Presentation on theme: "Warm-up: Centripetal Acceleration Practice Assume a satellite in low Earth orbit has an acceleration, caused by gravity, of 9.81 m/s 2. If the radius of."— Presentation transcript:
Warm-up: Centripetal Acceleration Practice Assume a satellite in low Earth orbit has an acceleration, caused by gravity, of 9.81 m/s 2. If the radius of the orbit is approximately the radius of the Earth, 6370 km, find the velocity of the satellite and the time for one orbit.
Review: Centripetal acceleration A particle undergoing circular motion is accelerating, even if its speed is not changing. The acceleration of an object moving in a circle is toward the center of the circle. It is thus referred to as centripetal acceleration. The magnitude of the centripetal acceleration depends on the velocity of the particle tangent to the circle.
We will begin our study of rotational motion by considering the motion of a rigid body about a fixed axis, or an axis that is moving parallel to itself like a rolling ball.
Angular Position Because the object is rigid, and the axis is fixed, all the points in the object maintain the same relative position. When the object rotates, every point rotates through the same angle. The angle θ, measured relative to some starting position, gives the angular position of every point in the object.
Angular Displacement When the object rotates on its axis, each point undergoes the same angular displacement Δ θ. If we measure the angular displacement in radians, we can easily find the distance traveled by each point, since by definition of radian measure where r is the radial distance from the center.
Angular Displacement Example Two people ride on a carousel. One rides on a horse located 5 meters from the center. The other rides on a swan located 3 meters from the center. When the carousel goes around ¼ of a revolution, how far does each person travel?
Angular velocity We define the angular velocity as the rate of change of angular position. Since radians are dimensionless, the dimensions of the angular velocity are T -1. The magnitude of the angular velocity is the angular speed. Counterclockwise rotation corresponds to a positive velocity.
Angular velocity practice What is the angular speed, in radians/second, of a motor that spinning with 6000 rpm?
Angular acceleration The angular acceleration is the rate of change of angular velocity. Since radians are dimensionless, the dimensions of angular acceleration velocity are T -2. This acceleration refers to the increase or decrease in rotational speed of the particle.
Angular and Tangential Quantities Each angular quantity has a corresponding tangential quantity. The arc length s is the distance travelled along the circle, in m. The tangential velocity is the speed of the particle in the direction tangent to the circle, in m/s. The tangential acceleration is the acceleration of the particle tangent to the circle, in m/s 2.
Angular equations of motion for constant acceleration Begin with the definition of angular acceleration. If the acceleration is constant, what is the angular velocity? Then what is the displacement?
Rotational motion of a rigid body is analogous to linear motion. Straight line motionRotation about a fixed axis Linear position x. Linear displacement x Linear velocity v Linear acceleration a Angular position Angular displacement Angular velocity ω Angular acceleration α
Equations of motion for constant acceleration Straight line motionRotation about a fixed axis
Problem solving with rotational motion. Convert tangential velocities, speeds etc. to their angular counterparts by dividing by the radius. Solve for the angular motion by using the angular equations of motion just as we did with linear motion. Convert quantities to their tangential counterparts if needed.
Centripetal acceleration in angular form. We can write the equation for centripetal acceleration in terms of the angular velocity.
Angular kinematics examples Toast falling off a table usually starts to fall when it makes an angle of 30 degrees with the horizontal, and falls with an initial angular speed of where l is the length of one side. On what side of the bread will the toast land if it falls from a table 0.5 m high? If it falls from a table 1.0 m high? Assume l = 0.10 m, and ignore air resistance. When a turntable rotating at 33 rev/min is turned off, it comes to rest in 26 s. Assuming constant angular acceleration, find the angular acceleration and the angular displacement. If the turntable is 0.20 m in radius, how far would an ant riding on the outside edge have moved in that time?
Rotational Dynamics: Equilibrium
Rotational Equilibrium What causes angular acceleration? F 1 and F 2 are forces equal in size but acting in opposite directions. Will both situations result in angular acceleration?
Torque A force F acts on mass m located with a rigid body. Only the tangential component affects the rotation of the disk. The farther away from the axis of rotation the force is, the greater the effect on the rotation.
Torque The torque that acts on mass m in the rigid body is the tangential force F T times the distance from the axis of rotation r. Τ =F T r = (F sin φ ) r
Torque Another way of looking at the same thing is to say the torque is caused by the entire force, but the distance is only the perpendicular distance of the force to the line of action of the force. This distance, which is r sin φ, is known as the lever arm l. So our formula becomes Τ =F l = F (r sin φ ) = Fr sin φ
Conditions of Equilibrium We learned last year that there are two conditions for equilibrium. The net external force acting on the body must be zero. The net external torque about any point must be zero.
Torque and Rotational Equilibrium Examples A 700 N window washer is standing on a uniform scaffold supported by a vertical rope at each end. The scaffold weights N and is 3.00 m long. What is the force in each rope when the window washer stands 1.00 m from one end? Does the Sun exert a torque on the Earth relative to the Earth’s rotational axis? Discuss.
Rotational equilibrium example A 700 N window washer is standing on a uniform scaffold supported by a vertical rope at each end. The scaffold weights N and is 3.00 m long. What is the force in each rope when the window washer stands 1.00 m from one end? Answers: 333 N and 567 N
Rotational Dynamics: Newton’s 2 nd Law for Rotation
Newton’s Second Law for Rotational Motion Consider a mass m in a rigid body that is free to rotate. The mass is acted on by a force F. What does the radial component do? What does the tangential force do? The mass will undergo a tangential acceleration a t. This acceleration is in addition to the constant centripetal acceleration.
Newton’s Second Law for Rotational Motion We can add up the torques on every mass m i in the object to derive an equation for the entire rigid object. Since this is a rigid object, α is the same for every particle. Thus we can remove it from the sum.
Newton’s Second Law for Rotational Motion The internal torques add up to zero, so the left hand side of the above is just the net external torque. The quantity in brackets above depends on the mass and radius of each particle, both of which are fixed for a rigid object and choice of axis. So, that sum is also fixed for a given object and rotational axis. It is called the rotational inertia.
Newton’s Second Law for Rotational Motion Where I = rotational inertia Rotational inertia is also called the moment of inertia.
Parallel Axis Theorem The parallel-axis theorem states Where I CM is the rotational inertia about an axis containing the Center of Mass I is the rotational inertia about an axis parallel to I CM M = total mass and h = distance between axes.
Rotational Kinetic Energy We can apply the formula for kinetic energy to rotation as well. In a rotating rigid body, each particle moves with a velocity v t =r ω.
Rotational Kinetic Energy Example Calculate the rotational kinetic energy of the Earth, and compare that to the kinetic energy of the Earth’s motion around the Sun. Assume the Earth is a sphere of radius 6.4 x 10 6 m and mass 6.0 x kg. The radius of the Earth’s orbit is 1.5 x m.
Consider an object with a circular cross section that rolls without slipping across a flat surface. The point of contact between the wheel and the ground moves a distance s = R φ. Since the center of mass remains above the point of contact, it also moves s=R φ.
For an object that rolls without slipping, the motion of the center of mass is described by the same equations as an object rotating about a fixed axis. Newton’s Second Law for rotation also holds, if the torques are calculated about an axis containing the center of mass. τ NET, CM = I CM α
Rolling Objects The total kinetic energy of a moving object is the kinetic energy of the center of mass plus the kinetic energy of the particles relative to the CM. Here, that means the rotation kinetic energy plus the kinetic energy of the CM motion.
A bowling ball of radius 11 cm rolls without slipping up a ramp. If the initial speed of the ball is 2 m/s find the height h that the ball reaches when it momentarily comes to rest.
A cue stick hits a cue ball horizontally a distance d above the center of the ball. Find the value of d for which the cue ball will roll without slipping from the beginning. Express your answer in terms of the radius R of the ball.