ap rotational dynamics lessons 91, 94

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AP Rotational Dynamics
Lessons 91 and 94

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

If
◦ a =Δv/t

And
◦ ΣF = ma
◦ We can think of mass as a measurement of an
object’s linear 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 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

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?
F
3m

110Nm

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?

1.4Nm

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

1.5rad/s2


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

Long, thin rod spinning about its center
◦ I = 1/12 mL2

Long, thin rod spinning about one end
◦ I = 1/3mL2

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

5rad/s2

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

6.8Nm

Kinetic Energy
◦ KE = 1/2mv2

KE Rotational Energy (Energy when an object
is spinning)
◦ KE = ½ Iω2
This has linear and rotational KE
KE = 1/2mv2 +1/2 Iω2
We get the KElinear from PEinitial!

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?

1.5J

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?

4.6m/s

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

Momentum = mvr
 v=rω
◦ Momentum = mr2 ω
 mr2=I
◦ Momentum = I ω
◦ L=Iω

Angular momentum
◦ L=Iω (kgm2/s)
◦ τΔt = Δ(I ω)
 angular impulse creates a change in angular
momentum

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.

0.00070kgm2/s

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

3.0Nm

Linear momentum in a closed system is
conserved
◦ Absence of friction

Same goes for angular momentum
◦ Li = Lf

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.

0.30rad/s

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?
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