Bellringer: What would be the net acceleration of a 15 g toy car down a 30 degree incline if the...

15
Bellringer: What would be the net acceleration of a 15 g toy car down a 30 degree incline if the acceleration due to friction is 1.8 m/s 2 ? Include a free-body diagram of all forces.

description

Angular Motion  What is angular motion? How is it different from linear motion?

Transcript of Bellringer: What would be the net acceleration of a 15 g toy car down a 30 degree incline if the...

Page 1: Bellringer: What would be the net acceleration of a 15 g toy car down a 30 degree incline if the acceleration due to friction is 1.8 m/s 2 ? Include a.

Bellringer:

What would be the net acceleration of a 15 g toy car down a 30 degree incline if

the acceleration due to friction is 1.8 m/s2? Include a free-body diagram of all

forces.

Page 2: Bellringer: What would be the net acceleration of a 15 g toy car down a 30 degree incline if the acceleration due to friction is 1.8 m/s 2 ? Include a.

Chapter 8 – Angular motion, torque, and moment of inertia

Page 3: Bellringer: What would be the net acceleration of a 15 g toy car down a 30 degree incline if the acceleration due to friction is 1.8 m/s 2 ? Include a.

Angular Motion What is angular motion?

How is it different from linear motion?

Page 4: Bellringer: What would be the net acceleration of a 15 g toy car down a 30 degree incline if the acceleration due to friction is 1.8 m/s 2 ? Include a.

Vocabulary Revolution Radian Angular displacement Angular velocity Angular acceleration

Page 5: Bellringer: What would be the net acceleration of a 15 g toy car down a 30 degree incline if the acceleration due to friction is 1.8 m/s 2 ? Include a.

Equations 1 revolution = 2π radians Angular displacement (Θ, in rad)

Θ = d/r (linear distance in m / radius) Angular velocity (ω, in rad/s)

ω = Θ/t (also equal to v/r) Angular acceleration (α, in rad/s2)

α = ω/t (also equal to a/r)

Page 6: Bellringer: What would be the net acceleration of a 15 g toy car down a 30 degree incline if the acceleration due to friction is 1.8 m/s 2 ? Include a.

Practice p.200 (1-3) Keep answers to number 1 in terms of π

Page 7: Bellringer: What would be the net acceleration of a 15 g toy car down a 30 degree incline if the acceleration due to friction is 1.8 m/s 2 ? Include a.

Bellringer What is the equation to convert from linear distance to

angular displacement? What is the equation to convert from linear velocity to

angular velocity? What is the equation to convert from linear

acceleration to angular acceleration? WHAT IS SIMILAR IN ALL OF THESE

EQUATIONS? Write out the equations to find angular velocity and to

find angular acceleration.WHAT DO YOU NOTICE ABOUT THESE

EQUATIONS?

Page 8: Bellringer: What would be the net acceleration of a 15 g toy car down a 30 degree incline if the acceleration due to friction is 1.8 m/s 2 ? Include a.

Torque What is torque?

DefinitionEquationLever arm (perpendicular distance from the axis

of rotation to the point where the force is exerted)○ L = r sin Θ when the angle is not perpendicular

Page 9: Bellringer: What would be the net acceleration of a 15 g toy car down a 30 degree incline if the acceleration due to friction is 1.8 m/s 2 ? Include a.

Practice p.203 (11-15)11. τ = 35 Nm, L = 25 cm, F = ?

12. 0.407 m13. 36.6o

14. 94 Nm15.

Page 10: Bellringer: What would be the net acceleration of a 15 g toy car down a 30 degree incline if the acceleration due to friction is 1.8 m/s 2 ? Include a.

Net Torque To find net torque,

use the equation:τ = Fg r

Use net torque when you have a situation with a center fulcrum and 2 lever arms

Page 11: Bellringer: What would be the net acceleration of a 15 g toy car down a 30 degree incline if the acceleration due to friction is 1.8 m/s 2 ? Include a.

Practice p.205 (16-20)16.mA = 43 kg, rA = 1.8 m, mS = 52 kg, rS

= ?

17. 2.7 Nm18. 0.056 kg19. 0.042 kg20. 789 N

Page 12: Bellringer: What would be the net acceleration of a 15 g toy car down a 30 degree incline if the acceleration due to friction is 1.8 m/s 2 ? Include a.

Moment of Inertia Definition Equation Compare and contrast the different

moments of inertia for various objects in Table 8-2. What object would have the highest?

Lowest?What does this mean about this object?

Page 13: Bellringer: What would be the net acceleration of a 15 g toy car down a 30 degree incline if the acceleration due to friction is 1.8 m/s 2 ? Include a.

Practice p.208 (21 and 24)Units for moment of inertia are kgm2.21. What is the change in I when r is

increased from 0.3 m to 0.6 m?

24.

Page 14: Bellringer: What would be the net acceleration of a 15 g toy car down a 30 degree incline if the acceleration due to friction is 1.8 m/s 2 ? Include a.

Newton’s Second Law for Rotational Motion What is Newton’s Second Law?

How does it change with rotational motion?

Page 15: Bellringer: What would be the net acceleration of a 15 g toy car down a 30 degree incline if the acceleration due to friction is 1.8 m/s 2 ? Include a.

Practice p.210 (27)