Rotational Equilibrium Dr Walker Torque Torque is produced

Rotational Equilibrium Dr. Walker

Torque • Torque is produced by this turning force and tends to produce rotational acceleration. – Product of the Rotational Force and the Length of a Lever Arm – The effect a force will have on rotational motion

Torque • Torque is required to turn objects – Loosening a lug nut on a tire – Opening that stubborn jar – Opening a door – Torque = Force x Lever arm (length)

https: //www. grc. nasa. gov/www/k-12/airplane/Images/torque. gif

Torque • Direction – Torque is a vector quantity – (+) is torque applied clockwise – (-) is torque applied counterclockwise • Units – Force = N – Torque = N. m

Application of Torque • Torque is most effective when applied perpendicular to the radius (left) • Torque can be applied when not perpendicular, but the difference in angle must be accounted for (right)

Torque…At An Angle • Previously, it was stated that torque = force x length of lever arm – t = Fr • When torque is applied at an angle, this must be accounted for to become – t = Fr(sinq) – When the angle is 90 degrees, sin(90 o) = 1, so it reduces out https: //qph. ec. quoracdn. net/main-qimg-dff 33 b 4 fd 66 aa 8039 db 734 d 56 b 859043

Example 1 • What is the torque on a bolt produced by a 15 N force exerted perpendicular to a wrench that is 25 cm long?

Example 1 • What is the torque on a bolt produced by a 15 N force exerted perpendicular to a wrench that is 25 cm long? • t = Fr(sinq) • t = (15 N)(0. 25 m)(sin 90) • t = 3. 75 N. m

Example 2 • A person slowly lowers a 3. 3 kg crab trap over the side of a dock. What torque does the trap exert about the person’s shoulder (length of arm = 0. 70 m)?

Example 2 • A person slowly lowers a 3. 3 kg crab trap over the side of a dock. What torque does the trap exert about the person’s shoulder (length of arm = 0. 70 m)? • t = Fr(sinq) • t = (3. 3 kg)(9. 8 m/s 2)(0. 7 m)(sin 90) • t = 23 N. m

Rotational Equilibrium • Mechanical Equilibrium – SF = 0 • Rotational Equilibrium – St = 0 – When a system has multiple “torques”, the object is in rotational equilibrium (no net rotation) when the sum of the torques is zero • Remember, different directions have different signs! • Torque is a vector!!

Example 1 • My 27 kg daughter is sitting on a teeter-totter 2. 0 m from the pivot point. How much force do I need to place on the opposite side at the same distance to balance the teeter-totter? How much do I need to make her go up?

Example 1 • My 27 kg daughter is sitting on a teeter-totter 2. 0 m from the pivot point. How much force do I need to place on the opposite side at the same distance to balance the teeter-totter? How much do I need to make her go up? • Frchild = Fradult • (27 kg)(9. 8 m/s 2)(2. 0 m) = (F)(2. 0 m) • F = 264. 6 N – This is the force required to balance the teeter-totter – If I apply more force (>264. 6 N), the board will go down further, and my daughter goes up

Example 2 • At the local playground, a 16 kg child sits on the end of a horizontal teeter-totter, 1. 5 m from the pivot point. On the other side of the pivot an adult pushes straight down on the teeter-totter with a force of 95 N. In which direction does the teeter-totter rotate if the adult applies the force at a distance of 3. 0 m from the pivot?

Example 2 • At the local playground, a 16 kg child sits on the end of a horizontal teetertotter, 1. 5 m from the pivot point. On the other side of the pivot an adult pushes straight down on the teeter-totter with a force of 95 N. In which direction does the teeter-totter rotate if the adult applies the force at a distance of 3. 0 m from the pivot? • Remember, St = 0 • For the teeter totter to stand still tchild= tadult • Frchild = Fradult • (16 kg)(9. 8 m/s 2)(1. 5 m) = (95 N)(radult) • radult = 2. 47 – The adult only needs to be 2. 47 m away to balance the force – Since the adult is further away, he has MORE torque than the child…. so the teeter totter will be pushed down and the child will go UP!

Example 3 • How does this work? Why doesn’t it fall?

Example 3 • Notes – The bar is 45 lbs (roughly 20 kg). The weight on the 18 in (45. 7 cm) end is 20 kg. – We’ll assume the bars mass is evenly distributed across it’s 84 inch (213 cm) length – The lever arm on the left is 45. 6 cm. The mass of the smaller lever arm is (45. 6 cm/213 cm) x 20 kg = 4. 29 kg + 20 kg (the weight) = 24. 29 kg – The mass of the larger lever arm is 15. 71 kg, it’s length is 66 in (167. 6 cm)

Example 3 • Are the torques equal? – Frsmall = (24. 29 kg)(9. 8 m/s 2)(0. 456 m) = 108. 5 N m – Frlong = (15. 71 kg)(9. 8 m/s 2)(1. 676 m) = 258. 1 N m – The end without the weight on it still has significantly more torque, so it’s not going anywhere! – The bar doesn’t tip because of the 2 nd support

Example 3 - Extension • How much weight COULD you put on the short side before it topples?

Example 3 - Extension • How much weight COULD you put on the short side before it topples? • Frlong = (15. 71 kg)(9. 8 m/s 2)(1. 676 m) = 258. 1 N m • Frsmall = (4. 29 kg + x)(9. 8 m/s 2)(0. 456 m) = 258. 1 N m – Solve for x = 53. 5 kg = 117. 9 lbs – Hypothetically, you could put another 70 lbs on that side and it wouldn’t topple…. but I wouldn’t try it…

• There is 100 lbs (~45 kgs) on the left hand side of the bar. The longer lever arm on the right side allows the barbell to be stable, even with the extra 100 lbs on the left (and no, my hand isn’t in contact with the bar!).

Static Equilibrium • No net force AND no net torque – SF = 0 AND St = 0

Center Of Mass A baseball thrown into the air follows a smooth parabolic path. A baseball bat thrown into the air does not follow a smooth path. The bat wobbles about a special point. This point stays on a parabolic path, even though the rest of the bat does not. The motion of the bat is the sum of two motions: • a spin around this point, and • a movement through the air as if all the mass were concentrated at this point. This point, called the center of mass, is where all the mass of an object can be considered to be concentrated

Center Of Mass • The centers of mass of the baseball and of the spinning baseball bat each follow parabolic paths.

Center Of Mass • For a symmetrical object, such as a baseball, the center of mass is at the geometric center of the object. • For an irregularly shaped object, such as a baseball bat, the center of mass is toward the heavier end. • The center of mass for each object is shown by the red dot

Center Of Mass • Objects that are not homogeneous will have a center of mass which is NOT near the center – Notice the bottle is only part full – most of the mass is towards the lower half of the bottle https: //media. musely. com/u/d 7 cc 4 ce 5 -935 d-4 e 6 c-bf 44 -f 30 acdf 2 f 73 d. jpg

Center Of Mass and Stability • If the center of mass is outside the base of an object, it is unstable and will roll over without additional torque (right) • If the center of mass is above the base of the object, it is stable. (left) • If the base of the object is very narrow and the center of mass is high, then the object is stable, but the slightest force will cause it to tip over. http: //www. bbc. co. uk/schools/gcsebitesize/science/triple_aqa/using_physics_make_things_work/moments/revision/ 4/

Center Of Mass - People https: //acewebcontent. azureedge. net/blogs/blog-examprep-032813 -1. jpg

Center of Gravity • Center of gravity – Average position of all particles of weight that make up an object • These terms are interchangeable for objects on or near Earth – There can be a small difference between center of gravity and center of mass when an object is large enough for gravity to vary from one part to another. – The center of gravity of the Sears Tower in Chicago is about 1 mm below its center of mass because the lower stories are pulled a little more strongly by Earth’s gravity than the upper stories

The Vocabulary of Rotation • I = Rotational Inertia – Similar to inertia (or mass) for linear kinematics • w = Angular Velocity – Similar to velocity for linear kinematics • L = Angular Momentum – Similar to momentum for linear kinematics • t = Torque • Similar to force for linear kinematics • All of the old kinematics equations have a rotational version!!

Moment of Inertia • Moment of Inertia (I) – Also called rotational inertia – I = ½ mr 2 for a solid cylinder – Valid for a “point” mass • “point” mass assumes no volume , concentrated mass – Different shapes have different formulas

http: //hyperphysics. phy-astr. gsu. edu/hbase/mi. html

Rotational Inertia • By holding a long pole, the tightrope walker increases his rotational inertia – Remember, inertia is the resistance to change (a force). More inertia means he’s harder to knock off course, or off the tightrope

Example 1 • What is the rotational inertia of a 0. 22 m radius cylinder with a mass of 33 kg?

Example 1 • What is the rotational inertia of a 0. 22 m radius cylinder with a mass of 33 kg? • I = ½ x 33 x. 222 = 0. 80 kgm 2

Example 2 • A solid cylinder with a radius of 4 m has a rotational inertia of 50 kgm 2. What is the mass of this cylinder?

Example 2 • A cylinder with a radius of 0. 4 m has a rotational inertia of 50 kgm 2. What is the mass of this cylinder? • I = mr 2 • 50 kgm 2 = ½ m(4 m)2 • m = 6. 25 kg

Example 3 • A spinning figure skater has a mass of 50 kg and a wingspan (diameter) of 1. 63 m. What is her rotational inertia (assuming we can use the same equation)?

Example 3 • A spinning figure skater has a mass of 50 kg and a wingspan (diameter) of 1. 63 m. What is her rotational inertia? • I = ½ mr 2 • I = ½ (50 kg)(0. 815 m)2 • I = 66. 5 kgm 2

Angular Velocity • Linear Velocity – V = d/t • Angular Velocity – Change in angle (not distance!) – over time – w = q/t – Usually given in radians (rad) per second (rad/s) – Remember, there are 2 p radians (angle, not distance!) in a complete circle Circumference = 2 pr https: //courses. lumenlearning. com/boundless-physics/chapter/angular-acceleration/

Example 1 • You have a toy plane on a string that goes around three complete circles in 9. 0 s. What is its angular velocity?

Example 1 • You have a toy plane on a string that goes around three complete circles in 9. 0 s. What is its angular velocity? • Then w = q/t • If 1 circle is 2 p, then 3 circles is 6 p • Then w = q/t – w = 6 p / 9. 0 s = 2. 09 rad/s

Further Thought… • If the angular velocity is 2. 09 rad/s, what is the linear velocity? ? ?

Further Thought… • If the angular velocity is 2. 09 rad/s, what is the linear velocity? ? ? – Without more information (the radius/diameter/circumference), we can’t know!!! – The change in angle over time (our angular velocity) is the same no matter what the size of the circle • The linear or circular velocity changes with the circumference!

Example 2 • The second hand of a clock takes 30 seconds to move through an arc of 180 degrees. What is the angular velocity?

Example 2 • The second hand of a clock takes 30 seconds to move through an arc of 180 degrees. What is the angular velocity? • 180 degrees is ½ of a circle • Moves ½ (2 p radians) = p radians • w = q/t = p/30 s = 0. 105 rad/s

Example 3 • The moon goes around Earth in about 27. 3 days. What is its angular velocity?

Example 3 • The moon goes around Earth in about 27. 3 days. What is its angular velocity? • First, we have to convert 27. 3 days to seconds – 27. 3 days = 655. 2 hrs = 39312 min = 2. 36 x 106 s • Then w = q/t • 2 p/2. 36 x 106 s = 2. 66 x 10 -6 rad/s

Example 4 • A LP-album has a diameter of 30 cm and a mass of 0. 18 kg. LP albums revolves at a rate of 33 1/3 revolutions per minute (rpm). What is the angular velocity of a LP?

Example 4 • A LP-album has a diameter of 30 cm and a mass of 0. 18 kg. LP albums revolves at a rate of 33 1/3 revolutions per minute (rpm). What is the angular velocity of a LP? • w = q/t = (2 p x 33 1/3)/60 s = 3. 49 radians/s

Angular Momentum • A rotating object will continue to rotate until acted on by force • Angular momentum (L) – Product of rotational inertia (I) and rotational velocity (w) – “Rotational” version of mometum • I similar to mass, w similar to velocity – L = I x w

Angular Momentum • An object of concentrated mass m whirling in a circular path of radius r with a speed v has angular momentum mvr. • Remember mass is related to inertia (I) and velocity (v) is related to angular velocity (w)

Angular Momentum • In a linear system, momentum is conserved (doesn’t change) in the absence of an external force • In a rotational system, angular momentum is conserved in the absence of an external torque – Remember, torque is force x lever arm, so an external force is required for an external torque

Conservation of Angular Momentum • Remember, angular momentum is conserved in the absence of an external force • The inertia on the left in larger because of the larger radius • The inertia on the right is far less – – L = I x w L is the same for both Left skater – large I, small w Right skater – small I, large w http: //ffden-2. phys. uaf. edu/webproj/211_fall_2014/Ariel_Ellison/angmom. jpg

Example 1 • A DVD disc has a radius of 0. 0600 m, and a mass of 0. 0200 kg. When a DVD in a certain machine starts playing, it has an angular velocity of 160. 0 radians/s. What is the angular momentum of this disc? (Note: The rotational inertia for a flat disk is 1/4 mr 2)

Example 1 • A DVD disc has a radius of 0. 0600 m, and a mass of 0. 0200 kg. When a DVD in a certain machine starts playing, it has an angular velocity of 160. 0 radians/s. What is the angular momentum of this disc? (Note: The rotational inertia for a flat disk is 1/4 mr 2) • • L = I x w L = (1/4 mr 2) w L = (1/4 x 0. 0200 kg x (0. 0600 m)2)(160 radians/s) L = 0. 00288 kg∙m 2/s

Example 2 • The mass of the moon is 7. 35 × 1022 kilograms. If the moon revolves around the Earth in 27. 3 days, what is the moon’s angular momentum? The equation for the rotational inertia of a solid sphere is I = 2/5 mr 2 and the radius of the moon is 1. 737 x 106 m.

Example 2 • The mass of the moon is 7. 35 × 1022 kilograms. If the moon revolves around the Earth in 27. 3 days, what is the moon’s angular momentum? The equation for the rotational inertia of a solid sphere is I = 2/5 mr 2 and the radius of the moon is 1. 737 x 106 m. • First, we have to convert 27. 3 days to seconds – 27. 3 days = 655. 2 hrs = 39312 min = 2. 36 x 106 s • Then w = q/t • 2 p/2. 36 x 106 s = 2. 66 x 10 -6 rad/s • Then L = Iw • • • For a sphere, I = 2/5 mr 2 I = 2/5(7. 35 × 1022 kg)(1. 737 x 106 m)2 I = 8. 87 x 1034 kgm 2 • L = Iw = (8. 87 x 1034 kgm 2)(2. 66 x 10 -6 rad/s) = 2. 35 x 1029 kgm 2/s

Example 3 • A LP-album has a diameter of 30 cm and a mass of 0. 18 kg. LP albums revolves at a rate of 33 1/3 revolutions per minute (rpm). What is the angular velocity of a LP? Assume the rotational inertia of a flat disk is ¼ mr 2

Example 3 • A LP-album has a diameter of 30 cm and a mass of 0. 18 kg. LP albums revolves at a rate of 33 1/3 revolutions per minute (rpm). What is the angular velocity of a LP? Assume the rotational inertia of a flat disk is ¼ mr 2 • • L = I w w = q/t = (2 p x 33 1/3)/60 s = 3. 49 radians/s I = ¼ (0. 18 kg)(0. 15 m)2 = 1. 01 x 10 -3 kgm 2 L = (1. 01 x 10 -3 kgm 2)(3. 49 radians/s) = 3. 53 x 10 -3 kgm 2/s

Angular Acceleration • Linear acceleration based on Newton’s 2 nd law – F = ma – a = F/m • Angular (rotational) acceleration – a = t/I or t = I a just like Newton’s 2 nd law! – t is torque, analogous to force – I is rotation inertia, analogous to mass https: //courses. lumenlearning. com/boundless-physics/chapter/dynamics/

Example 1 • What is the angular acceleration of a 8 kg cylinder with a radius of 0. 18 m if 74 N. m of torque is applied?

Example 1 • What is the angular acceleration of a 8 kg cylinder with a radius of 0. 18 m if 74 N. m of torque is applied? • a = t/I • I = ½ mr 2 = ½ (8 kg)(0. 18 m)2 = 0. 1296 kgm 2 • a = 74 N. m / 0. 1296 kgm 2 = 571 rad/s 2

Angular Acceleration • Linear Acceleration – a = v/t or a = Dv/Dt • Angular Acceleration – a = w/t or a = Dw/Dt • Remember, any linear equation can have a rotational version

Example • A disc in a DVD player starts from rest, and then begins spinning when the user presses "Play". After 4. 00 s, the disc is spinning at 160 radians/s. What was the average angular acceleration of the disc?

Example • A disc in a DVD player starts from rest, and then begins spinning when the user presses "Play". After 4. 00 s, the disc is spinning at 160 radians/s. What was the average angular acceleration of the disc? • a = Dw/Dt • a = (160 radians/s – 0 radian/s) = 4 s • a = 40 rad/s 2

Rotational Kinetic Energy • For straight line motion – KE = ½ mv 2 • For rotation – KE = ½ I w 2 – Remember, rotational inertia and angular momentum are rotational versions of mass and velocity

Example 1 • A round mill stone with a moment of inertia of I = 1500 kg∙m 2 is rotating at an angular velocity of 8. 00 radians/s. What is the stone's rotational kinetic energy?

Example 1 • A round mill stone with a moment of inertia of I = 1500 kg∙m 2 is rotating at an angular velocity of 8. 00 radians/s. What is the stone's rotational kinetic energy? • KE = ½ I w 2 • KE = ½ (1500 kg∙m 2)(8. 00 radians/s)2 = 48000 J

Example 2 • When a 0. 54 kg can of soup with a diameter of 11 cm is rolled down an incline, it has a angular velocity of 4. 7 rad/s. What is the rotational kinetic energy of the can? (Assume rotational inertia for a cylinder is I = ½ mr 2)

Example 2 • When a 0. 54 kg can of soup with a diameter of 11 cm is rolled down an incline, it has a angular velocity of 4. 7 rad/s. What is the rotational kinetic energy of the can? • KE = ½ I w 2 • KE = ½( ½ x 0. 54 kg x (0. 055 m)2) x (4. 7 rad/s)2 • KE = 0. 0090 J

Example 3 • The mass of the moon is 7. 35 × 1022 kilograms. If the moon revolves around the Earth in 27. 3 days, what is the moon’s rotational kinetic energy? The equation for the rotational inertia of a solid sphere is I = 2/5 mr 2 and the radius of the moon is 1. 737 x 106 m.

Example 3 • The mass of the moon is 7. 35 × 1022 kilograms. If the moon revolves around the Earth in 27. 3 days, what is the moon’s angular momentum? The equation for the rotational inertia of a solid sphere is I = 2/5 mr 2 and the radius of the moon is 1. 737 x 106 m. • First, we have to convert 27. 3 days to seconds – 27. 3 days = 655. 2 hrs = 39312 min = 2. 36 x 106 s • Then w = q/t • 2 p/2. 36 x 106 s = 2. 66 x 10 -6 rad/s • Then L = Iw • • • For a sphere, I = 2/5 mr 2 I = 2/5(7. 35 × 1022 kg)(1. 737 x 106 m)2 I = 8. 87 x 1034 kgm 2 • KE = ½ Iw 2 = ½ (8. 87 x 1034 kgm 2)(2. 66 x 10 -6 rad/s)2 = 3. 13 x 1023 J

Equation Review Quantity Equation Linear Equivalent Torque t = Fr Rotational Inertia I = 1/2 mr 2 (for cylinder) Angular Velocity w = q/t V = d/t Angular Momentum L = I x w p = mv Angular Acceleration a = t/I or a = w/t Rotational Kinetic Energy KE = ½ I w 2 a = F/m KE = ½ mv 2