Center of Gravity. We touched on this briefly in chapter 7! x 2

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Richard Walters
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1 Center of Gravity We touched on this briefly in chapter 7! x 1 x 2 cm m 1 m 2 This was for what is known as discrete objects. Discrete refers to the fact that the two objects separated and individual. In the above case, this was used to determine the point in space two objects rotated about one another.

Discrete refers to the fact that the two objects separated and individual.

2 Center of Gravity (2) This is an important concept for two reasons! This is the point in an extended body where you may consider all the mass to be located and then use this single point for further calculations. The center of gravity is the point where the object is in rotational equilibrium. If you recall from the last lecture, this is the point where the sum of all torques acting on the object exactly equal zero.

this single point for further calculations.

3 Real World Example In part a), the center of gravity is between the front and rear landing gear. The torque produced if we take the pivot point to be the rear wheels is counter-acted by the front wheels. Once the center of gravity is moved behind the rear wheels, there is nothing to counter this torque and the plane rotates its nose off the ground. See the book for a great picture of this really happening!

4 Can You Explain this Picture?

5 Newton's 2 nd Law for Rotation about a Fixed Axis. Recall Newton's 2 nd law: F = m a Remember that forces cannot cause rotation, only torques can do this. Recall the formula for torque: τ = F T r Define: Substitute this into the torque formula: F T = m a T τ = m a T r Recall from Ch. 8: a T = r α Finally we find: m r r m r 2 m r 2 moment of Inertia

6 Moment of Inertia The real difference between an object moving in a linear fashion and an object rotating is the fact that mass associated with different parts of the rotating system may be rotating faster or slower than other parts. Take for example a compact disc: You can think of the mass of the disc as the summation of tiny blocks of mass all added up (m 1 and m 2 are two of many such masses). m 2 m 1 The entire disk spins at one angular speed. This means mass closer to the center of the disk has a tangential velocity which is slower than mass near the outer edge. The moment of inertial takes this all into account and allows you to ignore the differences in tangential speeds of the small bits of mass that make up the entire object.

Take for example a compact disc: You can think of the mass of the disc as the summation of tiny blocks of mass all added up (m 1 and m 2 are two of many such masses).

7 Newton's 2 nd Law F m a objects moving in a linear motion I objects moving in a rotating fashion

8 Moment of Inertia for Different Objects Moments of Inertia for various rigid objects is shown on the right. Here are some interesting things to notice: 1) A solid cylinder which has the same mass as a hollow cylinder will have ½ the moments of inertia (look at the values of the moment of inertia for both objects). This is because all the mass of the hollow cylinder is located far from the axis of rotation. 2) A solid sphere which has the same mass as a hollow sphere will also have a smaller moment of inertia. This is because the hollow sphere also has all the mass located far from the axis of rotation

at the values of the moment of inertia for both objects). This is because all the mass of the hollow cylinder is located far from the axis of rotation.

9 Example Problem R A 1 kg mass is hanging from a flywheel. The flywheel has a radius of 5 cm (0.05 m) and a mass of 2 kg. What is the angular acceleration of the flywheel once the mass is released from rest? The torque acting on the flywheel is produced by the hanging mass and is: (F T R) = (1 kg 9.80 m/s m) = 0.49 N-m Using Newton's 2 nd law for rotation we know: Στ = Iα 1 kg Looking up the moment of inertial for a solid disk we find: I disk =½MR 2 =(0.5)(2 kg)(0.05 m) 2 = kg-m 2 therefore: α = Στ / Ι = 0.49 N-m / kg-m 2 = 196 rad. / s 2

The torque acting on the flywheel is produced by the hanging mass and is: (F T R) = (1 kg 9.80 m/s 2 0.05 m) = 0.

10 Rotational Work Recall work is given by: Work = Force distance (or F Cos θ distance if the force and the motion are not in the same direction) Remember for rotational motion, we substitute torques for forces and θs for distances. This gives us a new definition of rotational work: W R = τ θ where the angle of rotation is measured in radians and the unit of work is the joule (same unit as the work done along a linear path) For the previous example problem, the overhanging mass does work on the pulley as it rotates it faster and faster.

This gives us a new definition of rotational work: W R = τ θ where the angle of rotation is measured in radians and the unit of work

11 Kinetic Energy of a Rotating Object We are now ready to define the kinetic energy associated with a rotating object. Recall the kinetic energy formula of an object moving in a linear fashion: KE = ½ mv 2 Transforming linear velocity v to angular velocity ω and making the identification that mass m now becomes a moment of inertia I, we find: KE R = ½ I ω 2 Keep in mind that an object rolling on a surface has both rotational motion and translational motion (if it is not completely slipping on the surface). This forces us to consider both motions to calculate total kinetic energy! In this case, the total kinetic energy is given as: KE total = ½ mv 2 + ½ Iω 2 The potential energy term contains no information about rotation so it remains unchanged.

becomes a moment of inertia I, we find: KE R = ½ I ω 2 Keep in mind that an object rolling on a surface has both rotational motion and translational motion (if it is not completely

12 Angular Momentum of a Rotating Object Recall that we defined momentum of an object moving in a linear fashion to be: p = mv Now, substituting I for m and ω for v, we can define an angular momentum to be: L = Iω Just like the case with linear motion, if there are no external forces acting on a system then conservation of linear momentum may be used to solve the problem. In the case of rotational motion, we do not consider forces but torques and the momentum conserved is an angular momentum. Therefore, if there is no external torque (or the sum of all external torque adds to zero) then we may write: L i = L f

conservation of linear momentum may be used to solve the problem.

Chapter 11. h = 5m. = mgh + 1 2 mv 2 + 1 2 Iω 2. E f. = E i. v = 4 3 g(h h) = 4 3 9.8m / s2 (8m 5m) = 6.26m / s. ω = v r = 6.

Chapter 11. h = 5m. = mgh + 1 2 mv 2 + 1 2 Iω 2. E f. = E i. v = 4 3 g(h h) = 4 3 9.8m / s2 (8m 5m) = 6.26m / s. ω = v r = 6. Chapter 11 11.7 A solid cylinder of radius 10cm and mass 1kg starts from rest and rolls without slipping a distance of 6m down a house roof that is inclined at 30 degrees (a) What is the angular speed