Rotational Motion

Axis of rotation Angular displacement 𝜃= 𝑥 𝑟 θ is measured in radians 1 rev = 360° = 2π rad Radians is a “ghost” unit

Angular velocity 𝜔= 𝑣 𝑟 𝜔 is measured in rad/s 𝜔= 𝑑𝜃 𝑑𝑡

Angular acceleration 𝛼= 𝜔 𝑟 𝛼 is measured in rad/s2 𝛼= 𝑑𝜔 𝑑𝑡

Rotation with constant angular acceleration 𝑣= 𝑣 0 +𝑎𝑡 𝜔= 𝜔 0 +𝛼𝑡 𝑥= 𝑣 0 𝑡+ 1 2 𝑎 𝑡 2 𝜃= 𝜔 0 𝑡+ 1 2 𝛼 𝑡 2 𝑣 2 = 𝑣 0 2 +2𝑎𝑥 𝜔 2 = 𝜔 0 2 +2𝛼𝜃 𝑥= 1 2 𝑣 0 +𝑣 𝑡 𝜃= 1 2 𝜔 0 +𝜔 𝑡

Period of rotation Determine the period of a car travelling at 35 m/s around a track with a radius of 75 m. X s 1 rev = 1 s 35 m × 2π 75 m 1 rev 𝑇= 2𝜋 75 35 𝑇= 2𝜋𝑟 𝑣 𝑇=13.5 s

Determine the period of a rotating disc with an angular speed of 4 Determine the period of a rotating disc with an angular speed of 4.3 rad/s X s 1 rev = 1 s 4.3 rad × 2π rad 1 rev 𝑇= 2𝜋 4.3 𝑇= 2𝜋 𝜔 𝑇=1.46 s

Acceleration A rotating object 𝑎 𝑡 =𝛼𝑟 𝑎 𝑟 = 𝑣 2 𝑟 = 𝜔 2 𝑟 Does have centripetal acceleration Might have linear (tangential) acceleration 𝑎 𝑡 =𝛼𝑟 𝑎 𝑟 = 𝑣 2 𝑟 = 𝜔 2 𝑟

Kinetic energy If we treat a rotating object as a series of particles 𝐾= 1 2 𝑚 1 𝑣 1 2 + 1 2 𝑚 2 𝑣 2 2 + 1 2 𝑚 3 𝑣 3 2 +… 𝐾= 1 2 𝑚 𝑣 𝑖 2 𝐾= 1 2 𝑚 𝜔 𝑟 𝑖 2 𝐾= 1 2 𝑚 𝑟 𝑖 2 𝜔 2

𝐼= 𝑚 𝑟 𝑖 2 I is the moment of inertia or rotational inertia I is the rotational analog for mass 𝐾= 1 2 𝐼 𝜔 2

𝐼= 𝑚 𝑟 𝑖 2 If a body is continuous 𝐼= 𝑟 2 𝑑𝑚

Objects for which you need to be able to determine rotational inertia Thin hoop rotated about its center Thin rod rotated about an axis through its center and perpendicular to the rod Solid disk rotated about its center

Thin hoop rotated about its center 𝐼= 𝑟 2 𝑑𝑚 𝐼= 𝑅 2 𝑑𝑚 𝐼=𝑚 𝑅 2

ALL moments of inertia will have the same form: 𝐼=𝑥𝑀 𝑅 2 Where x ≤ 1

Thin rod rotated about an axis through its center and perpendicular to the rod 𝐼= 𝑟 2 𝑑𝑚

Consider a thin slice of width dr The mass of this slice, dm, is 𝑑𝑚=𝜇𝑑𝑟 Where μ is linear density 𝜇= 𝑀 𝐿 𝑑𝑚= 𝑀 𝐿 𝑑𝑟

𝐼= 𝑀 3𝐿 𝐿 3 4 𝐼= 𝑀 𝐿 2 12 or 𝐼= 1 12 𝑀 𝐿 2 𝐼= 𝑟 2 𝑀 𝐿 𝑑𝑟 𝐼= 𝑀 𝐿 𝑟 2 𝑑𝑟 𝐼= 𝑟 2 𝑀 𝐿 𝑑𝑟 𝐼= 𝑀 𝐿 𝑟 2 𝑑𝑟 𝐼= 𝑀 𝐿 − 𝐿 2 𝐿 2 𝑟 2 𝑑𝑟 𝐼= 𝑀 𝐿 𝑟 3 3 − 𝐿 2 𝐿 2 𝐼= 𝑀 3𝐿 𝐿 2 3 − − 𝐿 2 3 𝐼= 𝑀 3𝐿 𝐿 3 8 − − 𝐿 3 8 𝐼= 𝑀 3𝐿 𝐿 3 4 𝐼= 𝑀 𝐿 2 12 or 𝐼= 1 12 𝑀 𝐿 2

Solid disk (or cylinder) rotated about its center 𝐼= 𝑟 2 𝑑𝑚

Consider a thin ring a distance r from the center with a width of dr The mass of this slice, dm, is 𝑑𝑚=𝜎𝑑𝑟 Where σ is areal density 𝜎= 𝑀 𝐴 𝑑𝑚= 𝑀 𝐴 𝑑𝐴

𝑑𝑚= 𝑀 𝐴 𝑑𝐴 𝑑𝑚= 𝑀 𝜋 𝑅 2 𝑑𝐴 Area = ? “Uncurl” the slice 𝑑𝐴=2𝜋𝑟𝑑𝑟 𝑑𝑚= 𝑀 𝜋 𝑅 2 2𝜋𝑟𝑑𝑟 𝐼= 𝑟 2 𝑑𝑚 𝐼= 𝑟 2 𝑀 𝜋 𝑅 2 2𝜋𝑟𝑑𝑟 dr 2πr

𝐼= 𝑟 2 𝑀 𝜋 𝑅 2 2𝜋𝑟𝑑𝑟 𝐼= 2𝑀 𝑅 2 𝑟 3 𝑑𝑟 𝐼= 2𝑀 𝑅 2 0 𝑅 𝑟 3 𝑑𝑟 𝐼= 2𝑀 𝑅 2 𝑟 4 4 0 𝑅 𝐼= 2𝑀 𝑅 2 𝑅 4 4 𝐼= 1 2 𝑀 𝑅 2 dr 2πr

A rectangular plate about an axis through its center 𝐼= 𝑟 2 𝑑𝑚 𝑑𝑚= 𝑀 𝐴 𝑑𝐴

Consider a small piece of the plate of size da × db 𝑑𝑚= 𝑀 𝐴 𝑑𝐴 𝑑𝑚= 𝑀 𝑎𝑏 𝑑𝐴 𝑑𝑚= 𝑀 𝑎𝑏 𝑑𝑎𝑑𝑏 𝐼= 𝑟 2 𝑀 𝑎𝑏 𝑑𝑎𝑑𝑏 𝐼= 𝑀 𝑎𝑏 𝑟 2 𝑑𝑎𝑑𝑏 𝐼= 𝑀 𝑎𝑏 𝑟 𝑏 2 𝑟 𝑎 2 𝑑𝑎𝑑𝑏

𝐼= 𝑀 𝑎𝑏 𝑟 𝑏 2 𝑟 𝑎 2 𝑑𝑎𝑑𝑏 𝐼= 𝑀 𝑎𝑏 − 𝑏 2 𝑏 2 𝑟 𝑏 2 − 𝑎 2 𝑎 2 𝑟 𝑎 2 𝑑𝑎𝑑𝑏 𝐼= 𝑀 𝑎𝑏 − 𝑏 2 𝑏 2 𝑟 𝑏 2 𝑟 𝑎 3 3 𝑏 − 𝑎 2 𝑎 2 𝑑𝑎𝑑𝑏 𝐼= 𝑀 𝑎𝑏 − 𝑏 2 𝑏 2 𝑟 𝑏 2 𝑎 3 24 𝑏− − 𝑎 3 24 𝑏 𝑑𝑎𝑑𝑏 𝐼= 𝑀 𝑎𝑏 − 𝑏 2 𝑏 2 𝑟 𝑏 2 𝑎 3 𝑏 12 𝑑𝑎𝑑𝑏

𝐼= 𝑀 𝑎𝑏 − 𝑏 2 𝑏 2 𝑟 𝑏 2 𝑎 3 𝑏 12 𝑑𝑎𝑑𝑏 𝐼= 𝑀 𝑎𝑏 𝑎 3 𝑏 12 𝑟 𝑏 3 3 𝑎 − 𝑏 2 𝑏 2 𝐼= 𝑀 𝑎𝑏 𝑎 3 𝑏 12 𝑏 3 24 𝑎− − 𝑏 3 24 𝑎 𝐼= 𝑀 𝑎𝑏 𝑎 3 𝑏 12 + 𝑏 3 𝑎 12 𝐼= 1 12 𝑀 𝑎 2 + 𝑏 2

Annular cylinder about an axis through its center 𝐼= 𝑟 2 𝑑𝑚 𝑑𝑚= 𝑀 𝐴 𝑑𝐴 𝐼= 𝑀 𝐴 𝑟 2 𝑑𝐴 𝐼= 𝑀 𝜋 𝑟 2 2 −𝜋 𝑟 1 2 𝑟 2 𝑑𝐴 𝐼= 𝑀 𝜋 𝑟 2 2 −𝜋 𝑟 1 2 𝑟 2 2𝜋𝑟 𝑑𝑟 𝐼= 2𝜋𝑀 𝜋 𝑟 2 2 −𝜋 𝑟 1 2 𝑟 3 𝑑𝑟

𝐼= 2𝜋𝑀 𝜋 𝑟 2 2 −𝜋 𝑟 1 2 𝑟 3 𝑑𝑟 𝐼= 2𝑀 𝑟 2 2 − 𝑟 1 2 𝑟 1 𝑟 2 𝑟 3 𝑑𝑟 𝐼= 2𝑀 𝑟 2 2 − 𝑟 1 2 𝑟 4 4 𝑟 1 𝑟 2 𝐼= 𝑀 2 𝑟 2 2 − 𝑟 1 2 𝑟 2 4 − 𝑟 1 4 𝐼= 𝑀 2 𝑟 2 2 − 𝑟 1 2 𝑟 2 2 − 𝑟 1 2 𝑟 2 2 + 𝑟 1 2 𝐼= 1 2 𝑀 𝑟 1 2 + 𝑟 2 2

Thin rod rotated about an axis through its center and perpendicular to the rod 𝐼= 1 12 𝑀 𝐿 2 end ?

Consider a thin slice of width dr The mass of this slice, dm, is 𝑑𝑚=𝜇𝑑𝑟 Where μ is linear density 𝜇= 𝑀 𝐿 𝑑𝑚= 𝑀 𝐿 𝑑𝑟

𝐼= 𝑟 2 𝑀 𝐿 𝑑𝑟 𝐼= 𝑀 𝐿 𝑟 2 𝑑𝑟 𝐼= 𝑀 𝐿 0 𝐿 𝑟 2 𝑑𝑟 𝐼= 𝑀 𝐿 𝑟 3 3 0 𝐿 𝐼= 𝑀 3𝐿 𝐿 3 − 0 3 𝐼= 𝑀 3𝐿 𝐿 3 𝐼= 𝑀 𝐿 2 3 or 𝐼= 1 3 𝑀 𝐿 2

The rotational inertia for a thin rod about and axis through its center and perpendicular to the rod was 1 12 𝑀 𝐿 2 . Determine the rotational inertia about a parallel axis through the end of the rod. 𝐼= 𝐼 𝑐𝑜𝑚 +𝑀 𝑑 2 (Parallel axis theorem) 𝐼= 1 12 𝑀 𝐿 2 +𝑀 𝑑 2 𝐼= 1 12 𝑀 𝐿 2 +𝑀 𝐿 2 2 𝐼= 1 12 𝑀 𝐿 2 + 1 4 𝑀 𝐿 2 𝐼= 1 12 𝑀 𝐿 2 + 3 12 𝑀 𝐿 2 𝐼= 4 12 𝑀 𝐿 2 = 1 3 𝑀 𝐿 2

Torque is the ability of a force to cause rotation in an object 𝜏=𝑟𝐹 sin 𝜑 r is the distance from the axis of rotation to the point of application of the force φ is the smaller of the two angles made by r and F

Torques are additive Torques in the clockwise direction are negative, counterclockwise are positive

Newton’s second law for rotation Σ𝜏=𝐼𝛼 Unit for torque 𝑀 𝑅 2 𝑟𝑎𝑑 𝑠 2 =𝑘𝑔∙ 𝑚 2 1 𝑠 2 𝑘𝑔∙𝑚 1 𝑠 2 =𝑁 𝑁∙𝑚 (NOT joules!!!!!!!)