# L24-s1,8 Physics 114 – Lecture 24 §8.5 Rotational Dynamics Now the physics of rotation Using Newton’s 2 nd Law, with a = r α gives F = m a = m r α τ =

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L24-s1,8 Physics 114 – Lecture 24 §8.5 Rotational Dynamics Now the physics of rotation Using Newton’s 2 nd Law, with a = r α gives F = m a = m r α τ = r F sin θ = r ┴ F = r F ┴ gives with θ = 90 0 or with r ┴ = r, τ = m r 2 α m F r

L24-s2,8 Physics 114 – Lecture 24 Generalizing this result to a rigid body with point masses, m 1, m 2, m 3, …, at distances from the axis of rotation of, r 1, r 2, r 3, …, gives τ 1 = m 1 r 1 2 α 1, etc., A rigid body rotates such that ω 1 = ω 2 = ω 3 = … = ω and likewise α 1 = α 2 = α 3 = … = α Summing, τ net = Σ τ i = Σ (m i r i 2 α i ) = Σ (m i r i 2 ) α, where we have used α 1 = α 2 = α 3 = … = α

L24-s3,8 Physics 114 – Lecture 22 We define the moment of inertia of a body about a particular axis to be, I = Σ (m i r i 2 ) = m 1 r 1 2 + m 2 r 2 2 + m 3 r 3 2 + … Finally, τ net = I α Examples: Wheel of radius,r, – all the mass on the rim: I = mr 2 Cylinder or disc of radius, r, about its axis: I =½mr 2 Sphere of radius, r, about an axis through its center: I =(2/5)mr 2

L24-s4,8 Physics 114 – Lecture 24 §8.6 Solving Problems in Rotational Dynamics Study box, Problem Solving – Rotational Motion on p 209 Lets look at some examples

L24-s5,8 Physics 114 – Lecture 24 §8.7 Rotational Kinetic Energy Consider in a rigid body of finite size a particle of mass, m i, located at a perpendicular distance, r i, from the axis of rotation, with this body rotating about this axis with an angular velocity, ω Its linear velocity, v i, which is in the tangential direction, is then v i = r i ω and its kinetic energy of rotation, KE rot, is KE rot = ½ m i v i 2 = ½ m i (r i ω) 2 = ½ (m i r i 2 ) ω 2

L24-s6,8 Physics 114 – Lecture 24 Summing for all particles in the body, we obtain, KE rot = Σ ½ (m i r i 2 ) ω 2 = ½ (Σ m i r i 2 ) ω 2, since ω = const Using the definition, moment of inertia, I = Σ (m i r i 2 ), this becomes, KE rot = ½ I ω 2 For rolling motion, KE total = KE translational + KE rotational Giving, KE tot = ½ m v 2 + ½ I ω 2

L24-s7,8 Physics 114 – Lecture 24 For a body rolling down an incline, where the work done by the non-conservative forces, W NC, is 0 and where I ≡ I CM, since CM moves with velocity v = rω E = KE tot + PE = ½ m v 2 + ½ I ω 2 + mgh = constant This quantity is conserved just as we had earlier that the quantity, in the absence of rotational motion, E = KE tot + PE = ½ m v 2 + mgh = constant Lets see some examples: Wheel → KE tot = ½ m v 2 + ½ I ω 2 = ½ m v 2 + ½ mr 2 ω 2 = ½ m v 2 + ½ m(r ω) 2 = ½ m v 2 + ½ m v 2 = m v 2

L24-s8,8 Physics 114 – Lecture 24 Disc or cylinder → KE tot = ½ m v 2 + ½ I ω 2 = ½ m v 2 + ½(½ mr 2 ) ω 2 = ½ m v 2 + ¼ m(r ω) 2 = ½ m v 2 + ¼ m v 2 = ¾ m v 2 = 0.75 m v 2 Sphere → KE tot = ½ m v 2 + ½ I ω 2 = ½ m v 2 + ½(( 2/5) mr 2 ) ω 2 = ½ m v 2 + ( 1/5) m(r ω) 2 = ½ m v 2 + ( 1/5) m v 2 = ( 7/10) m v 2 = 0.7 m v 2 After starting from rest, at the same level on an incline, which will reach the foot of an incline first? A wheel, a cylinder or a sphere? Examples

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