Moment of Inertia and Rotational Dynamics: Calculations and Applications, Assignments of Physics

The concept of moment of inertia, or rotational inertia, which is the rotational analog of mass. formulas for calculating the moment of inertia for various objects, including a single revolving point mass, a set of point masses, and common shapes like disks and spheres. It also discusses the relationship between torque, moment of inertia, and angular acceleration, and provides examples of calculating the moment of inertia and the required torque to make an object rotate. suitable for students studying classical mechanics, physics, or engineering.

Typology: Assignments

2019/2020

Uploaded on 11/12/2020

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Physics
Section 8-5 to 8-
6
Rotational Dynamics
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Physics

• Section 8-5 to 8-

Rotational Dynamics

Moment of Inertia We saw previously that mass can be defined by an object’s ability to resist motion. This is called inertial mass. The rotational analog of mass is called moment of inertia, or rotational inertia.

The symbol is I.

The units are kg•m 2 .

Moments for various objects

I = ⅟ ML

2 12

I = MR

2

I = ⅖ MR

2

I = ⅓ ML

2

I = ½ MR

2

I = ⅔ MR

2

I = ⅟ Ma

2 12

I = ⅟ M(a

2

  • b 2 12 )

Rotational dynamics

r m point mass We know that the angular acceleration is proportional to the net torque applied to a system.

α ∝ ∑^ τ

This is analogous to the statement of Newton’s second law for constant mass: a ∑^ F For a single revolving point mass: F = ma Since a T = r α , we can say: (^) F = mr α

Newton’s Second Law Newton’s second law applies to rotating objects.

τ replaces F, I replaces m, and α replaces a:

τnet = I α

  1. A Japanese quarterstaff (bō) has a mass of 1.20 kg, and measures 183 cm (length) by 3.00 cm (diameter). (a) Calculate the moment of inertia if twirled while holding the center of the staff. (b) Calculate the moment of inertia if swung while holding the end of the staff.

I = ⅟ ML

2 12

I =

(1.20 kg) (1.83 m) 2 12 = 0.335 kg•m 2 This assumes a rod a lot longer than its diameter. A real 3-D rod should use the equation: I = 1/12 ML 2

  • 1/4 MR 2

. In this problem, this would only add an additional 0.0000675 kg•m 2 !

I = ⅓ ML

2

I =

(1.20 kg) (1.83 m) 2 3 1.34 kg•m 2

I =

(b) If you apply this torque by wrapping a strap around the outside of the wheel, how much force should you exert on the strap? Solve for torque:

τnet = I α

τnet = (0.363 kg•m

2 )(3.35 rad/s 2 )

τnet = 1.21 Nm

τnet = rFsinθ θ = 90°), we can

τnet = rF

F = =

τnet

r 1.21 Nm 0.220 m

= 5.50 N