Table of Contents g. # 1 1/11/16 Momentum & Impulse (Bozemanscience Videos) 2 1/13/16 Conservation of Momentum 3 1/19/16 Elastic and Inelastic Collisions 4 1/19/16 Lab 1 Momentum 5 1/26/16 Rotational tatics (i.e., τ = 0) 6 2/8/16 Lab 2 Determination of Rotational Inertia 7 2/8/16 Rotational Dynamics (i.e., τ= Iα)
Table of Contents g. # 8 1/11/16 9 1/11/16 Rotational Inertia of a ingle article ( I = mr 2 ) Rotational Inertia of a ingle article ( I = mr 2 ) 10 2/18/16 arallel xis Theorem 11 2/18/16 12 2/18/16 13 2/18/16 Definition of the Center of Mass (cm) Definition of the Center of Mass (cm) Motion of the Center of Mass (cm)
Table of Contents g. # 13 2/18/16 Motion of the Center of Mass (cm) 14 2/19/16 erpendicular xis Theorem 15 2/19/16 Center of Mass Frame of Reference
Warm Up (2/16/16) Rotational Inertia 15.0- force (represented by F T ) is applied to a cord wrapped around a wheel of mass M = 4.00 kg and radius R = 33.0 cm. The wheel is observed to accelerate uniformly from rest to reach an angular speed of 30.0 rad/s in 3.00 s. If there is a frictional torque τ fr = 1.10 m, determine the moment of inertia of the wheel and its radius of gyration. The wheel is assumed to rotate about its center. Its freebody diagram is shown in the figure. Given: F T = 15, M = 4 kg, R =.33 m, ω o = 0, ω = 30 rad/s, t = 3 s, τ fr = 1.10 m, τ = I ext I =?, radius of gyration, k =? α ; ystem: wheel of mass, M
Warm Up (2/16/16) Rotational Inertia Given: F T = 15, M = 4 kg, R =.33 m, ω o = 0, ω = 30 rad/s, t = 3 s, τ fr = 1.10 m, τ = I ext I =?, radius of gyration, k =? α ; ystem: wheel of mass, M τ = I ext τ FT τ Ffr = I ω Δt l F T l F fr = I ω Δt ω ω o Δt 0, from rest! I = l F T l F fr ω t = 0.33 m (15 m) 1.10 m [ (30 rad/s) (3 s) ] = 0.385 kg m 2
Chapter 9 Moment of Inertia When discussing moments of inertia, especially for unusual or irregularly shaped objects, it is often convenient to work with the radius of gyration, k, which is a sort of average radius. Radius of Gyration, k = I M For a uniform cylinder, k = ½MR 2 M = R ½ 0.71 R
Warm Up (2/16/16) Rotational Inertia Given: F T = 15, M = 4 kg, R =.33 m, ω o = 0, ω = 30 rad/s, t = 3 s, τ fr = 1.10 m, τ = I ext I =?, radius of gyration, k =? α ; ystem: wheel of mass, M τ = I ext τ FT τ Ffr = I ω Δt l F T l F fr = I ω Δt ω ω o Δt 0, from rest! k = I M k = 0.310 m = 0.386 kg m 2 4.0 kg I = l F T l F fr ω t = 0.33 m (15 m) 1.10 m [ (30 rad/s) (3 s) ] = 0.385 kg m 2
Lab 2 Determination of Rotational Inertia roblem: What is the rotational inertia ( I ) of this device? tring pulls and rotates VC. Mass falls to accelerate system. Measurements: mass of falling weight, linear acceleration of falling weight, lever arm, τ = I ext τ FT τ Ffr = I α ; ystem: rotating device a r
Lab 2 Determination of Rotational Inertia Table 1 Trial a (m/s 2 ) 1 2 tring pulls and rotates VC. Mass falls to accelerate system. 3 4 verage:
Lab 2 Determination of Rotational Inertia τ FT τ Ffr = I a r a (m/s 2 ) Lever rm (m) α (rad/s 2 ) FT () τ ( m) I (kg m 2 )
Warm Up (2/17/16) Rotating Rod uniform rod of mass M and length l can pivot freely (i.e., we ignore friction) about a hinge or pin attached to the case of a large machine. The rod is held horizontally and then released. t the moment of release (when you are no longer exerting a force holding it up), determine (a) the angular acceleration of the rod and (b) the linear acceleration of the tip of the rod. ssume the force of gravity acts at the center of mass of the rod, as shown. a = r α = l α τ = I α ; ystem: Uniform Rod, M ext gf τ = I α ½l (Mg) = 2 ( ⅓Ml ) α Radius of Circle a = l [3 g (2 l)]= 3g 2 α = ½ (g) ⅓ l = 3 g 2 l
Determining Moments of Inertia By Experiment Lab 2 Using Calculus
Warm Up (2/18/16) arallel xis Theorem There are two simple theorems that are helpful in obtaining moments of inertia. The first is called the parallelaxis theorem. It relates the moment of inertia I of an object of total mass M about any axis, and its moment of inertia I cm about an axis passing through the center of mass and parallel to the first axis. If the axes are a distance h apart, then I = I cm + Mh 2 Determine the moment of inertia of a solid cylinder of radius R o and mass M about an axis tangent to its edge and parallel to its symmetry axis.
Warm Up (2/18/16) arallel xis Theorem I = I cm + Mh 2 Determine the moment of inertia of a solid cylinder of radius R o and mass M about an axis tangent to its edge and parallel to its symmetry axis. Given: h = R o, I cm = ½MR o2, I =? I = ½MR o 2 + MR o 2 I = 3 2 MR o 2
xis Definition of the Center of Mass (cm) τ = r F = ( r inθ )F = l F m 4 m3 m r 2 2 3 mj r j r j,cm 1 r 1 m 1 (a) = m j ( r j r cm ) = m j r j m j r cm = 0 O r cm Torque balance about the center of mass of a rigid body exerted by Fg. 0 τ = τ jcm r jcm m j g = m j r jcm g = 0 m j r jcm = r j = r cm + r j,cm cm 2 m j r cm r cm = = m j r j m j r j m j
Definition of the Center of Mass (cm) Four swimmers with masses 35 kg, 41 kg, 54 kg, and 63 kg sit on the corners of a square diving raft with 2.0 m sides. Where is the cm of the group of swimmers?
Definition of the Center of Mass (cm) Four swimmers with masses 35 kg, 41 kg, 54 kg, and 63 kg sit on the corners of a square diving raft with 2.0 m sides. Where is the cm of the group of swimmers? x cm y cm = 1 M = 1 M m j x j = 54kg(0 m) + 63kg(0 m) + 35kg( 2 m ) + 41kg(2 m ) m j y j = 152 kg m 193 kg 35kg + 41 kg + 54 kg + 63 kg = 0.79 m. r cm = 1 M m j r j = (54kg + 41 kg )(0 m) +63kg + 35kg)2m = 1 m. 35kg + 41 kg + 54 kg + 63kg r cm = 0.79m ^ x + 1 m y^
Motion of the Center of Mass xis m 4 m3 r 2 m 2 3 r j mj r j,cm 1 r 1 m 1 v cm = r cm t (a) = t r j = r cm + r j,cm m j r j M = 1 M O m j r j t r cm = 1 M cm m j v j = 1 M = M m j p j = p tot M Thus, p tot = M v cm Therefore, F = p tot ext t = M v cm t = M v cm t = M a cm
Warm Up (2/19/16) erpendicular xis Theorem The parallel axis theorem applies to any object. The perpendicular axis theorem, can be applied to only plane (flat) objects - that is to two-dimensional objects, or objects of uniform thickness whose thickness can be neglected. It states that the sum of the moments of inertia of a plane object about any two perpendicular axes in the plane of the object, is equal to the moment of inertia about an axis through their point of intersection perpendicular to the plane of the object. That is, if the object is in the xy-plane I z = I x + I y
Center of Mass Frame of Reference hysical situations are always described with respect to some frame of reference.
Center of Mass Frame of Reference hysical situations are always described with respect to some frame of reference. mj xis m 4 m3 r 2 m 2 3 r j r j,cm 1 r 1 m 1 (a) r j = r cm + r j,cm O r cm cm F = p tot ext t = M v cm t = M v cm t = M a cm What if the particles in a system are moving? The motion of particles within a system are best described in a frame of reference with its origin stationary at the system s center of mass.
Center of Mass Frame of Reference hysical situations are always described with respect to some frame of reference. xis Recall that p tot = M v cm mj m 4 m3 r 2 m 2 3 r j r j,cm 1 p tot,cm r 1 m 1 r j = r cm + r j,cm (a) v j = v cm + v j,cm The total linear momentum in the system s center of mass frame is: = p j, cm = m j v j,cm = m j ( v j v cm ) = m j v j m j v cm O r cm p tot,cm = 0 The total momentum in the center of mass frame vanishes! cm p tot M v cm