Connection between angular and linear speed If a point-like object is in motion on a circular path of radius R at an instantaneous speed v, then its instantaneous angular speed ω is v = ω R Example: A satellite is in orbit around the Earth at a height h = 800 km over the equator. If it remains above the same point all the time, what are its speed and centripetal acceleration? Earth spins about its axis at an angular speed ω = π 4 hrs = π 4 3600s =.74 0 3 rad/s Top View In order to stay above the same point all the time, satellite must travel at the same angular speed. The radius of the satellite s circular orbit is R = (6400 + 800) km Thus, the speed of the satellite is v = ω R = (.74 0 3 )(7. 0 6 ) m/s =.3 0 4 m/s The centripetal acceleration of the satellite is ac = = ω R = (.74 0 3 ) (7. 0 6 ) m/s =.8 m/s v R
Angular and tangential acceleration For a point-like object is in motion on a circular path of constant radius R, the angular acceleration α corresponds to a change of angular speed over a time Δt. α = Δω Δt Correspondingly, the speed of the object moving on the circular path of constant radius R changes by Δv = R Δω, i.e., there is a tangential acceleration at = Δv Δt =R Δω Δt = α R Example: A compact disk, which has a diameter D = cm, speeds up uniformly from 0.00 to 4.00 rev/s in 3.00 s. What is the tangential acceleration of a point on the disk, located halfway between the centre and the rim of the disk? D One revolution corresponds to an angle of π rads Thus, after 3s the angular speed of the disk is (4 π) rad/s Top View The angular acceleration of the disk is α = Δω Δt = 8π 3 rad/s The point of interest on the disk is at a distance D/4 from the centre Thus, the tangential acceleration of that point of the disk is at = α D/4 = 8π 0. m/s 3 4 = 0.5 m/s
Kinetic Energy of a rigid body Definition: The kinetic energy of a many-body system of total mass M, comprising N point-like objects of masses m, m,... mn, equals the sum of two distinct contributions K = Mv + Σi mi ui v is the velocity of the center of mass of the system ui = vi v relative velocity of the ith point with respect to the center of mass The first contribution is that of a point-like object with the mass of the whole system, moving at a velocity equal to that of the center of mass of the system. It describes the translational motion of the system as a whole, with respect to an inertial observer The second contribution is the sum of the kinetic energy of each and every a point in the system, moving as seen from an observer sitting at the center of mass Special case: Rigid Body For a rigid body, the above expression is very useful and takes on a simple form Example: hockey puck of mass M slides on ice... ω v K = Mv + I ω while spinning about axis through centre rotational kinetic energy I is the moment of inertia of the puck The two motions are generally unrelated
Rotation of rigid bodies Example: A dumbbell consists of two spheres of mass m =.3 kg each, connected by a rod of negligible mass. This object can spin on a plane about an axis through its centre. The two spheres are held at a distance L =. m. Side View Top View L/ =. m v The dumbbell is set on rotation at an angular speed of 6π rad/s. What are the speed and kinetic energy of each sphere? Angular speed ω = 6π rad/s each sphere does a full circle (an angle of π rads ) 3 times in s. In turn, this means that each sphere travels a distance 3 π (L/) = 3πL in s. Thus, v = 3π (.) m/s = ωl/ = 0.7 m/s The kinetic energy of each sphere is mv = mω (L /4) = mω L 8 = [.3 (36 π )(.) /8] J = 79 J The kinetic energy of the whole dumbbell is therefore I = ml moment of inertia of the rigid dumbbell mω 8 L = ( ml ) ω = Iω
Moment of Inertia Definition: The rotational kinetic energy of a rigid body spinning about a fixed axis going through its center of mass is generally expressed as y Krot = Iω r x I = Σi mi (xi + yi ) y = m (x + y ) + m (x + y ) +... + mn (xn + yn ) x moment of inertia of the body with respect to specific axis rn Top View The moment of inertia is a scalar quantity, with dimensions [M][L ]. Its calculation is simple only in few cases, otherwise it requires calculus. Luckily, there are tables of moments of inertia for important objects. Important: The moment of inertia depends on the axis with respect to which it is computed. Example: Rigid dumbbell with connecting rod of negligible mass Case : axis through centre Case : axis through one end m m L I = ml /4 + ml /4 = ml / I = m 0 + ml = ml
Moment of Inertia (cont d) Example: What is the moment of inertia of a bicycle wheel with respect to an axis going through its centre? Assume that the wheel has a radius R = 30cm, a mass M =.75 kg (including the tire) and its spokes have negligible mass. Δm R Top View each element of mass Δm contributes to the moment of inertia the same amount, i.e., ΔmR Hence, Ih = MR = (.75)(0.3) kg m = 0.58 kg m If the wheel is spinning about its axis with a kinetic energy Krot = 4 J, what is the angular speed of the wheel? Krot = Since Iω it is ω = K rot = 8 Ih 0.58 rad /s = 59 rad /s ω = 7 rad/s Question: If a disk of equal mass and radius had the same rotational kinetic energy, what would be its angular speed? R The moment of inertia for a disk going through the centre, is Id = MR (one half of that of the hoop) Thus, Id = 0.079 kg m ω = 0 rad/s and ω = K rot Id = 8 0.079 rad /s = 0405 rad /s lower I greater ω for the same K
Moment of Inertia (cont d) Hoop or cylindrical shell I = MR Disk (solid cylinder) I = MR Disk (axis at rim) I = 3MR Thin rod (axis through midpoint) I = ML / Thin rod (axis through one end) I = ML /3 Hollow sphere I = MR 3 Solid sphere I = MR 5 Solid sphere (axis at rim) I = 7 MR 5 Solid plate (axis through center, in plane of plate) I =ML / Solid plate (axis perpendicular to plane of plate) I =M(L + W )/ For equal mass and characteristic size, hollow objects have higher inertia, because more of the mass is further away from the axis moments of
Rolling Motion In the general case, rotational and translational motions of a rigid body are disconnected and independent of one another. An important exception is constituted by rolling motion, in which translational and rotational motion are related. Rolling is made possible by the particular shape (spherical or circular) of some objects Smooth wedge If the wedge is perfectly smooth, the box slides down A ball, a cylinder, any other object would slide in the same way. Shape is immaterial on a smooth surface Rough wedge If wedge is rough (μs > tan θ), box remains at rest. A cylinder or a sphere, on the other hand, comes down rolling (without sliding). θ Rolling motion requires a rough surface. No rolling is possible on a smooth surface. Friction prevents sliding on the one hand, causing rolling on the other.
Rolling Motion (cont d) πr A rolling object is instantaneously spinning about an axis going through its centre, perpendicular to the direction of motion Concurrently, the object is moving forward as a result of its spin If the center of mass of the system advances by πr in a time T, then the speed of the center of mass is πr vcm = vcm = ωr rolling condition T Kinetic energy of rolling object K = Mvcm + I ω = I vcm Mvcm + R = I Mvcm ( + MR ) Rolling motion conserves mechanical energy. It must always be remembered, however, that part of the kinetic energy is going to be rotational.
Example: A.0-kg disk with a radius r =0 cm rolls without slipping. If the linear speed of the disk is vcm =.4 m/s, find the translational, rotational and total kinetic energy of the disk Translational: Kt = mvcm = (0.5)(.0)(.4) J =. J Total: Kt + Kr =.8 J I vcm Rotational: Kr = I ω = because r vcm = ω r for the disk, I = mr / Kr = mvcm i.e., = 0.6 J Question: What would change for a hoop with identical mass and radius? Translational kinetic energy: unchanged Rotational: Since for a hoop I = mr it is Kr = mv cm =. J same as Kt Total: Kt + Kr =.4 J Question: What would change for a solid sphere with identical mass and radius? Translational kinetic energy: unchanged Rotational: Since for a solid sphere I = mr /5 Kr = mvcm = 0.48 J 5 Total: Kt + Kr =.68 J Given two objects with the same mass and radius of rotation, more work is needed to impart a given translational speed to the one with the greater moment of inertia
Which object rolls down faster? Rolling objects with different moments of inertia with respect to the instantaneous axis of rotation, roll down a wedge in different times, even though mechanical energy is always conserved. Example: Solid sphere of mass M and radius R Potential energy at the top of the wedge: Mgh By the time sphere reaches the bottom, energy is entirely kinetic Since rolling sphere spins about axis through its centre h I = MR 5 Energy conservation: Mgh = yielding vcm = 0 7 gh at the bottom I Mvcm ( + MR ) = 7 0 Mvcm For a point-like object of mass M sliding frictionlessly, it would be gh Question: what if the sphere had identical mass and radius, but were hollow? For a hollow sphere, the moment of inertia with respect to the same axis is I = MR 3 Mvcm ( + 5 Energy conservation: Mgh = ) Mvcm 3 = 6 i.e., 6 vcm = gh at the bottom, solid sphere gets to the bottom sooner 5 Objects with a lower moment of inertia roll faster, for equal mass and radius
Non-ideal pulleys A pulley is a disk, spinning about an axis through its centre. A real pulley has a finite mass m, a radius r, and therefore a finite moment of inertia I = mr / m M M h Suppose two weights, one of mass M =. kg and the other of mass M, are initially suspended from a height h = 0 cm. The pulley is non-ideal, m = 0gr. The blocks are suddenly released. The heavy one falls while the lighter ons is lifted upward. We are interested in the speed of both blocks before the heavy one hits the floor. Energy conservation: 3Mgh is the initial potential energy Potential energy when block hits the floor: Mgh (lighter block is at height h) Kinetic energy when block hits the floor: (M + M)v + I ω = (M + M)v + v mr Angular speed of the pulley must be consistent with speed of blocks, i.e., v = ωr Hence, 3Mgh = Mgh ( 3M + + m 4 ) v (.)(9.8)(0.) J = (.65 + 0.) kg v yielding v =. m /s or v =. m/s r