Rotational Dynamics Slide 2 / 34 l Moment of Inertia To - - PowerPoint PPT Presentation
Slide 1 / 34 Rotational Dynamics Slide 2 / 34 l Moment of Inertia To determine the moment of inertia we divide the object into tiny masses of m i a distance r i from the center. is the sum of all the tiny masses which makes up the object.
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Moment of Inertia
To determine the moment of inertia we divide the object into tiny masses of mi a distance ri from the center. is the sum of all the tiny masses which makes up the object. This quantity is known as the moment of inertia and is denoted by I.
l
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Moment of Inertia Calculations: Rotating a Rod about its Center
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Parallel Axis Theorem
Figure A Figure B
Figure A shows the moment of inertia of a rod rotated about its center and Figure B shows its rotation about one of its
the integration to find it at the end, the parallel axis theorem can be used. Icm is the moment of inertia about the rods center and d is the distance away from Icm's rotational axis.
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1 A rod of length L when rotated about its center has a moment of inertia . What is the moment of inertia about one of its ends?
A
B
C D
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2 A uniform stick has length L. The moment of inertia about the center of the stick is Io. A particle
moment of inertia of the combined system about the center of the stick is
A Io + 1/16 ML2
B
Io + 1/9 ML2
C
Io + 1/4 ML2
D
Io + 1/2 ML2
E
Io + ML2
L L/4
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Rigid-Body Rotation about a Moving Axis
When an object moves it can have both rotational kinetic energy and translational kinetic energy. Translational Velocity Rotational Velocity For an object rolling without slipping the velocity for the center
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3
A sphere rolls down the incline at height h without slipping from rest. What is the velocity of the sphere at the bottom? The moment of intertia of a sphere is . A
B
C D
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4 A hollow sphere initially at rest rolls down the incline at height h without slipping. What is the velocity of the hollow sphere at the bottom? The moment of intertia of a sphere is .
A
B C D
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5 A solid cylinder of mass m and radius r rolls across the floor with a velocity v. Which of the following would be the best estimate of the ring's total kinetic energy as it rolls across the floor?
A mv2
B
1/4 mv2
C
1/2 mv2
D
3/4 mv2
E
1/2 mv2 + (mv2)/r
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Work and Power in Rotational Motion
We learned before that: Therefore:
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6 A conical pendulum makes 2 complete revolutions
acceleration of α. What is the work done on the bob whose moment of inertia is mr2
?
A
B
C D
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Angular Momentum
The analog of linear momentum is angular momentum. Angular momentum is represented by an L, and is the product of the position vector and the momentum vector.
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Angular Momentum of a rigid body rotating about a symmetrical axis
We said before that angular momentum is defined as L=mvr. If we insert v as ωr:
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Conservation of Angular Momentum
When the net external torque acting on a system is zero, the total angular momentum of the system is constant (conserved).
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a
b
7
In the diagram below what is the objects angular momentum around the origin? A B
C D
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8 An person is spinning about a vertical axis with their arms fully extended. If the arms are pulled in closer to the body, in which of the following ways are the kinetic energy and angular momentum changed?
A
B C D E
Kinetic Energy Angular Momentum increase increase remains constant decrease remains constant increase increase remains constant remains constant remains constant
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9
An person is spinning about a vertical axis with their arms fully extended. If the arms are pulled in closer to the body, in which of the following ways are the angular velocity and moment of inertia changed? A B
C D Angular Velocity Moment of Inertia increase increase increase decrease decrease increase decrease decrease
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10 A tire of mass M and radius R rolls on a flat track without slipping. If the angular velocity of the wheel is , what is its linear momentum?
A M R
B
M 2R
C
M R2
D
M 2R2/2
E
Zero
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11 The system above rotates with a speed . If the mass of the supports is negligible, what is the ratio
spheres to the two lower spheres?
A 16/1
B
4/1
C
1/16
D
1/4
E
1/1
m/2 2m 2m m/2
L/2 L/2 L L
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Torque
r F
Torque is the rotational equivalent to force. r is the length of the lever arm and it is the shortest distance from the center of rotation to the point where the force is being applied. For any type of torque we will consider the clockwise direction to be a positive torque, whereas the counter- clockwise direction is considered to be a negative torque.
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12
What is the translational analogue of torque?
A
Kinetic Energy
B
Acceleration
C
Force D Mass
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13 A B C D
An object of mass m is placed a distance r away from the center of a balance. If another object of mass 4m is to be placed on the balance, what distance does it have to be placed away from the center for the system to be at equilibrium?
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Calculate torque (magnitude and direction) about the pivot point of a rod of length 4m due to a force F = 10N.
90o F 30o F 45o F F
zero zero 40 N*m, out of page 20 N*m, out of page
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Torque and Angular Acceleration
z
y
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Net Torque
The net torque acting on a rotating body is equal to the derivative of the object's angular momentum. Since L can be described in terms of moment of inertia and angular velocity, the net torque can be shown as:
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A yo-yo is released with the condition that the string does not slip on the cylinder. The moment of inertia for the yo-yo is . a) Draw the free-body diagram for the yo-yo. b) Find the downward acceleration of the yo-yo. c) Find the tension in the string.
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Στ = Iα Στ = FTR = Iα = (1/2MR2)α α = a/R FTR = (1/2MR2)(a/R) FTR = 1/2MRa FT = 1/2Ma
ΣF = Ma ΣFy = FT - Mg = -Ma ΣFy = FT = Mg - Ma FT = Mg - Ma and FT = 1/2Ma
1/2Ma = Mg - Ma 3/2Ma = Mg 3/2a = g a = 2/3g a)
FT Mg
a
b) c) FT = 1/2Ma a = 2/3g FT = (1/2M)(2/3g) FT = 1/3Mg
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A solid sphere rolls down a ramp without slipping. The ramp has an angle of θ. The solid sphere has the moment of inertia of . a) Draw the free-body diagram for the solid sphere. b) Find the sphere's acceleration. c) Evaluate for the force of friction.
θ
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a)
f
FN Mg
θ
b) Στ = Iα Στ = fR = Iα = (2/5MR2)α α = a/R fR = (2/5MR2)(a/R) fR = 2/5MRa f = 2/5Ma
ΣF = Ma ΣFy = Fn - Mgcosθ = Ma ΣFx = Mgsinθ - f = Ma f = Mgsinθ - Ma f = Mgsinθ - Ma and f = 2/5Ma Mgsinθ - Ma = 2/5Ma gsinθ = 7/5a a = 5/7gsinθ
c)
f = 2/5Ma a= 5/7gsinθ f = (2/5M) (5/7gsinθ) f = 2/7Mgsinθ
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14 Forces F1 = 10.0 N and F2 = 4.5 N are applied tangentially to a wheel with a radius of 1.5 m, as shown below. What is the net torque on the wheel due to the forces?
A 6.75 N*m
B
8.25 N*m
C
15 N*m
D
21.75 N*m
F1 F2
1.5m
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15
An object with moment of inertia I is originally at rest and begins to undergo a constant angular
angular velocity of ω. What is the net torque applied to the object?
A B
C D
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16 If a ladder of mass m and length L is propped
up against a wall at an angle of 45o, what is the minimum coefficient of static friction in order for the ladder not to slip?
A
B C
D L
45o
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