AP Rotational DynamicsLessons 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 Inertia of a point mass

◦ I = mr2

A point mass is based off of an imaginary object having mass and no volume

◦ Therefore τ = Iα , similar to F=ma τ = Σ(mr2)α

Example I A 2.0kg point mass is attached to the end of

a mass less rod 3.0m 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 6.0rad/s2?

3m

F

Answer 110Nm

Example II Three point masses are affixed to the surface of a

massless table that can turn without friction about its center. The masses(m1, m2, m3) are 1.0, 2.0, and 3.0kg respectively. The lengths(r1, r2, r3) are 0.10m, 0.20m, and 0.30m. What torque is required to give the system an angular acceleration of 4.0rad/s2?

Answer 1.4Nm

Example III A 4.0kg mass and a 7.0kg 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 150Nm is applied, what will be its angular acceleration?

3.0m 3.0m

Answer 1.5rad/s2

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 = mr2

◦ Cylinder I = ½ mr2

◦ Solid Sphere I = 2/5 mr2

Moment of Inertia Long, thin rod spinning about its center

◦ I = 1/12 mL2

Long, thin rod spinning about one end◦ I = 1/3mL2

Example IV A torque of 200Nm 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 20kg and its radius is 2m. Find the angular acceleration produced by the torque.

2m

Answer 5rad/s2

Example V A long thin homogeneous rod has an

angular acceleration of 2.0rad/s2 about its center point, as shown. The length of the rod is 1.5m and its mass is 18kg. Find the torque which produces this angular acceleration.

L

Answer 6.8Nm

Lesson 94 KE and Momentum

Kinetic Energy◦ KE = 1/2mv2

KE Rotational Energy (Energy when an object is spinning)◦ KE = ½ Iω2

What if you are rolling downhill?

This has linear and rotational KEKE = 1/2mv2

+1/2 Iω2

We get the KElinear from PEinitial!

Example VI A cylinder whose mass is 6.0kg has a radius

of 10cm. It is rotating at 10rad/s. What is the rotational KE of the cylinder?

Answer 1.5J

Example VII A homogeneous 6.3kg cylinder whose

radius is 11cm rolls, without slipping, down an inclined plane a vertical distance of 1.6m. What is its speed at the bottom if it starts from rest?

Answer 4.6m/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 Momentum = mvr

v=rω◦ Momentum = mr2 ω

mr2=I◦ Momentum = I ω◦ L = I ω

Angular Momentum Angular momentum

◦ L=Iω (kgm2/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 0.051m, a mass of 0.16kg, and an angular speed about its center of 4.2rad/s.

Answer 0.00070kgm2/s

Example IX A point mass of 0.50kg is mounted on the

end of a very light rod 2.0m long, as shown. The point mass moves with an angular speed of 5.0rad/s. What torque should be applied for 10s to increase the angular speed to 20rad/s?

0.50kg

Answer 3.0Nm

Conservation of Angular Momentum

Linear momentum in a closed system is conserved◦ Absence of friction

Same goes for angular momentum◦ Li = Lf

Example X A 30kg boy runs at 3.0m/s, tangent to a

merry go round which is not moving. The merry go round has a moment of inertia of 480kgm2, and a radius of 2.0m. Find the angular speed of the system after the boy jumps on the merry go round.

Answer 0.30rad/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 1.5s when in the tuck position, what is her angular speed (in rev/s) when her body is in the straight position?