Rotational Motion AP Physics

Angular Motion ∆𝜽=𝟐𝝅 𝒓𝒂𝒅 ∆𝜽=?? 𝑾𝒉𝒚 𝒓𝒂𝒅𝒊𝒂𝒏𝒔 𝒂𝒏𝒅 𝒏𝒐𝒕 𝒅𝒆𝒈𝒓𝒆𝒆𝒔? 𝜽 End Slide Angular Motion         ∆𝜽=𝟐𝝅 𝒓𝒂𝒅 ∆𝜽=?? 𝜽 𝑾𝒉𝒚 𝒓𝒂𝒅𝒊𝒂𝒏𝒔 𝒂𝒏𝒅 𝒏𝒐𝒕 𝒅𝒆𝒈𝒓𝒆𝒆𝒔? "𝑹𝒂𝒅𝒊𝒂𝒏𝒔" 𝒂𝒓𝒆 𝒂 𝒓𝒂𝒕𝒊𝒐 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝒕𝒉𝒆 𝒂𝒓𝒄𝒍𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒂 𝒄𝒊𝒓𝒄𝒍𝒆 𝒂𝒏𝒅 𝒕𝒉𝒆 𝒄𝒊𝒓𝒄𝒍 𝒆 ′ 𝒔 𝒓𝒂𝒅𝒊𝒖𝒔. 𝟐 𝒔𝒆𝒄𝒐𝒏𝒅𝒔 𝟎 𝒔𝒆𝒄𝒐𝒏𝒅𝒔 𝟑 𝒔𝒆𝒄𝒐𝒏𝒅𝒔 𝟒 𝒔𝒆𝒄𝒐𝒏𝒅𝒔 𝟓 𝒔𝒆𝒄𝒐𝒏𝒅𝒔 1 𝒔𝒆𝒄𝒐𝒏𝒅𝒔 𝑬𝒙𝒂𝒎𝒑𝒍𝒆… ∆𝜽= ∆𝒙 𝒓 ⇒ ∆𝒙=∆𝜽∗𝒓 𝑪𝒐𝒏𝒔𝒆𝒒𝒖𝒆𝒏𝒕𝒍𝒚, 𝒂 𝒇𝒖𝒍𝒍 𝒄𝒊𝒓𝒄𝒍𝒆 𝒊𝒔 𝟔.𝟐𝟖 𝒓𝒂𝒅𝒊𝒂𝒏𝒔 𝟐𝝅 . 𝑻𝒉𝒖𝒔… 𝒓 ∆𝒙 ∆𝜽 𝒄𝒊𝒓𝒄𝒖𝒎𝒇𝒆𝒓𝒆𝒏𝒄𝒆=𝟐𝝅𝒓

Angular Motion 𝒇= # 𝒄𝒚𝒄𝒍𝒆𝒔 𝒕 ∆𝜽=𝟐𝝅 𝒓𝒂𝒅 𝝎= ∆𝜽 𝒕 = 𝟐𝝅 𝟓.𝟎 𝒇= 𝟏 𝟓.𝟎 End Slide Angular Motion         𝒇= # 𝒄𝒚𝒄𝒍𝒆𝒔 𝒕 ∆𝜽=𝟐𝝅 𝒓𝒂𝒅 𝜽 𝝎= ∆𝜽 𝒕 = 𝟐𝝅 𝟓.𝟎 𝒇= 𝟏 𝟓.𝟎 𝒇=𝟎.𝟐𝟎 𝑯𝒛 𝑨𝒏𝒈𝒖𝒍𝒂𝒓 𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝝎=𝟏.𝟐𝟓𝟕 𝒓𝒂𝒅 𝒔𝒆𝒄 𝟓 𝒔𝒆𝒄𝒐𝒏𝒅𝒔 𝒇=𝟎.𝟐𝟎 𝒓𝒆𝒗 𝒔𝒆𝒄 𝑾𝒉𝒂 𝒕 ′ 𝒔 𝒕𝒉𝒆 𝒓𝒆𝒍𝒂𝒕𝒊𝒐𝒏𝒔𝒉𝒊𝒑 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝝎 𝒂𝒏𝒅 𝒇? 𝑯𝒐𝒘 𝒊𝒔 "𝝎" different than "𝒇"? 𝑺𝒊𝒏𝒄𝒆 𝒕𝒉𝒆𝒓𝒆 𝒂𝒓𝒆 𝟐𝝅 𝒓𝒂𝒅𝒊𝒂𝒏𝒔 𝒇𝒐𝒓 𝒆𝒗𝒆𝒓𝒚 𝒓𝒆𝒗𝒆𝒍𝒖𝒕𝒊𝒐𝒏, 𝒕𝒉𝒆𝒏 = 𝟐𝝅 𝑻 𝝎=𝟐𝝅𝒇

Linear vs. Angular Motion End Slide Linear vs. Angular Motion Linear Motion . Angular Motion . Measured as change in position (∆𝑥) and in units of meters Linear Velocity is change in position over time 𝑣= ∆𝑥 𝑡 Linear acceleration is change in linear velocity over time a= ∆𝑣 𝑡 Measured as change in angle (∆𝜃) and in units of radians Angular Velocity is change in angle over time 𝜔= ∆𝜃 𝑡 Angular acceleration is change in angular velocity over time 𝛼= ∆𝜔 𝑡

Linear vs. Angular Motion End Slide Linear vs. Angular Motion Linear Motion . Angular Motion . ∆𝒙= 𝟏 𝟐 𝐚 𝒕 𝟐 + 𝒗 𝒐 𝒕 𝒗 𝒇 = 𝒗 𝒐 +𝐚𝒕 𝒗 𝒇 𝟐 = 𝒗 𝒐 𝟐 +𝟐𝐚∆𝒙 ∆𝜽= 𝟏 𝟐 𝜶 𝒕 𝟐 + 𝝎 𝒐 𝒕 𝝎 𝒇 = 𝝎 𝒐 +𝜶𝒕 𝝎 𝒇 𝟐 = 𝝎 𝒐 𝟐 +𝟐𝜶∆𝜽 𝑹𝒐𝒕𝒂𝒕𝒊𝒐𝒏𝒂𝒍 𝑺𝒚𝒎𝒃𝒐𝒍𝒔 & 𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏𝒔 𝑾𝒌𝒔𝒕 𝑳𝒊𝒌𝒆𝒘𝒊𝒔𝒆, 𝑹𝒆𝒎𝒆𝒎𝒃𝒆𝒓, 𝒗=𝝎∗𝒓 ∆𝜽= ∆𝒙 𝒓 ⇒ ∆𝒙=∆𝜽∗𝒓 𝐚=𝜶∗𝒓 𝑵𝒐𝒕 𝒕𝒉𝒆 𝒔𝒂𝒎𝒆 𝒂𝒔 𝒄𝒆𝒏𝒕𝒓𝒊𝒑𝒆𝒕𝒂𝒍 𝒂𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏

End Slide Rotating Paper Disks The speed of a moving bullet can be determined by allowing the bullet to pass through two rotating paper disks mounted a distance 96 cm apart on the same axle. From the angular displacement 24.2o of the two bullet holes in the disks and the rotational speed 471 rev/min of the disks, what is the speed of the bullet? 𝒗= ∆𝒙 𝒕 = 𝟎.𝟗𝟔 𝒎 𝟎.𝟎𝟎𝟖𝟓𝟔 𝒔𝒆𝒄 ⇒ 𝒗=𝟏𝟏𝟐.𝟏 𝒎 𝒔 × 𝝎= ∆𝜽 𝒕 ⇒𝒕= ∆𝜽 𝝎 = 𝟐𝟒.𝟐° 𝟒𝟕𝟏 𝒓𝒆𝒗 𝒎𝒊𝒏 = 𝟐𝟒.𝟐 𝟒𝟕𝟏 ∗ °∗𝒎𝒊𝒏 𝒓𝒆𝒗 ∗ 𝟏 𝒓𝒆𝒗 𝟑𝟔𝟎° ∗ 𝟔𝟎 𝒔𝒆𝒄 𝟏 𝒎𝒊𝒏 × 𝒕=𝟎.𝟎𝟎𝟖𝟓𝟔 𝒔𝒆𝒄

Angular Speed of a Record End Slide Angular Speed of a Record A record player has a frequency of 26 rev/min. What is its angular speed? Through what angle does it rotate in 1.13 s? =𝟐𝝅 𝟐𝟔 𝒓𝒆𝒗 𝒎𝒊𝒏 ∗ 𝟏 𝒎𝒊𝒏 𝟔𝟎 𝒔𝒆𝒄 𝝎=𝟐𝝅𝒇 =𝟐.𝟕𝟐 𝒓𝒂𝒅 𝒔𝒆𝒄 𝝎= ∆𝜽 𝒕 ⇒∆𝜽=𝝎𝒕 =𝟐.𝟕𝟐∗𝟏.𝟏𝟑 =𝟑.𝟎𝟖 𝒓𝒂𝒅

Rotating Potter’s Wheel End Slide Rotating Potter’s Wheel A potter’s wheel of radius 14 cm starts from rest and rotates with constant angular acceleration until at the end of 25 s it is moving with angular velocity of 15 rad/s. What is its angular acceleration? What is the linear velocity of a point on the rim at the end of the 25 s? 𝜶= ∆𝝎 𝒕 = 𝟏𝟓−𝟎 𝟐𝟓 =𝟎.𝟔𝟎 𝒓𝒂𝒅 𝒔 𝟐 𝒗=𝝎𝒓 =𝟏𝟓∗𝟎.𝟏𝟒 =𝟐.𝟏𝟎 𝒎 𝒔

Rotating Potter’s Wheel End Slide Rotating Potter’s Wheel A potter’s wheel of radius 14 cm starts from rest and rotates with constant angular acceleration until at the end of 25 s it is moving with angular velocity of 15 rad/s. Through what angle did the wheel rotate in the 25 s? What is the average angular velocity of the wheel during the 25 s? ∆𝜽= 𝟏 𝟐 𝜶 𝒕 𝟐 + 𝝎 𝒐 𝒕 = 𝟏 𝟐 ∗𝟎.𝟔𝟎∗ 𝟐𝟓 𝟐 +𝟎∗𝟐𝟓 =𝟏𝟖𝟕.𝟓 𝒓𝒂𝒅 𝝎= ∆𝜽 𝒕 = 𝟏𝟖𝟕.𝟓 𝟐𝟓 =𝟕.𝟓𝟎 𝒓𝒂𝒅 𝒔

End Slide Spin Cycle In the spin cycle of a washing machine, the tub of radius 0.499 m develops a speed of 659 rpm. What is the maximum linear speed with which water leaves the machine? ??? =𝟔𝟓𝟗 𝒓𝒆𝒗 𝒎𝒊𝒏 ∗𝟎.𝟒𝟗𝟗 𝒎 𝒓𝒂𝒅 ∗𝟐𝝅 𝒓𝒂𝒅 𝒓𝒆𝒗 ∗ 𝟏 𝒎𝒊𝒏 𝟔𝟎 𝒔𝒆𝒄 𝒗=𝝎𝒓 ⇒ 𝒗=𝟑𝟒.𝟒 𝒎 𝒔

Nothing to do with radius End Slide Merry Go Round Jason and Isaac are riding on a merry-go-round. Jason rides on a horse at the outer rim of the circular platform, twice as far from the center of the circular platform as Isaac, who rides on an inner horse. When the merry- go-round is rotating at a constant angular speed, what is Jason’s angular speed relative to Isaac’s? The angular speed is the same as Isaac’s Their translational velocities will be different because of the difference in radius. Jason will have twice the translational velocity. 𝝎= ∆𝜽 𝒕 𝒗=𝝎∗𝒓 Nothing to do with radius Includes radius

End Slide Space Station Design You want to design a large, permanent space station so that no artificial gravity is necessary. You decide to shape it like a large coffee can of radius 198 m and rotate it about its central axis. What rotational speed would be required to simulate gravity? If an astronaut jogged in the direction of the rotation at 4.7 m/s, what simulated gravitational acceleration would the astronaut feel? = 𝟗.𝟖 𝟏𝟗𝟖 𝐚 𝒄 = 𝒗 𝟐 𝒓 = 𝝎𝒓 𝟐 𝒓 ⇒ 𝝎= 𝐚 𝒓 = 𝝎 𝟐 𝒓 ⇒ 𝝎=𝟎.𝟐𝟐𝟐 𝒓𝒂𝒅 𝒔𝒆𝒄 = 𝝎 𝒐 + 𝒗 𝑨 𝒓 =𝟎.𝟐𝟐𝟐+ 𝟒.𝟕 𝟏𝟗𝟖 𝝎= 𝝎 𝒐 + 𝝎 𝑨 =𝟎.𝟐𝟒𝟔 𝒓𝒂𝒅 𝒔𝒆𝒄 𝐚 𝒄 = 𝝎 𝟐 𝒓 = 𝟎.𝟐𝟒𝟔 𝟐 ∗𝟏𝟗𝟖 ⇒ 𝐚=𝟏𝟐. 𝟎 𝒎 𝒔 𝟐

End Slide Turtle Named Dizzy A small turtle, appropriately named “Dizzy”, is placed on a horizontal, rotating turntable at a distance of 15 cm from its center. Dizzy’s mass is 50 g, and the coefficient of static friction between his feet and turntable is 0.2. Find the maximum number of radians per second the turntable can have if Dizzy is to remain stationary relative to the turntable. What’s the frequency? ⇒ 𝝎< 𝝁 𝒔 𝒈 𝒓 𝑭 𝒄 < 𝑭 𝒇 ⇒ 𝒎 𝝎 𝟐 𝒓< 𝝁 𝒔 𝑭 𝑵 ⇒ 𝒎 𝝎 𝟐 𝒓< 𝝁 𝒔 𝒎𝒈 𝝎< 𝟎.𝟐∗𝟗.𝟖 𝟎.𝟏𝟓 ⇒ 𝝎<𝟑.𝟔𝟏 𝒓𝒂𝒅 𝒔 ⇒ 𝒇< 𝟑.𝟔𝟏 𝟐𝝅 𝝎=𝟐𝝅𝒇 ⇒ 𝒇<𝟎.𝟓𝟕𝟓 𝒓𝒆𝒗 𝒔

End Slide Turtle Named Dizzy A small turtle, appropriately named “Dizzy”, is placed on a horizontal, rotating turntable at a distance of 15 cm from its center. Dizzy’s mass is 50 g, and the coefficient of static friction between his feet and turntable is 0.2. The turntable starts from rest at t = 0, and has a uniform acceleration of 1.8 rad/s2. Find the time at which Dizzy begins to slip. 𝝎=𝟑.𝟔𝟏 𝒓𝒂𝒅 𝒔 ⇒ 𝒕= 𝝎 𝒇 𝜶 = 𝟑.𝟔𝟏 𝟏.𝟖 𝝎 𝒇 = 𝝎 𝒐 +𝜶𝒕 ⇒ 𝒕=𝟐.𝟎𝟏 𝒔𝒆𝒄

Newton’s 2nd Law For Rotation End Slide Newton’s 2nd Law For Rotation 𝑭𝒐𝒓 𝒓𝒐𝒕𝒂𝒕𝒊𝒐𝒏… 𝚺𝑭=𝒎𝐚 𝐚=𝜶∗𝒓 𝚺𝑭=𝒎𝜶𝒓 𝑻𝒐𝒓𝒒𝒖𝒆 𝒊𝒔 𝒇𝒐𝒓𝒄𝒆 𝒕𝒉𝒂𝒕 𝒑𝒓𝒐𝒅𝒖𝒄𝒆𝒔 𝒓𝒐𝒕𝒂𝒕𝒊𝒐𝒏 𝚺𝑭∗𝒓=𝒎𝜶𝒓∗𝒓 𝚺𝝉=𝚺𝑭∗𝒓 𝚺𝝉=𝒎 𝒓 𝟐 𝜶 𝚺𝝉=𝑵𝒆𝒕 𝑻𝒐𝒓𝒒𝒖𝒆 𝒎 𝒓 𝟐 =𝑹𝒐𝒕𝒂𝒕𝒊𝒐𝒏𝒂𝒍 𝑰𝒏𝒆𝒓𝒕𝒊𝒂 𝜶=𝑹𝒐𝒕𝒂𝒕𝒊𝒐𝒏𝒂𝒍 𝑨𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏

End Slide Rotational Inertia Different objects have their masses distributed differently. This distribution of mass will cause one object’s rotation to be harder to change than another.

End Slide Rotational Inertia Newton’s 1st Law for Rotation – an object rotating with a constant rotational inertia will continue with the same rotation unless acted on by an outside net torque. Rotational Inertia (𝑰) measures the tendency for a rotating object to continue to rotate. The more rotational inertia an object has, the harder it is to change its rotation. The basic equation for rotational inertia is… …where “k” is a constant that depends on the distribution of mass. 𝑰=𝒌𝒎 𝒓 𝟐

Newton’s 2nd Law For Rotation End Slide Newton’s 2nd Law For Rotation 𝚺𝝉=𝑰𝜶 …where the basic equation for 𝑰=𝒌𝒎 𝒓 𝟐

Examples of Rotational inertia End Slide Examples of Rotational inertia 𝒓 𝒎 Axis of Rotation For a single particle 𝑰 𝑷𝒂𝒓𝒕𝒊𝒄𝒍𝒆 =𝒎 𝒓 𝟐 𝒌=𝟏 𝒓 𝒎 Axis of Rotation 𝑰 𝑺𝒐𝒍𝒊𝒅 𝑺𝒑𝒉𝒆𝒓𝒆 = 𝟐 𝟓 𝒎 𝒓 𝟐 For a solid sphere 𝒌= 𝟐 𝟓 𝒓 𝒎 Axis of Rotation For a hollow sphere 𝑰 𝑯𝒐𝒍𝒍𝒐𝒘 𝑺𝒑𝒉𝒆𝒓𝒆 = 𝟐 𝟑 𝒎 𝒓 𝟐 𝒌= 𝟐 𝟑

Examples of Rotational inertia End Slide Examples of Rotational inertia Axis of Rotation 𝑳 𝒎 𝑰 𝑹𝒐𝒅 𝑪𝒆𝒏𝒕𝒆𝒓 = 𝟏 𝟏𝟐 𝒎 𝑳 𝟐 For a rod from center 𝒌= 𝟏 𝟏𝟐 Axis of Rotation 𝑳 𝒎 𝑰 𝑹𝒐𝒅 𝑬𝒏𝒅 = 𝟏 𝟑 𝒎 𝑳 𝟐 For a rod from end 𝒌= 𝟏 𝟑 For a disk or cylinder (central axis) Axis of Rotation 𝒓 𝒎 𝑰 𝑫𝒊𝒔𝒌 = 𝟏 𝟐 𝒎 𝒓 𝟐 𝒌= 𝟏 𝟐

Examples of Rotational inertia End Slide Examples of Rotational inertia Axis of Rotation 𝒎 For a thin hoop (central axis) 𝑰 𝑯𝒐𝒐𝒑 𝑪𝒆𝒏𝒕𝒆𝒓 =𝒎 𝒓 𝟐 𝒌=𝟏 Axis of Rotation 𝒎 For a thin hoop (diameter) 𝑰 𝑯𝒐𝒐𝒑 𝑫𝒊𝒂𝒎𝒆𝒕𝒆𝒓 = 𝟏 𝟐 𝒎 𝒓 𝟐 𝒌= 𝟏 𝟐

End Slide Rim of a Bicycle A 1.28 kg bicycle wheel, which can be thought of as a thin hoop, has a radius of 42 cm. The gear attached to the central axis of the wheel has a radius of 6.8 cm and a chain is pulling on the gear with a constant force of 300 N. What is the angular acceleration of the wheel? Starting from rest, what is the angular velocity of the wheel after 1.80 sec? 𝑰 𝑯𝒐𝒐𝒑 𝑪𝒆𝒏𝒕𝒆𝒓 =𝒎 𝒓 𝟐 =𝟏.𝟐𝟖∗ 𝟎.𝟒𝟐 𝟐 =𝟎.𝟐𝟐𝟓𝟖 𝒌𝒈∙ 𝒎 𝟐 ⇒ 𝜶= 𝟑𝟎𝟎∗𝟎.𝟎𝟔𝟖 𝟎.𝟐𝟐𝟓𝟖 𝚺𝝉=𝑰𝜶 ⇒ 𝑭∗𝒓=𝑰𝜶 ⇒ 𝜶=𝟗𝟎.𝟑 𝒓𝒂𝒅 𝒔𝒆𝒄 𝟐 𝝎 𝒇 = 𝝎 𝒐 +𝜶𝒕 =𝟗𝟎.𝟑∗𝟏.𝟖𝟎 ⇒ 𝝎 𝒇 =𝟏𝟔𝟐.𝟔 𝒓𝒂𝒅 𝒔𝒆𝒄

End Slide Rim of a Merry Go Round A 150 kg merry-go-round in the shape of a horizontal disk of radius 1.5 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.5 rev/s in 2 s? ⇒ 𝑭= 𝟏𝟔𝟖.𝟕𝟓∗𝟏.𝟓𝟕𝟏 𝟏.𝟓 𝚺𝝉=𝑰𝜶 ⇒ 𝑭∗𝒓=𝑰𝜶 ⇒ 𝑭=𝟏𝟕𝟔.𝟕 𝑵 𝑰 𝑫𝒊𝒔𝒌 = 𝟏 𝟐 𝒎 𝒓 𝟐 = 𝟏 𝟐 ∗𝟏𝟓𝟎∗ 𝟏.𝟓 𝟐 =𝟏𝟔𝟖.𝟕𝟓 𝒌𝒈∙ 𝒎 𝟐 𝜶= 𝝎 𝒇 − 𝝎 𝒐 𝒕 = 𝟎.𝟓−𝟎 𝟐 =𝟎.𝟐𝟓 𝒓𝒆𝒗 𝒔 𝟐 ∗𝟐𝝅 𝒓𝒂𝒅 𝒓𝒆𝒗 =𝟏.𝟓𝟕𝟏 𝒓𝒂𝒅 𝒔 𝟐

End Slide Pivoting Rod A long uniform rod of length 1.11 m and mass 4.37 kg is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest in a vertical position as in the figure. At the instant the rod is horizontal, find the magnitude of its angular acceleration. At the same instant, find the magnitude of the acceleration of its center of mass. 𝑰 𝑹𝒐𝒅 𝑬𝒏𝒅 = 𝟏 𝟑 𝒎 𝑳 𝟐 = 𝟏 𝟑 ∗𝟒.𝟑𝟕∗ 𝟏.𝟏𝟏 𝟐 =𝟏.𝟕𝟗𝟓 𝒌𝒈∙ 𝒎 𝟐 ⇒ 𝜶= 𝟒.𝟑𝟕∗𝟗.𝟖∗𝟎.𝟓𝟓𝟓 𝟏.𝟕𝟗𝟓 𝚺𝝉=𝑰𝜶 ⇒ 𝑭 𝒈 ∗ 𝑳 𝟐 =𝑰𝜶 ⇒ 𝜶=𝟏𝟑.𝟐 𝒓𝒂𝒅 𝒔𝒆𝒄 𝟐 𝐚=𝜶𝒓 =𝜶∗ 𝑳 𝟐 =𝟏𝟑.𝟐∗ 𝟏.𝟏𝟏 𝟐 ⇒𝐚=𝟕.𝟑𝟓 𝒎 𝒔 𝟐

End Slide Pivoting Rod A long uniform rod of length 1.11 m and mass 4.37 kg is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest in a vertical position as in the figure. At the same instant, find the force exerted on the end of the rod by the axis. 𝑭 𝚺𝑭=𝒎𝐚 ⇒ 𝑭− 𝑭 𝒈 =−𝒎𝐚 Pivot 𝑭=𝒎𝒈−𝒎𝐚 =𝟒.𝟑𝟕∗𝟗.𝟖−𝟒.𝟑𝟕∗𝟕.𝟑𝟓 𝑭 𝒈 𝑭=𝟏𝟎.𝟕 𝑵

End Slide Atwood Machine An Atwood machine is constructed using a disk of mass 2.1 kg and radius 24.9 cm. The mass hanging on one side of the pulley is 1.61 kg and the mass on the other side is 1.38 kg. The pulley is free to rotate and the string connecting the masses does not slip. What is the acceleration of the system? Free Body Diagrams 𝑭 𝑻𝟏 𝑭 𝒈𝟏 𝑭 𝒈𝟏 − 𝑭 𝑻𝟏 = 𝒎 𝟏 𝐚 𝑭 𝑻𝟐 𝑭 𝒈𝟐 𝑭 𝑻𝟐 − 𝑭 𝒈𝟐 = 𝒎 𝟐 𝐚 𝒎 𝟏 𝒈− 𝑭 𝑻𝟏 = 𝒎 𝟏 𝐚 𝑭 𝑻𝟐 − 𝒎 𝟐 𝒈= 𝒎 𝟐 𝐚 𝟏.𝟔𝟏∗𝟗.𝟖− 𝑭 𝑻𝟏 =𝟏.𝟔𝟏𝐚 𝑭 𝑻𝟐 −𝟏.𝟑𝟖∗𝟗.𝟖=𝟏.𝟑𝟖𝐚 𝐚 𝐚 𝟏𝟓.𝟖− 𝑭 𝑻𝟏 =𝟏.𝟔𝟏𝐚 𝑭 𝑻𝟐 −𝟏𝟑.𝟓=𝟏.𝟑𝟖𝐚 𝑭 𝑻𝟏 =𝟏𝟓.𝟖−𝟏.𝟔𝟏𝐚 𝑭 𝑻𝟐 =𝟏𝟑.𝟓+𝟏.𝟑𝟖𝐚

Atwood Machine 𝚺𝝉=𝑰𝜶 𝑭 𝑻𝟏 𝒓− 𝑭 𝑻𝟐 𝒓=𝑰𝜶 End Slide Atwood Machine 𝒓 𝑰 𝑫𝒊𝒔𝒌 = 𝟏 𝟐 𝒎 𝒓 𝟐 𝑭 𝑻𝟏 𝑭 𝑻𝟐 𝑰 𝑫𝒊𝒔𝒌 = 𝟏 𝟐 ∗𝟐.𝟏∗ 𝟎.𝟐𝟒𝟗 𝟐 𝚺𝝉=𝑰𝜶 𝑰 𝑫𝒊𝒔𝒌 =𝟎.𝟎𝟔𝟓𝟏 𝒌𝒈∙ 𝒎 𝟐 𝑭 𝑻𝟏 𝒓− 𝑭 𝑻𝟐 𝒓=𝑰𝜶 ⟹ 𝟏𝟓.𝟖−𝟏.𝟔𝟏𝐚−𝟏𝟑.𝟓−𝟏.𝟑𝟖𝐚 𝟎.𝟐𝟒𝟗=𝟎.𝟎𝟔𝟓𝟏∗ 𝐚 𝒓 𝑭 𝑻𝟏 − 𝑭 𝑻𝟐 𝒓=𝑰𝜶 𝟐.𝟑−𝟐.𝟗𝟗𝐚 𝟎.𝟐𝟒𝟗=𝟎.𝟎𝟔𝟓𝟏∗ 𝐚 𝟎.𝟐𝟒𝟗 ⟹ 𝟎.𝟓𝟕𝟑−𝟎.𝟕𝟒𝟓𝐚=𝟎.𝟐𝟔𝟏𝐚 𝟎.𝟓𝟕𝟑=𝟏.𝟎𝟎𝟔𝐚 ⟹ 𝐚=𝟎.𝟓𝟕𝟎 𝒎 𝒔 𝟐 𝑭 𝑻𝟏 =𝟏𝟓.𝟖−𝟏.𝟔𝟏𝐚 𝑭 𝑻𝟐 =𝟏𝟑.𝟓+𝟏.𝟑𝟖𝐚

Rolling Down the Ramp 𝑳𝒆 𝒕 ′ 𝒔 𝒇𝒊𝒏𝒅 𝒕𝒉𝒊𝒔 𝒆𝒙𝒑𝒆𝒓𝒊𝒎𝒆𝒏𝒕𝒂𝒍𝒍𝒚 𝒇𝒊𝒓𝒔𝒕. End Slide Rolling Down the Ramp Two masses roll down an incline. One is a “hoop” and the other is a solid disk. Each have about the same mass (0.467 kg) and radius (0.076 m). Both will be released to roll 140 cm down an 8.0o incline. Which will get to the bottom first? What will be the difference in time between the two? 𝑻𝒉𝒆 𝑺𝒐𝒍𝒊𝒅 𝑫𝒊𝒔𝒌 ⇒ 𝑭 𝒇 𝒓=𝒌𝒎 𝒓 𝟐 𝐚 𝒓 𝚺𝝉=𝑰𝛂 𝚺𝑭=𝒎𝐚 ⇒ 𝑭 𝒈\\ − 𝑭 𝒇 =𝒎𝐚 𝑳𝒆 𝒕 ′ 𝒔 𝒇𝒊𝒏𝒅 𝒕𝒉𝒊𝒔 𝒆𝒙𝒑𝒆𝒓𝒊𝒎𝒆𝒏𝒕𝒂𝒍𝒍𝒚 𝒇𝒊𝒓𝒔𝒕. 𝑭 𝒇 =𝒎𝒈 𝐬𝐢𝐧 𝜽 −𝒎𝐚 𝑭 𝒇 =𝒌𝒎𝐚 𝑭 𝒇 =𝒎 𝒈 𝐬𝐢𝐧 𝜽 −𝐚 𝒌𝒎𝐚=𝒎 𝒈 𝐬𝐢𝐧 𝜽 −𝐚 𝒌𝐚=𝒈 𝐬𝐢𝐧 𝜽 −𝐚 𝐚= 𝒈 𝐬𝐢𝐧 𝜽 𝒌+𝟏

End Slide Rolling Down the Ramp Two masses roll down an incline. One is a “hoop” and the other is a solid disk. Each have about the same mass (0.467 kg) and radius (0.076 m). Both will be released to roll 140 cm down an 8.0o incline. Which will get to the bottom first? What will be the difference in time between the two? 𝑻𝒉𝒆 𝑺𝒐𝒍𝒊𝒅 𝑫𝒊𝒔𝒌 ⇒ 𝒕= 𝟐∆𝒙 𝐚 𝐚= 𝒈 𝐬𝐢𝐧 𝜽 𝒌+𝟏 ∆𝒙= 𝟏 𝟐 𝐚 𝒕 𝟐 𝑺𝒐𝒍𝒊𝒅 𝑫𝒊𝒔𝒌 𝑯𝒐𝒐𝒑 𝑺𝒐𝒍𝒊𝒅 𝑫𝒊𝒔𝒌 𝑯𝒐𝒐𝒑 𝒕 𝑫 = 𝟐∗𝟏.𝟒 𝟎.𝟗𝟎𝟗 𝒕 𝑯 = 𝟐∗𝟏.𝟒 𝟎.𝟔𝟖𝟐 𝐚 𝑫 = 𝟗.𝟖 𝐬𝐢𝐧 𝟖.𝟎 𝟎.𝟓+𝟏 𝐚 𝑯 = 𝟗.𝟖 𝐬𝐢𝐧 𝟖.𝟎 𝟏+𝟏 𝐚 𝑫 =𝟎.𝟗𝟎𝟗 𝒎 𝒔 𝟐 𝐚 𝑯 =𝟎.𝟔𝟖𝟐 𝒎 𝒔 𝟐 𝒕 𝑫 =𝟏.𝟕𝟓𝟓 𝒔𝒆𝒄 𝒕 𝑯 =𝟐.𝟎𝟐𝟔 𝒔𝒆𝒄 ∆𝒕=𝟎.𝟐𝟕𝟏 𝒔𝒆𝒄