AP Rotational Dynamics Lessons 91 and 94 Rotational

AP Rotational Dynamics Lessons 91 and 94

Rotational Inertia � Matter tends to resist changes in motion ◦ Resistance to a change in velocity is inertia ◦ Resistance to a change in angular velocity is rotational inertia, I

Rotational Inertia � If ◦ a =Δv/t � And ◦ ΣF = ma ◦ We can think of mass as a measurement of an object’s linear inertia

Rotational Inertia � If we want to give an object rotational acceleration, we have to apply torque. ◦ The distribution of an objects mass relative to a point of rotation resists change!

Rotational Inertia � Rotational ◦ I = mr 2 Inertia of a point mass �A point mass is based off of an imaginary object having mass and no volume ◦ Therefore �τ = Iα , similar to F=ma �τ = Σ(mr 2)α

Example I �A 2. 0 kg point mass is attached to the end of a mass less rod 3. 0 m long. The rod is pinned so that it is free to rotate about the pinned end. What torque is required to give the mass an angular acceleration of 6. 0 rad/s 2? F 3 m

Answer � 110 Nm

Example II � Three point masses are affixed to the surface of a massless table that can turn without friction about its center. The masses(m 1, m 2, m 3) are 1. 0, 2. 0, and 3. 0 kg respectively. The lengths(r 1, r 2, r 3) are 0. 10 m, 0. 20 m, and 0. 30 m. What torque is required to give the system an angular acceleration of 4. 0 rad/s 2?

Answer � 1. 4 Nm

Example III �A 4. 0 kg mass and a 7. 0 kg mass are mounted on the ends of a light rod, as shown. The rod is pinned so it is free to rotate about its center point. A torque of 150 Nm is applied, what will be its angular acceleration? 3. 0 m

Answer � 1. 5 rad/s 2

Moment of Inertia � Mass of an object is its linear inertia � Moment of Inertia is calculated differently for different shapes (we use calculus to determine most of them) ◦ Thin Hoop �I = mr 2 ◦ Cylinder �I = ½ mr 2 ◦ Solid Sphere �I = 2/5 mr 2

Moment of Inertia � Long, thin rod spinning about its center � Long, thin rod spinning about one end ◦ I = 1/12 m. L 2 ◦ I = 1/3 m. L 2

Example IV �A torque of 200 Nm is applied to a solid cylinder as shown. The torque is applied such that the cylinder rotates about its center. The mass of the cylinder is 20 kg and its radius is 2 m. Find the angular acceleration produced by the torque. 2 m

Answer � 5 rad/s 2

Example V �A long thin homogeneous rod has an angular acceleration of 2. 0 rad/s 2 about its center point, as shown. The length of the rod is 1. 5 m and its mass is 18 kg. Find the torque which produces this angular acceleration. L

Answer � 6. 8 Nm

Lesson 94 KE and Momentum � Kinetic Energy ◦ KE = 1/2 mv 2 � KE Rotational Energy (Energy when an object is spinning) ◦ KE = ½ Iω2

What if you are rolling downhill? This has linear and rotational KE KE = 1/2 mv 2 +1/2 Iω2 We get the KElinear from PEinitial!

Example VI �A cylinder whose mass is 6. 0 kg has a radius of 10 cm. It is rotating at 10 rad/s. What is the rotational KE of the cylinder?

Answer � 1. 5 J

Example VII �A homogeneous 6. 3 kg cylinder whose radius is 11 cm rolls, without slipping, down an inclined plane a vertical distance of 1. 6 m. What is its speed at the bottom if it starts from rest?

Answer � 4. 6 m/s

Angular Momentum � Linear momentum is the product of the mass of an object and its linear speed ◦ p = mv ◦ Forces cause a changed in the momentum ◦ FΔt = Δp (impulse creates a change in momentum) �F = Δp / Δt

Angular Momentum �v=rω = mvr ◦ Momentum = mr 2 ω �mr 2=I ◦ Momentum = I ω ◦ L=Iω

Angular Momentum � Angular momentum ◦ L=Iω (kgm 2/s) ◦ τΔt = Δ(I ω) �angular impulse creates a change in angular momentum

Example VIII � Find the angular momentum of a sphere with a radius of 0. 051 m, a mass of 0. 16 kg, and an angular speed about its center of 4. 2 rad/s.

Answer � 0. 00070 kgm 2/s

Example IX �A point mass of 0. 50 kg is mounted on the end of a very light rod 2. 0 m long, as shown. The point mass moves with an angular speed of 5. 0 rad/s. What torque should be applied for 10 s to increase the angular speed to 20 rad/s? 0. 50 kg

Answer � 3. 0 Nm

Conservation of Angular Momentum � Linear momentum in a closed system is conserved ◦ Absence of friction � Same goes for angular momentum ◦ Li = L f

Example X �A 30 kg boy runs at 3. 0 m/s, tangent to a merry go round which is not moving. The merry go round has a moment of inertia of 480 kgm 2, and a radius of 2. 0 m. Find the angular speed of the system after the boy jumps on the merry go round.

Answer � 0. 30 rad/s

Exit Ticket �A cliff diver reduces her moment of inertia by a factor of 3. 5 when she moves between the straight position and the “tuck position”. If she makes two complete revolutions in 1. 5 s when in the tuck position, what is her angular speed (in rev/s) when her body is in the straight position?