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Rotational Motion AP Physics.

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Presentation on theme: "Rotational Motion AP Physics."β€” Presentation transcript:

1 Rotational Motion AP Physics

2 Angular Motion βˆ†πœ½=πŸπ… 𝒓𝒂𝒅 βˆ†πœ½=?? π‘Ύπ’‰π’š π’“π’‚π’…π’Šπ’‚π’π’” 𝒂𝒏𝒅 𝒏𝒐𝒕 π’…π’†π’ˆπ’“π’†π’†π’”? 𝜽
End Slide Angular Motion βˆ†πœ½=πŸπ… 𝒓𝒂𝒅 βˆ†πœ½=?? 𝜽 π‘Ύπ’‰π’š π’“π’‚π’…π’Šπ’‚π’π’” 𝒂𝒏𝒅 𝒏𝒐𝒕 π’…π’†π’ˆπ’“π’†π’†π’”? "π‘Ήπ’‚π’…π’Šπ’‚π’π’”" 𝒂𝒓𝒆 𝒂 π’“π’‚π’•π’Šπ’ π’ƒπ’†π’•π’˜π’†π’†π’ 𝒕𝒉𝒆 π’‚π’“π’„π’π’†π’π’ˆπ’•π’‰ 𝒐𝒇 𝒂 π’„π’Šπ’“π’„π’π’† 𝒂𝒏𝒅 𝒕𝒉𝒆 π’„π’Šπ’“π’„π’ 𝒆 β€² 𝒔 π’“π’‚π’…π’Šπ’–π’”. 𝟐 𝒔𝒆𝒄𝒐𝒏𝒅𝒔 𝟎 𝒔𝒆𝒄𝒐𝒏𝒅𝒔 πŸ‘ 𝒔𝒆𝒄𝒐𝒏𝒅𝒔 πŸ’ 𝒔𝒆𝒄𝒐𝒏𝒅𝒔 πŸ“ 𝒔𝒆𝒄𝒐𝒏𝒅𝒔 1 𝒔𝒆𝒄𝒐𝒏𝒅𝒔 π‘¬π’™π’‚π’Žπ’‘π’π’†β€¦ βˆ†πœ½= βˆ†π’™ 𝒓 β‡’ βˆ†π’™=βˆ†πœ½βˆ—π’“ π‘ͺπ’π’π’”π’†π’’π’–π’†π’π’•π’π’š, 𝒂 𝒇𝒖𝒍𝒍 π’„π’Šπ’“π’„π’π’† π’Šπ’” πŸ”.πŸπŸ– π’“π’‚π’…π’Šπ’‚π’π’” πŸπ… . 𝑻𝒉𝒖𝒔… 𝒓 βˆ†π’™ βˆ†πœ½ π’„π’Šπ’“π’„π’–π’Žπ’‡π’†π’“π’†π’π’„π’†=πŸπ…π’“

3 Angular Motion 𝒇= # π’„π’šπ’„π’π’†π’” 𝒕 βˆ†πœ½=πŸπ… 𝒓𝒂𝒅 𝝎= βˆ†πœ½ 𝒕 = πŸπ… πŸ“.𝟎 𝒇= 𝟏 πŸ“.𝟎
End Slide Angular Motion 𝒇= # π’„π’šπ’„π’π’†π’” 𝒕 βˆ†πœ½=πŸπ… 𝒓𝒂𝒅 𝜽 𝝎= βˆ†πœ½ 𝒕 = πŸπ… πŸ“.𝟎 𝒇= 𝟏 πŸ“.𝟎 𝒇=𝟎.𝟐𝟎 𝑯𝒛 π‘¨π’π’ˆπ’–π’π’‚π’“ π‘½π’†π’π’π’„π’Šπ’•π’š 𝝎=𝟏.πŸπŸ“πŸ• 𝒓𝒂𝒅 𝒔𝒆𝒄 πŸ“ 𝒔𝒆𝒄𝒐𝒏𝒅𝒔 𝒇=𝟎.𝟐𝟎 𝒓𝒆𝒗 𝒔𝒆𝒄 𝑾𝒉𝒂 𝒕 β€² 𝒔 𝒕𝒉𝒆 π’“π’†π’π’‚π’•π’Šπ’π’π’”π’‰π’Šπ’‘ π’ƒπ’†π’•π’˜π’†π’†π’ 𝝎 𝒂𝒏𝒅 𝒇? π‘―π’π’˜ π’Šπ’” "𝝎" different than "𝒇"? π‘Ίπ’Šπ’π’„π’† 𝒕𝒉𝒆𝒓𝒆 𝒂𝒓𝒆 πŸπ… π’“π’‚π’…π’Šπ’‚π’π’” 𝒇𝒐𝒓 π’†π’—π’†π’“π’š π’“π’†π’—π’†π’π’–π’•π’Šπ’π’, 𝒕𝒉𝒆𝒏 = πŸπ… 𝑻 𝝎=πŸπ…π’‡

4 Linear vs. Angular Motion
End Slide Linear vs. Angular Motion Linear Motion . Angular Motion . Measured as change in position (βˆ†π‘₯) and in units of meters Linear Velocity is change in position over time 𝑣= βˆ†π‘₯ 𝑑 Linear acceleration is change in linear velocity over time a= βˆ†π‘£ 𝑑 Measured as change in angle (βˆ†πœƒ) and in units of radians Angular Velocity is change in angle over time πœ”= βˆ†πœƒ 𝑑 Angular acceleration is change in angular velocity over time 𝛼= βˆ†πœ” 𝑑

5 Linear vs. Angular Motion
End Slide Linear vs. Angular Motion Linear Motion . Angular Motion . βˆ†π’™= 𝟏 𝟐 𝐚 𝒕 𝟐 + 𝒗 𝒐 𝒕 𝒗 𝒇 = 𝒗 𝒐 +πšπ’• 𝒗 𝒇 𝟐 = 𝒗 𝒐 𝟐 +πŸπšβˆ†π’™ βˆ†πœ½= 𝟏 𝟐 𝜢 𝒕 𝟐 + 𝝎 𝒐 𝒕 𝝎 𝒇 = 𝝎 𝒐 +πœΆπ’• 𝝎 𝒇 𝟐 = 𝝎 𝒐 𝟐 +πŸπœΆβˆ†πœ½ π‘Ήπ’π’•π’‚π’•π’Šπ’π’π’‚π’ π‘Ίπ’šπ’Žπ’ƒπ’π’π’” & π‘¬π’’π’–π’‚π’•π’Šπ’π’π’” π‘Ύπ’Œπ’”π’• π‘³π’Šπ’Œπ’†π’˜π’Šπ’”π’†, π‘Ήπ’†π’Žπ’†π’Žπ’ƒπ’†π’“, 𝒗=πŽβˆ—π’“ βˆ†πœ½= βˆ†π’™ 𝒓 β‡’ βˆ†π’™=βˆ†πœ½βˆ—π’“ 𝐚=πœΆβˆ—π’“ 𝑡𝒐𝒕 𝒕𝒉𝒆 π’”π’‚π’Žπ’† 𝒂𝒔 π’„π’†π’π’•π’“π’Šπ’‘π’†π’•π’‚π’ π’‚π’„π’„π’†π’π’†π’“π’‚π’•π’Šπ’π’

6 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= βˆ†π’™ 𝒕 = 𝟎.πŸ—πŸ” π’Ž 𝟎.πŸŽπŸŽπŸ–πŸ“πŸ” 𝒔𝒆𝒄 β‡’ 𝒗=𝟏𝟏𝟐.𝟏 π’Ž 𝒔 Γ— 𝝎= βˆ†πœ½ 𝒕 ⇒𝒕= βˆ†πœ½ 𝝎 = πŸπŸ’.𝟐° πŸ’πŸ•πŸ 𝒓𝒆𝒗 π’Žπ’Šπ’ = πŸπŸ’.𝟐 πŸ’πŸ•πŸ βˆ— Β°βˆ—π’Žπ’Šπ’ 𝒓𝒆𝒗 βˆ— 𝟏 𝒓𝒆𝒗 πŸ‘πŸ”πŸŽΒ° βˆ— πŸ”πŸŽ 𝒔𝒆𝒄 𝟏 π’Žπ’Šπ’ Γ— 𝒕=𝟎.πŸŽπŸŽπŸ–πŸ“πŸ” 𝒔𝒆𝒄

7 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? =πŸπ… πŸπŸ” 𝒓𝒆𝒗 π’Žπ’Šπ’ βˆ— 𝟏 π’Žπ’Šπ’ πŸ”πŸŽ 𝒔𝒆𝒄 𝝎=πŸπ…π’‡ =𝟐.πŸ•πŸ 𝒓𝒂𝒅 𝒔𝒆𝒄 𝝎= βˆ†πœ½ 𝒕 β‡’βˆ†πœ½=πŽπ’• =𝟐.πŸ•πŸβˆ—πŸ.πŸπŸ‘ =πŸ‘.πŸŽπŸ– 𝒓𝒂𝒅

8 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? 𝜢= βˆ†πŽ 𝒕 = πŸπŸ“βˆ’πŸŽ πŸπŸ“ =𝟎.πŸ”πŸŽ 𝒓𝒂𝒅 𝒔 𝟐 𝒗=πŽπ’“ =πŸπŸ“βˆ—πŸŽ.πŸπŸ’ =𝟐.𝟏𝟎 π’Ž 𝒔

9 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? βˆ†πœ½= 𝟏 𝟐 𝜢 𝒕 𝟐 + 𝝎 𝒐 𝒕 = 𝟏 𝟐 βˆ—πŸŽ.πŸ”πŸŽβˆ— πŸπŸ“ 𝟐 +πŸŽβˆ—πŸπŸ“ =πŸπŸ–πŸ•.πŸ“ 𝒓𝒂𝒅 𝝎= βˆ†πœ½ 𝒕 = πŸπŸ–πŸ•.πŸ“ πŸπŸ“ =πŸ•.πŸ“πŸŽ 𝒓𝒂𝒅 𝒔

10 End Slide Spin Cycle In the spin cycle of a washing machine, the tub of radius m develops a speed of 659 rpm. What is the maximum linear speed with which water leaves the machine? ??? =πŸ”πŸ“πŸ— 𝒓𝒆𝒗 π’Žπ’Šπ’ βˆ—πŸŽ.πŸ’πŸ—πŸ— π’Ž 𝒓𝒂𝒅 βˆ—πŸπ… 𝒓𝒂𝒅 𝒓𝒆𝒗 βˆ— 𝟏 π’Žπ’Šπ’ πŸ”πŸŽ 𝒔𝒆𝒄 𝒗=πŽπ’“ β‡’ 𝒗=πŸ‘πŸ’.πŸ’ π’Ž 𝒔

11 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= βˆ†πœ½ 𝒕 𝒗=πŽβˆ—π’“ Nothing to do with radius Includes radius

12 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? = πŸ—.πŸ– πŸπŸ—πŸ– 𝐚 𝒄 = 𝒗 𝟐 𝒓 = πŽπ’“ 𝟐 𝒓 β‡’ 𝝎= 𝐚 𝒓 = 𝝎 𝟐 𝒓 β‡’ 𝝎=𝟎.𝟐𝟐𝟐 𝒓𝒂𝒅 𝒔𝒆𝒄 = 𝝎 𝒐 + 𝒗 𝑨 𝒓 =𝟎.𝟐𝟐𝟐+ πŸ’.πŸ• πŸπŸ—πŸ– 𝝎= 𝝎 𝒐 + 𝝎 𝑨 =𝟎.πŸπŸ’πŸ” 𝒓𝒂𝒅 𝒔𝒆𝒄 𝐚 𝒄 = 𝝎 𝟐 𝒓 = 𝟎.πŸπŸ’πŸ” 𝟐 βˆ—πŸπŸ—πŸ– β‡’ 𝐚=𝟏𝟐. 𝟎 π’Ž 𝒔 𝟐

13 End Slide Turtle Named Dizzy A small turtle, appropriately named β€œDizzy”, 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? β‡’ 𝝎< 𝝁 𝒔 π’ˆ 𝒓 𝑭 𝒄 < 𝑭 𝒇 β‡’ π’Ž 𝝎 𝟐 𝒓< 𝝁 𝒔 𝑭 𝑡 β‡’ π’Ž 𝝎 𝟐 𝒓< 𝝁 𝒔 π’Žπ’ˆ 𝝎< 𝟎.πŸβˆ—πŸ—.πŸ– 𝟎.πŸπŸ“ β‡’ 𝝎<πŸ‘.πŸ”πŸ 𝒓𝒂𝒅 𝒔 β‡’ 𝒇< πŸ‘.πŸ”πŸ πŸπ… 𝝎=πŸπ…π’‡ β‡’ 𝒇<𝟎.πŸ“πŸ•πŸ“ 𝒓𝒆𝒗 𝒔

14 End Slide Turtle Named Dizzy A small turtle, appropriately named β€œDizzy”, 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=πŸ‘.πŸ”πŸ 𝒓𝒂𝒅 𝒔 β‡’ 𝒕= 𝝎 𝒇 𝜢 = πŸ‘.πŸ”πŸ 𝟏.πŸ– 𝝎 𝒇 = 𝝎 𝒐 +πœΆπ’• β‡’ 𝒕=𝟐.𝟎𝟏 𝒔𝒆𝒄

15 Newton’s 2nd Law For Rotation
End Slide Newton’s 2nd Law For Rotation 𝑭𝒐𝒓 π’“π’π’•π’‚π’•π’Šπ’π’β€¦ πšΊπ‘­=π’Žπš 𝐚=πœΆβˆ—π’“ πšΊπ‘­=π’ŽπœΆπ’“ 𝑻𝒐𝒓𝒒𝒖𝒆 π’Šπ’” 𝒇𝒐𝒓𝒄𝒆 𝒕𝒉𝒂𝒕 𝒑𝒓𝒐𝒅𝒖𝒄𝒆𝒔 π’“π’π’•π’‚π’•π’Šπ’π’ πšΊπ‘­βˆ—π’“=π’ŽπœΆπ’“βˆ—π’“ πšΊπ‰=πšΊπ‘­βˆ—π’“ πšΊπ‰=π’Ž 𝒓 𝟐 𝜢 πšΊπ‰=𝑡𝒆𝒕 𝑻𝒐𝒓𝒒𝒖𝒆 π’Ž 𝒓 𝟐 =π‘Ήπ’π’•π’‚π’•π’Šπ’π’π’‚π’ π‘°π’π’†π’“π’•π’Šπ’‚ 𝜢=π‘Ήπ’π’•π’‚π’•π’Šπ’π’π’‚π’ π‘¨π’„π’„π’†π’π’†π’“π’‚π’•π’Šπ’π’

16 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.

17 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 β€œk” is a constant that depends on the distribution of mass. 𝑰=π’Œπ’Ž 𝒓 𝟐

18 Newton’s 2nd Law For Rotation
End Slide Newton’s 2nd Law For Rotation πšΊπ‰=π‘°πœΆ …where the basic equation for 𝑰=π’Œπ’Ž 𝒓 𝟐

19 Examples of Rotational inertia
End Slide Examples of Rotational inertia 𝒓 π’Ž Axis of Rotation For a single particle 𝑰 π‘·π’‚π’“π’•π’Šπ’„π’π’† =π’Ž 𝒓 𝟐 π’Œ=𝟏 𝒓 π’Ž Axis of Rotation 𝑰 π‘Ίπ’π’π’Šπ’… 𝑺𝒑𝒉𝒆𝒓𝒆 = 𝟐 πŸ“ π’Ž 𝒓 𝟐 For a solid sphere π’Œ= 𝟐 πŸ“ 𝒓 π’Ž Axis of Rotation For a hollow sphere 𝑰 π‘―π’π’π’π’π’˜ 𝑺𝒑𝒉𝒆𝒓𝒆 = 𝟐 πŸ‘ π’Ž 𝒓 𝟐 π’Œ= 𝟐 πŸ‘

20 Examples of Rotational inertia
End Slide Examples of Rotational inertia Axis of Rotation 𝑳 π’Ž 𝑰 𝑹𝒐𝒅 π‘ͺ𝒆𝒏𝒕𝒆𝒓 = 𝟏 𝟏𝟐 π’Ž 𝑳 𝟐 For a rod from center π’Œ= 𝟏 𝟏𝟐 Axis of Rotation 𝑳 π’Ž 𝑰 𝑹𝒐𝒅 𝑬𝒏𝒅 = 𝟏 πŸ‘ π’Ž 𝑳 𝟐 For a rod from end π’Œ= 𝟏 πŸ‘ For a disk or cylinder (central axis) Axis of Rotation 𝒓 π’Ž 𝑰 π‘«π’Šπ’”π’Œ = 𝟏 𝟐 π’Ž 𝒓 𝟐 π’Œ= 𝟏 𝟐

21 Examples of Rotational inertia
End Slide Examples of Rotational inertia Axis of Rotation π’Ž For a thin hoop (central axis) 𝑰 𝑯𝒐𝒐𝒑 π‘ͺ𝒆𝒏𝒕𝒆𝒓 =π’Ž 𝒓 𝟐 π’Œ=𝟏 Axis of Rotation π’Ž For a thin hoop (diameter) 𝑰 𝑯𝒐𝒐𝒑 π‘«π’Šπ’‚π’Žπ’†π’•π’†π’“ = 𝟏 𝟐 π’Ž 𝒓 𝟐 π’Œ= 𝟏 𝟐

22 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 π‘ͺ𝒆𝒏𝒕𝒆𝒓 =π’Ž 𝒓 𝟐 =𝟏.πŸπŸ–βˆ— 𝟎.πŸ’πŸ 𝟐 =𝟎.πŸπŸπŸ“πŸ– π’Œπ’ˆβˆ™ π’Ž 𝟐 β‡’ 𝜢= πŸ‘πŸŽπŸŽβˆ—πŸŽ.πŸŽπŸ”πŸ– 𝟎.πŸπŸπŸ“πŸ– πšΊπ‰=π‘°πœΆ β‡’ π‘­βˆ—π’“=π‘°πœΆ β‡’ 𝜢=πŸ—πŸŽ.πŸ‘ 𝒓𝒂𝒅 𝒔𝒆𝒄 𝟐 𝝎 𝒇 = 𝝎 𝒐 +πœΆπ’• =πŸ—πŸŽ.πŸ‘βˆ—πŸ.πŸ–πŸŽ β‡’ 𝝎 𝒇 =πŸπŸ”πŸ.πŸ” 𝒓𝒂𝒅 𝒔𝒆𝒄

23 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? β‡’ 𝑭= πŸπŸ”πŸ–.πŸ•πŸ“βˆ—πŸ.πŸ“πŸ•πŸ 𝟏.πŸ“ πšΊπ‰=π‘°πœΆ β‡’ π‘­βˆ—π’“=π‘°πœΆ β‡’ 𝑭=πŸπŸ•πŸ”.πŸ• 𝑡 𝑰 π‘«π’Šπ’”π’Œ = 𝟏 𝟐 π’Ž 𝒓 𝟐 = 𝟏 𝟐 βˆ—πŸπŸ“πŸŽβˆ— 𝟏.πŸ“ 𝟐 =πŸπŸ”πŸ–.πŸ•πŸ“ π’Œπ’ˆβˆ™ π’Ž 𝟐 𝜢= 𝝎 𝒇 βˆ’ 𝝎 𝒐 𝒕 = 𝟎.πŸ“βˆ’πŸŽ 𝟐 =𝟎.πŸπŸ“ 𝒓𝒆𝒗 𝒔 𝟐 βˆ—πŸπ… 𝒓𝒂𝒅 𝒓𝒆𝒗 =𝟏.πŸ“πŸ•πŸ 𝒓𝒂𝒅 𝒔 𝟐

24 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 πŸ‘ π’Ž 𝑳 𝟐 = 𝟏 πŸ‘ βˆ—πŸ’.πŸ‘πŸ•βˆ— 𝟏.𝟏𝟏 𝟐 =𝟏.πŸ•πŸ—πŸ“ π’Œπ’ˆβˆ™ π’Ž 𝟐 β‡’ 𝜢= πŸ’.πŸ‘πŸ•βˆ—πŸ—.πŸ–βˆ—πŸŽ.πŸ“πŸ“πŸ“ 𝟏.πŸ•πŸ—πŸ“ πšΊπ‰=π‘°πœΆ β‡’ 𝑭 π’ˆ βˆ— 𝑳 𝟐 =π‘°πœΆ β‡’ 𝜢=πŸπŸ‘.𝟐 𝒓𝒂𝒅 𝒔𝒆𝒄 𝟐 𝐚=πœΆπ’“ =πœΆβˆ— 𝑳 𝟐 =πŸπŸ‘.πŸβˆ— 𝟏.𝟏𝟏 𝟐 β‡’πš=πŸ•.πŸ‘πŸ“ π’Ž 𝒔 𝟐

25 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 πšΊπ‘­=π’Žπš β‡’ π‘­βˆ’ 𝑭 π’ˆ =βˆ’π’Žπš Pivot 𝑭=π’Žπ’ˆβˆ’π’Žπš =πŸ’.πŸ‘πŸ•βˆ—πŸ—.πŸ–βˆ’πŸ’.πŸ‘πŸ•βˆ—πŸ•.πŸ‘πŸ“ 𝑭 π’ˆ 𝑭=𝟏𝟎.πŸ• 𝑡

26 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 π‘»πŸ 𝑭 π’ˆπŸ 𝑭 π’ˆπŸ βˆ’ 𝑭 π‘»πŸ = π’Ž 𝟏 𝐚 𝑭 π‘»πŸ 𝑭 π’ˆπŸ 𝑭 π‘»πŸ βˆ’ 𝑭 π’ˆπŸ = π’Ž 𝟐 𝐚 π’Ž 𝟏 π’ˆβˆ’ 𝑭 π‘»πŸ = π’Ž 𝟏 𝐚 𝑭 π‘»πŸ βˆ’ π’Ž 𝟐 π’ˆ= π’Ž 𝟐 𝐚 𝟏.πŸ”πŸβˆ—πŸ—.πŸ–βˆ’ 𝑭 π‘»πŸ =𝟏.πŸ”πŸπš 𝑭 π‘»πŸ βˆ’πŸ.πŸ‘πŸ–βˆ—πŸ—.πŸ–=𝟏.πŸ‘πŸ–πš 𝐚 𝐚 πŸπŸ“.πŸ–βˆ’ 𝑭 π‘»πŸ =𝟏.πŸ”πŸπš 𝑭 π‘»πŸ βˆ’πŸπŸ‘.πŸ“=𝟏.πŸ‘πŸ–πš 𝑭 π‘»πŸ =πŸπŸ“.πŸ–βˆ’πŸ.πŸ”πŸπš 𝑭 π‘»πŸ =πŸπŸ‘.πŸ“+𝟏.πŸ‘πŸ–πš

27 Atwood Machine πšΊπ‰=π‘°πœΆ 𝑭 π‘»πŸ π’“βˆ’ 𝑭 π‘»πŸ 𝒓=π‘°πœΆ
End Slide Atwood Machine 𝒓 𝑰 π‘«π’Šπ’”π’Œ = 𝟏 𝟐 π’Ž 𝒓 𝟐 𝑭 π‘»πŸ 𝑭 π‘»πŸ 𝑰 π‘«π’Šπ’”π’Œ = 𝟏 𝟐 βˆ—πŸ.πŸβˆ— 𝟎.πŸπŸ’πŸ— 𝟐 πšΊπ‰=π‘°πœΆ 𝑰 π‘«π’Šπ’”π’Œ =𝟎.πŸŽπŸ”πŸ“πŸ π’Œπ’ˆβˆ™ π’Ž 𝟐 𝑭 π‘»πŸ π’“βˆ’ 𝑭 π‘»πŸ 𝒓=π‘°πœΆ ⟹ πŸπŸ“.πŸ–βˆ’πŸ.πŸ”πŸπšβˆ’πŸπŸ‘.πŸ“βˆ’πŸ.πŸ‘πŸ–πš 𝟎.πŸπŸ’πŸ—=𝟎.πŸŽπŸ”πŸ“πŸβˆ— 𝐚 𝒓 𝑭 π‘»πŸ βˆ’ 𝑭 π‘»πŸ 𝒓=π‘°πœΆ 𝟐.πŸ‘βˆ’πŸ.πŸ—πŸ—πš 𝟎.πŸπŸ’πŸ—=𝟎.πŸŽπŸ”πŸ“πŸβˆ— 𝐚 𝟎.πŸπŸ’πŸ— ⟹ 𝟎.πŸ“πŸ•πŸ‘βˆ’πŸŽ.πŸ•πŸ’πŸ“πš=𝟎.πŸπŸ”πŸπš 𝟎.πŸ“πŸ•πŸ‘=𝟏.πŸŽπŸŽπŸ”πš ⟹ 𝐚=𝟎.πŸ“πŸ•πŸŽ π’Ž 𝒔 𝟐 𝑭 π‘»πŸ =πŸπŸ“.πŸ–βˆ’πŸ.πŸ”πŸπš 𝑭 π‘»πŸ =πŸπŸ‘.πŸ“+𝟏.πŸ‘πŸ–πš

28 Rolling Down the Ramp 𝑳𝒆 𝒕 β€² 𝒔 π’‡π’Šπ’π’… π’•π’‰π’Šπ’” π’†π’™π’‘π’†π’“π’Šπ’Žπ’†π’π’•π’‚π’π’π’š π’‡π’Šπ’“π’”π’•.
End Slide Rolling Down the Ramp Two masses roll down an incline. One is a β€œhoop” 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 π‘Ίπ’π’π’Šπ’… π‘«π’Šπ’”π’Œ β‡’ 𝑭 𝒇 𝒓=π’Œπ’Ž 𝒓 𝟐 𝐚 𝒓 πšΊπ‰=𝑰𝛂 πšΊπ‘­=π’Žπš β‡’ 𝑭 π’ˆ\\ βˆ’ 𝑭 𝒇 =π’Žπš 𝑳𝒆 𝒕 β€² 𝒔 π’‡π’Šπ’π’… π’•π’‰π’Šπ’” π’†π’™π’‘π’†π’“π’Šπ’Žπ’†π’π’•π’‚π’π’π’š π’‡π’Šπ’“π’”π’•. 𝑭 𝒇 =π’Žπ’ˆ 𝐬𝐒𝐧 𝜽 βˆ’π’Žπš 𝑭 𝒇 =π’Œπ’Žπš 𝑭 𝒇 =π’Ž π’ˆ 𝐬𝐒𝐧 𝜽 βˆ’πš π’Œπ’Žπš=π’Ž π’ˆ 𝐬𝐒𝐧 𝜽 βˆ’πš π’Œπš=π’ˆ 𝐬𝐒𝐧 𝜽 βˆ’πš 𝐚= π’ˆ 𝐬𝐒𝐧 𝜽 π’Œ+𝟏

29 End Slide Rolling Down the Ramp Two masses roll down an incline. One is a β€œhoop” 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 π‘Ίπ’π’π’Šπ’… π‘«π’Šπ’”π’Œ β‡’ 𝒕= πŸβˆ†π’™ 𝐚 𝐚= π’ˆ 𝐬𝐒𝐧 𝜽 π’Œ+𝟏 βˆ†π’™= 𝟏 𝟐 𝐚 𝒕 𝟐 π‘Ίπ’π’π’Šπ’… π‘«π’Šπ’”π’Œ 𝑯𝒐𝒐𝒑 π‘Ίπ’π’π’Šπ’… π‘«π’Šπ’”π’Œ 𝑯𝒐𝒐𝒑 𝒕 𝑫 = πŸβˆ—πŸ.πŸ’ 𝟎.πŸ—πŸŽπŸ— 𝒕 𝑯 = πŸβˆ—πŸ.πŸ’ 𝟎.πŸ”πŸ–πŸ 𝐚 𝑫 = πŸ—.πŸ– 𝐬𝐒𝐧 πŸ–.𝟎 𝟎.πŸ“+𝟏 𝐚 𝑯 = πŸ—.πŸ– 𝐬𝐒𝐧 πŸ–.𝟎 𝟏+𝟏 𝐚 𝑫 =𝟎.πŸ—πŸŽπŸ— π’Ž 𝒔 𝟐 𝐚 𝑯 =𝟎.πŸ”πŸ–πŸ π’Ž 𝒔 𝟐 𝒕 𝑫 =𝟏.πŸ•πŸ“πŸ“ 𝒔𝒆𝒄 𝒕 𝑯 =𝟐.πŸŽπŸπŸ” 𝒔𝒆𝒄 βˆ†π’•=𝟎.πŸπŸ•πŸ 𝒔𝒆𝒄

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