his tangential acceleration, (d) his centripetal acceleration (e) his total linear acceleration I do, We do, You do! 1. A rotating body makes 10 complete revolutions in 5 seconds. Find its average angular speed. 2. The blades of a sliding fan start from rest and attain a speed of 200 rev/min 10 seconds. Calculate its angular acceleration assuming it to be constant. 3. A body rotating from rest is accelerated uniformly by 16 rad/s2. Find the angular speed after 5 seconds. A rotating body makes 10 complete revolutions in 5 seconds. Find its average angular speed. Given: Asked: Formula : Solution: Convert θ to rads I do, You do, We do! 1. A rotating body makes 10 complete revolutions in 5 seconds. Find its average angular speed. 2. The blades of a sliding fan start from rest and attain a speed of 200 rev/min 10 seconds. Calculate its angular acceleration assuming it to be constant. 3. A body rotating from rest is accelerated uniformly by 16 rad/s2. Find the angular speed after 5 seconds. 2. The blades of a sliding fan start from rest and attain a speed of 200 rev/min in 10 seconds. Calculate its angular acceleration assuming it to be constant. Given: Asked: Formula: Solution: I do, You do, We do! 1. A rotating body makes 10 complete revolutions in 5 seconds. Find its average angular speed. 2. The blades of a sliding fan start from rest and attain a speed of 200 rev/min 10 seconds. Calculate its angular acceleration assuming it to be constant. 3. A body rotating from rest is accelerated uniformly by 16 rad/s2. Find the angular speed after 5 seconds. 3. A body rotating from rest is accelerated uniformly by 16 rad/s2. Find the angular speed after 5 seconds. Given: Asked: Formula: Solution: Uniformly Accelerated Rotational Motion  UARM - Uniformly Accelerated Rotational Motion In the previous modules, you have seen that analyzing straight-line motion with constant acceleration is fairly simple. This is also true for rotational motion that has constant acceleration on a fixed axis. You can derive the kinematic equations for uniformly accelerated rotational motion from those of translation motion.  Let us try this! After using your DVD player to watch a movie, you pressed the “stop” button, and the disc started to slow down. Initially, the disc is rotating at 30.0 rad/s, and then slows down at a constant rate of 10.0 rad/s2. At the same time, line AB on the disc is at the x-axis as shown in the figure provided. a. What is the angular velocity of the disc at t = 0.400 s? b. Find the angle that line AB makes with the x-axis at the given time. After using your DVD player to watch a movie, you pressed the “stop” button, and the disc started to slow down. Initially, the disc is rotating at 30.0 rad/s, and then slows down at a constant rate of 10.0 rad/s2. At the same time, line AB on the disc is at the x-axis as shown in the figure provide a. What is the angular velocity of the disc at t = 0.400 s? b. Find the angle that line AB makes with the x-axis at the given time. 𝑟𝑎𝑑 Given: 𝜔𝑜 = 30.0 𝑠 𝑟𝑎𝑑 𝛼 = 10.0 2 𝑠 𝑡 = 0.400 𝑠 Asked: a. 𝜔 b. 𝜃 Formula: After using your DVD player to watch a movie, you pressed the “stop” button, and the disc started to slow down. Initially, the disc is rotating at 30.0 rad/s, and then slows down at a constant rate of 10.0 rad/s2. At the same time, line AB on the disc is at the x-axis as shown in the figure provide a. What is the angular velocity of the disc at t = 0.400 s? b. Find the angle that line AB makes with the x-axis at the given time. Notice that the angular acceleration is negative (–10.0 rad/s2) because the disc is slowing down After using your DVD player to watch a movie, you pressed the “stop” button, and the disc started to slow down. Initially, the disc is rotating at 30.0 rad/s, and then slows down at a constant rate of 10.0 rad/s2. At the same time, line AB on the disc is at the x-axis as shown in the figure provide a. What is the angular velocity of the disc at t = 0.400 s? b. Find the angle that line AB makes with the x-axis at the given time. Knowledge Check ROTATIONAL DYNAMICS FORCE IS PUSH AND PULL TORQUE IS TWIST AND TURN TORQUE A quantity that measures the ability of a force to rotate an object around some axis. The effect that causes an object to rotate Also known as the turning effect of a force  What affects torque? Formula: Angular speed r axis of rotation 𝜏 = 𝑟𝐹 What is the unit of Torque? Unit: N∙m F Force applied at a certain distance from the axis of rotation produces a torque. This torque in turn, causes this object to rotate. Source: https://www.slideserve.com/barry-palmer/rotational-equilibrium-and-dynamics (Barry Palmer)  What affects torque? Force and Lever Arm Angular speed Formula: d axis of rotation 𝜏 = 𝑟𝐹 The PERPENDICULAR distance from the axis of rotation to a line drawn along the direction of the force is called the lever arm, or moment arm. The farther the force is from F Force applied to an extended the axis of rotation, the object can produce a torque. easier it is to rotate the object This torque in turn, causes this and the more torque is object to rotate. produced. Source: https://www.slideserve.com/barry-palmer/rotational-equilibrium-and-dynamics (Barry Palmer) Simulation: Torque Factors affecting the Torque Raise your cards! What can you say about the torque produced by the indicated forces? Knowledge Check  Conceptual Question… A student pushes with minimal force of 50.0 N on the middle of a door to open it. What minimum force must be applied at the edge of the door for the door to open? What minimum force must be applied to the hinged side of the door to open? Source: https://www.slideserve.com/barry-palmer/rotational-equilibrium-and-dynamics (Barry Palmer)  Conceptual Question… A student pushes with minimal force of 50.0 N on the middle of a door to open it. What minimum force must be applied at the edge of the door for the door to open? Ans: 25 Newtons (half as much because the distance in increased by 2) T=r*F What minimum force must be applied to the hinged side of the door to open? Ans: The door cannot be opened by a force at the hinge location. It can be broken but not opened normally. Source: https://www.slideserve.com/barry-palmer/rotational-equilibrium-and-dynamics (Barry Palmer) “I do, We do, You do” I do: What torque is produced by a force of 20 N upward applied from a perpendicular distance of 10 m from the axis of rotation? What torque is produced by a force of 20 N upward applied from a perpendicular distance of 10 m from the axis of rotation? Given: Asked: Formula : Solution: We do You exerted a downward 500-N force on a wrench as shown in the figure here. What is the magnitude and direction of the torque you exerted on the wrench at the position shown in the figure? Given: F = 500 N r = 0.25 m Solution: 𝜃 = 300 What is 𝜃? Asked: 𝜏 𝜏 = 500𝑁 0.25𝑚 𝑠𝑖𝑛60𝑜 Formula: 𝜏 = 𝑟 𝐹 𝑠𝑖𝑛𝜃 𝜏 = 108. 25 𝑁 ∙ 𝑚 You exerted a downward 500-N force on a wrench as shown in the figure here. What is the magnitude and direction of the torque you exerted on the wrench at the position shown in the figure? You do: For 5 minutes What torque results from a 250 N force applied 30° from the vertical axis on a horizontal wrench 0.28 meters away from the bolt? Torque can be (+) or (-) Torque is a vector quantity (i.e. has direction) To find net torque, simply add up the individual torques. Tnet = Sum of Torques = T1 + T2 = (F1d1) + (F2d2) The sign of the net torque value (+ or -) will tell you the direction an object rotates. Clockwise rotation = - torque Counterclockwise rotation = + torque  Torque A basketball is being pushed by two players during tip-off. One player exerts a downward force of 11 N at a distance of 0.07 m to the right of the axis of rotation. The second player applies an upward force of 15 N at a perpendicular distance of 0.14 m to the left of the axis of rotation. Find the net torque acting on the basketball. Givens: Formula: F1 = 15 N d1 = 0.14 m Tnet = T1 + T2 F2 = 11 N d2 = 0.07 m Source: https://www.slideserve.com/barry-palmer/rotational-equilibrium-and-dynamics (Barry Palmer)  Torque A basketball is being pushed by two players during tip-off. One player exerts a downward force of 11 N at a distance of 0.07 m to the right of the axis of rotation. The second player applies an upward force of 15 N at a perpendicular distance of 0.14 m to the left of the axis of rotation. Find the net torque acting on the basketball. F2 = 11 N Givens: Formula: F1 = 15 N d1 = 0.14 m d1 = 0.14 m Tnet = T1 + T2 F2 = 11 N d2 = 0.07 m F1 = 15 N d2 = 0.07 m Source: https://www.slideserve.com/barry-palmer/rotational-equilibrium-and-dynamics (Barry Palmer)  A basketball is being pushed by two players during tip-off. One player exerts a downward force of 11 N at a distance of 0.07 m to the right of the axis of rotation. The second player applies an upward force of 15 N at a perpendicular distance of 0.14 m to the left of the axis of rotation. Find the net torque acting on the basketball. Given: Formula: F1 = 15 N d1 = 0.14 m Tnet = T1 + T2 F2 = 11 N d2 = 0.07 m F2 = 11 N T1 = F1 d1 = -(15)(0.14) = -2.1 Nm d1 = 0.14 m T2 = F2 d2 = -(11) (0.07)= -0.77Nm (both forces produce a clockwise rotation Therefore both torques are negative) F1 = 15 N d2 = 0.07 m Source: https://www.slideserve.com/barry-palmer/rotational-equilibrium-and-dynamics (Barry Palmer)  Renovation (House With Sparkles) with solid fill A basketball is being pushed by two players during tip-off. One player exerts a downward force of 11 N at a distance of 0.07 m to the right of the axis of rotation. The second player applies an upward force of 15 N at a perpendicular distance of 0.14 m to the left of the axis of rotation. Find the net torque acting on the basketball. T1 = F1 d1 = -(15)(0.14) = -2.1 Nm T2 = F2 d2 = -(11) (0.07)= -0.77Nm F2 = 11 N Tnet = T1 + T2 = -2.1 – 0.77 d1 = 0.14 m Tnet = -2.9 N m Negative torque thus Ball rotates in a clockwise direction. F1 = 15 N d2 = 0.07 m Source: https://www.slideserve.com/barry-palmer/rotational-equilibrium-and-dynamics (Barry Palmer) MOMENT OF INERTIA  Moment of Inertia Moment of inertia: the tendency of a body rotating about a fixed axis to resist a change in rotational motion. The moment of inertia is a measure of the object’s resistance to a change in its rotational motion about some axis. Depends on the shape of the object. SI units: kg ∙ m2 ANGULAR MOMENTUM  Angular Momentum Momentum - it refers to the quantity of motion that an object has; mass in motion. The greater the momentum, the greater the effort needed to change its path or to stop the body from moving. In linear motion, p = mv (unit: kg.m/s) In rotational motion, L = I Unit: kg-m2/s Source: https://www.slideserve.com/barry-palmer/rotational-equilibrium-and-dynamics (Barry Palmer) We do Find the angular momentum of a hollow cylinder weighing 30 N rotating at the rate of 15 revs/s. The body's radius of gyration about the axis of rotation is 1m. Find the angular momentum of a hollow cylinder weighing 30 N rotating at the rate of 15 revs/s. The body's radius of gyration about the axis of rotation is 1m. Given: Hollow cylinder Asked: Formula : Solution: You do: For 7 minutes A solid cylinder weighing 25 N rotates at the rate of 10 revs/s. The radius of the cylinder is 1.5 m. What is its angular momentum? CONDITIONS FOR EQUILIBRIUM Using Newton’s laws, you have learned that the first condition for equilibrium states that the vector sum of all the external forces acting on the body is equal to zero. In rotational motion, equilibrium can be achieved if the sum of the external torques acting on the body is equal to zero. This is the second condition for equilibrium.  Second Condition for Equilibrium “In order that the rotational motion of a body will not change, the sum of the torques about any axis acting on the body must be equal to zero.”  = 0  Let us try this! What is the weight needed to maintain a seesaw at balance state given the following: 50 N at 15 m away from the center and the unknown weight at 5 m away from the center? We do: What is the weight needed to maintain a seesaw at balance state given the following: 50 N at 15 m away from the center and the unknown weight at 5 m away from the center? Given: 15 m 5m Asked: Formula: Solution: You do: For 5 minutes How far away must a 19.6 N weight be placed on the left of a fulcrum to balance a 9.8 N weight 0.4 meters on the right?