Moment of Inertia & Newton s Laws for Translation & Rotation

Size: px

Start display at page:

Download "Moment of Inertia & Newton s Laws for Translation & Rotation"

Caroline Martin
2 years ago
Views:

Moment of Inertia & Newton s Laws for Translation & Rotation In this training set, you will apply Newton s 2 nd Law for rotational motion: Στ = Σr i F i = Iα I is the moment of inertia of an object:

The first expression applies to a system of masses, the second and third for a whole body.

1 Moment of Inertia & Newton s Laws for Translation & Rotation In this training set, you will apply Newton s 2 nd Law for rotational motion: Στ = Σr i F i = Iα I is the moment of inertia of an object: I = Σm i r i 2 = r 2 dm= r 2 ρdv where r is distance perpendicular from the rotation axis, ρ is density and dv volume element. The first expression applies to a system of masses, the second and third for a whole body. Moments of inertia for common shapes about their centers of mass: point mass ring or hoop solid disk hollow sphere solid sphere stick MR 2 MR 2 ½ MR 2 2/3 MR 2 2/5 MR 2 1/12 ML 2 When the pivot point is offset from the center of mass, you can use the parallel axis theorem to calculate the offset moment of inertia: I = I COM + Md 2 where d is again the distance measured perpendicular to the rotation axis. Newton s Laws can be solved separately for the translational motion of the center of mass and the rotational motion about the center of mass: Στ = I COM α ΣF = ma with additional constraint equations as needed. Remember that the direction for rotation is along the axis of rotation and determined by the right hand rule. Part 1: Calculating Moments of Inertia Use the center of mass moments of inertia and parallel axis theorem to calculate the moments of inertia of the following objects about the labeled pivot point ( ). Your values should be in SI units. The first one has been done for you. (a) A stick of length 1.5 m and mass 300 g pivoted about one end. I COM = 1/12 ML 2 = 1/12 (0.30 kg)(1.5m) 2 = kg m 2 Offset axis distance d = 0.75 m => I = I COM + Md 2 = kg m 2 + (0.30 kg)(0.75 m) 2 = 0.23 kg m 2 1

(b) A stick of length 2.00 m and mass 300 g with a solid disk of mass 250 g and radius 0.

00 m and mass 300 g with a solid disk of mass 250 g and radius 0.

500 m attached to the other, pivoted about the center of the stick.

2 (b) A stick of length 2.00 m and mass 300 g with a solid disk of mass 250 g and radius m attached to one end, pivoted about the opposite end. (c) A stick of length 2.00 m and mass 300 g with a solid disk of mass 250 g and radius m attached to one end, and a ring of mass 150 g and radius m attached to the other, pivoted about the center of the stick. (d) A pendant of mass 120 g designed as a 4.0 cm diameter disk with a 2.0 cm diameter hole cut out of its edge, suspended on a 8.0 cm string of negligible mass, pivoted about the end of the string. 2

3 Part 2: Applying Newton s Laws of Translation and Rotation We can apply our Newtonian framework to solve for both translational and rotational motion by following a similar procedure as in Cycle 2, but with consideration of where forces act on an object and including torque and angular acceleration. Follow these steps for each of the following problems: (1) Make a prediction for both the translational and rotational motion; (2) Draw a force diagram, labeling the center of mass of an object and where the forces act; and a coordinate system, including the direction of positive rotation; (3) Write down ΣF = ma and Στ = I COM α; (4) If necessary, add constraint equations and make sure # unknowns = # equations; (5) Solve for the unknowns, carrying through your units; you may need to use kinematic relations for constant acceleration to solve for your target quantity The first example has been done for you. (a) A 6.5 kg bowling ball (solid sphere) is placed near the top of an inclined ramp with incline angle θ = 25º. Released from rest, the ball rolls down the ramp without slipping. What is the translational acceleration of the ball down the ramp? The ball will accelerate down the incline; at the same time it will increase its spin rate clockwise ( ); its acceleration will be less than g. 1 x: mgsinθ F f = ma x 2 y: F N mgcosθ = ma y 3 rot: τ = RF f = I COM α = 2/5 mr 2 α 4 a y = 0 (no movement off ramp) Equations: 5 5 Rα = a x (rolling without slipping) Unknowns: 5 3+5: F f = 2/5 mrα = 2/5 ma x +1: F f = mgsinθ - ma x = 2/5 ma x => 7/5ma x = mgsinθ => a x = 5/7(9.81 m/s 2 )(sin 25º) = 3.0 m/s 2 3

4 (b) A ribbon is wrapped around a solid disk of mass 0.55 kg and radius 10 cm. The disk is allowed to fall from rest while holding on to one of the ends of the ribbon, and the ribbon unwinds as the disk falls. What is the vertical acceleration of the disk, and how fast is it moving after 5.0 m of ribbon has unraveled? Equations: Unknowns: 4

5 (c) A hollow soccer ball of mass 430 grams and diameter 22 cm is set rolling with a speed of 15 m/s toward a ramp inclined by 50º. How much time does it spend going up and down the ramp before it rolls off in the other direction? Assume the ball rolls without slipping the entire time. Equations: Unknowns: 5

Chapters 10 & 11: Rotational Dynamics Thursday March 8 th

Chapters 10 & 11: Rotational Dynamics Thursday March 8 th Review of rotational kinematics equations Review and more on rotational inertia Rolling motion as rotation and translation Rotational kinetic energy