Chapter 5 Mechanics II (this one)

Mechanics II

Center of Mass and Gravity

Uniform Circular Motion

Torque

Equilibrium

Rotational Inertia

Center of Mass and Gravity

-when drawing a force diagram, objects are often represented by a dot. Each dot means the center of mass.

-the center is mass and center of gravity are terms that are used interchangeably.

-the center of mass on a hammer is where the hammer balances on your finger. This means that if you throw it with a glowing sticker on it in a dark room, the hammer thrown will make the pattern of a nice parabola with no loops.

The center of mass is the point where the object behaves as if it were a single particle and where all the mass in the object is concentrated.

homogenous object: density is constant, therefore, the center of mass is at the geometric center.

non-homogenous object: density is not constant, therefore, the center of mass is not at its geometric center.

homogenous object center of mass

A homogeneous object is an item or material that has identical physical properties, structure, and composition at every point within it.

non-homogenous object center of mass

Note that in some cases, the center of mass is not located within the body of the object (this is the case for non-homogenous objects), and because the density varies from point to point in a non-homogenous object, there is no single way to calculate the location of the center of mass.

Center of Mass for Point Masses

X_CM= m1x1+m2x2+m3x3+…./ m1 + m2 + m3…

the location of the center of the center of mass is denoted by (an 1. an x with a line over it or 2. “X_CM”

1. choose an origin (a reference point to call x=0). The locations of objects will be measured relative to this point. Often the easiest point to use will be at the location of the left-hand mass, but any point is fine; if a coordinate system is given in the problem, use it.)

2. Determine the locations (x1,x2,x3) of the objects

3. Multiply each mass by its location (m1x1, m2x2, m3x3, etc.) then add

4. Divide by the total mass (m1 + m2 + m3+…)

Example 5-1: In the figure below, three blocks hang below a massless meter stick. Block m, hangs from the 20 cm mark, block m₂ hangs from the 70 cm mark, and block m₃ hangs from the 80 cm mark. If m₁= 2 kg, m₂= 5 kg, and m₃= 3kg, at what mark on the meter stick should a string be attached so that this system would hang horizontally?

the question is giving you mass and distance, and asking you for the center of mass, because the center of mass is the point where this system would hang horizontally.

1. choose an origin, a reference point to call x=0. Since the question asks “how far from the left end”, the best place to choose our zero mark is at the left end. We now can write x₁= 20 cm, x₂= 70 cm, x₃= 80 cm.

2. use the formula

You do not need to convert distance into meters for center of mass calculations because units cancel out in the math formula.

1. what if we would have chosen the middle of the meter stick (the 50 cm mark)? In this case, numbers that are before the middle are negative (because they are to the left) and numbers that are after the middle are positive (because they are to the right of the reference point).

2. x=13 cm, this number is positive. the reference point is 50 cm, so 50 cm + 13 cm = 63 cm

example 5-2:

Falls are one of the most serious medical issues among the elderly. Falls result when the center of mass of a person’s body is not located over the base of support (largely determined by someone’s foot placement) and the person is unable to correct the imbalance with sufficient speed and coordination. Center of mass when standing is determined by body shape and weight distribution. It’s important to keep in mind that when people move, they redistribute their body mass and thus change the location of their centers of mass. What are some likely physical (as opposed to physiological, neurological, or environmental) risk factors for falling, and some possible avoidance strategies?

Solution:

1. Obesity: one risk factor for falling is obesity, obesity shifts the center of mass while standing still, obesity also affects walking motion, the obese individual may be more likely to experience a shift of the center of mass outside the base of support while moving and be less able to prevent the fall once it begins.

2. Posture: people often develop head protrusion and thoracic kyphosis (a hump in the upper back) as they age, shifting the center of mass forward.

3. Many ways of shifting the body’s center of mass closer to the feet and thus making it more likely that the center of mas will remain above the base of support (at least while both feet are planted) are impractical: heavy shoes and pants, for example. However, apart from exercises and physical therapies to avoid or alleviate the risk factors mentioned above, one common risk avoidance strategy is to increase the size of the support base with a cane or a walker. Both of these have the effect of providing a larger total area that the center of mass can occupy without causing imbalance.

Example 5-3: An ammonia molecule (NH₃) contain 3 hydrogen atoms that are positioned at the vertices of an equilateral triangle. The nitrogen atom lies 38 pm (1 pm= 1 picometer= 10^-12 m) directly above the center of this triangle. If the N:H mass ratio is 14:1, how far below the N atom is the center of mass of the molecule.

The question is asking you to find the center of mass, and then say how far below the N atom is the center of mass of this molecule.

Solution:

1. The objects in this system (the four atoms) are not arranged in a line, so how can we hope to determine the center of mass?

2. The key to answer the problem is to realize that we don’t need to include all four of these objects in a single calculation; we can divide the problem into stages. Since the three H atoms have equal masses and are symmetrically arranged at the corners of an equilateral triangle, the center of mass of just these 3 H’s is at their geometric center: namely, the center of the triangle. Therefore, by definition of center of mass, the 3 H atoms behave as if all their mass were concentrated at the center of the triangle. This now turns the problem into computing the center of mass of 2 objects (which obviously lie on a line): the 3 H atoms at the center of the triangle and the N atom:

3. let the position of the N atom be the zero mark.

4. use the formula:

y_cm= m_Nx_N+ m_3Hx_3H/ m_{N +} m_3H= (14)(0 cm) + (3) (38 ppm)/ 14+ 3

114/17= 7 pm (it was 6.7, but was rounded up to 7)

the center of mass of mass of the molecule is 7pm. Therefore, the center of mass of the NH3 molecule is 7/38. This makes sense because nitrogen is heavier and so we expect the center of mass to be closer to the nitrogen.

You do not need to convert distance into meters for center of mass calculations because units cancel out in the math formula.

you may not always be able to simplify the object into a single dimension (like what we just did with the nitrogen and hydrogens), in that case, break the multidimensional problems into x and y components and add the components separately. 1. find the center of mass in the x-direction, then find the center of mass in the y-direction.

A single dimension, or one-dimensional (1D) space, is a system where a location is specified using only a single coordinate or number.

Uniform Circular Motion

-Uniform Circular Motion abbreviated UCM.

-uniform circular motion describes a object moving in a circular path

-the speed in uniform circular motion is constant

-The acceleration of an object undergoing uniform circular motion always points toward the center of the circle. The term centripetal (from the Latin, meaning to seek the center) is therefore used to describe the acceleration of an object undergoing UCM. We’ll denote centripetal acceleration by a_c.

Magnitude of Centripetal Acceleration

a_c= v²/r

v is the speed of the object

r is the radius of the circular path

We now know the direction of centripetal acceleration at any point on the cirlce; what is the magnitude? If v is the speed of the object and r is the radius of the circular path, then the magnitude of the centripetal acceleration, a_c, is v²/r.

Magnitude of Centripetal Force

F_c= ma_c= mv²/r

If an object is accelerating, then it must be feeling a force (after all, F_net=ma, so you can’t have an acceleration without a force). Since F_net and a always point in the same direction, no matter what the path of the object, the net force on an object undergoing UCM must, like a, point toward the center. So, guess what we call it? Centripetal force (denoted F_c). This is the net force directed toward the center that acts on an object to make it execute circular motion. And since F_net= ma, we’ll have F_c=ma_c so the magnitude of the centripetal force is mv²/r, where m is the mass of the object that’s moving around the circle.

Example 5-4

Separating blood plasma from the solid bodies in blood (blood cells and platelets) by rapid sedimentation requires use of a centrifuge to produce a necessary accelerations on the order of 5000g. In one approach, the blood is placed in a bag inside a rigid container and mounted to the end of a horizontal rotor (so that it extends our beyond the rotor), which then spins up to several thousands of revolutions per minute. Suppose the rotor has a radius of 30 cm and rotates at a maximum rate of 5000 rpm, and that the bag is 10 cm long. Note that translational velocity v=rω, where w is in radians/second.

“Note that translational velocity v=rω, where is in radians/second.”

this is the equation that will be used to find v for a_c=v²/r

a) What will be the centripetal acceleration at the middle of the bag?

a_c= v²/r

1. convert rpm (revolutions per minute) “5000 rpm” to rad/s (radians per second) because ω is in rad/s. 5000 rpm x 2π/60= 500 rad/s

ω= 500 rad/s

2. read: they told you v=rω, so this is an equation you use to find the velocity for the a_c= v²/r. v is for velocity, r is for rad/s, ω is for angular velocity.

v=rω= (0.30m + 0.05m)+ 500 rad/s= (0.35m)(500 rad/s)= 175 m/s

3. use a_c= v²/r to find the centripetal acceleration: (175m/s)² / 0.35m= (200)²/0.5m= 8×10⁴ m/s²

“suppose the rotor has a radius of 30 cm”

Since the bag is 10 cm long, its midpoint is:

10 cm/2=5 cm

Convert 5 cm to meters:

5 cm=0.05 m

Example 5-5: If an object undergoing uniform circular motion is being acted upon by a constant force toward the center, why doesn’t the object fall into the center?

solution: actually, the object is falling toward the center, but because of its speed, the object remains in a circular orbit around the center. Remember, the direction of v is not necessarily the same as the direction of F_net. So, just because F_net points toward the center does not mean that v must point toward the center. It’s the direction of the acceleration, not the velocity that always matches the direction of F_net. Let’s look at the motion of the object at a certain point in its circular path:

look at the picture

1. the net force on the object at position 1 points downward toward the center of the circle

2. F_net is telling v₁ to move downward a little, so that at position 2, the velocity will point downward slightly.

3. Notice that this is just what we want in order to keep the object traveling in a circle!

Example 5-5: How would the net force on an object undergoing circular motion have to change if the object’s speed is doubled?

F_c=ma_c= mv²/r

if a_c has doubled, then the F_net has quadrupled.

Centripetal force is not some new kind of force like gravity or tension. It’s simply the name for the net force directed toward the center of the circular path. The vector sum of forces such as gravity and tension is what gets called centripetal force, when those forces, or components of them, are directed toward the center of the circle. When drawing a force diagram for an object undergoing UCM, here are a couple of tips:

1. Do not add a force called F_c in your picture; forces such as gravity, tension, normal force, etc. do go in your picture, but F_c doesn’t. Remember, F_c is what the forces toward the center have to add up to.

2. Always call toward the center the positive direction. Any forces toward the center are then positive forces, and any forces directed away from the center are negative. You’ll need this to find F_net and then set the result equal to F_c.

Example 5-7: “find an expression for the speed of the moon’s orbit”

1. we need to make an expression “x=xx”

2. start with “what provides the centripetal force”, the moon is orbiting the earth. So the centripetal force is provided by the gravity of the earth. F_{grav(of earth)}= mv²/r

3. Since we know F_grav= GMm/r², we get F_grav= F_c, GMm/r²= mv²/r, GM/r= v²→ v= √GM/r

notice that m cancels out, this means that the mass of the moon cancels out. So, any object orbiting at the same distance from the earth as the moon must move at the same speed as the moon.

Example 5-8: A string is tied around a rock of mass 0.2kg, and the rock is then whirled at a constant speed v in a horizontal circle of radius 0.4m, as shown in the figure below. If sinθ= 0.4 and cosθ= 0.9, what’s v?

(This figure also shows why the end of the string is slightly above the center of the cir

Torque

Torque is a measure of a force’s effectiveness at making an object spin or rotate.

All systems that can spin thor rotate have a “center” of turning. The “center” of turning is the point that does not move while the remainder of the object is rotating, becoming the “center” of the circle. There are many terms used to describe this point, it’s also called the pivot point or the fulcrum.

Torque is not a force, it’s the property of a force.

Radius vector, r: the vector from the pivot point (center of rotation) to the point of application.

theta: the angle between the vector r an dF

theta’: the angle between vectors r and F at the point where they actually meet, the angle they make when they start at the same point.

vector r starts at the pivot point, vector F starts at the end of vector r.

Formula for Torque

t=rFsinθ

the t is the greek letter for tau

Formula for Torque with lever arm

tau=lF

the “l” is a lower case L for “lever arm”

A lever arm (or moment arm) is the straight, perpendicular distance from a pivot or rotation point to the line where a force acts. Longer lever arms mean you need less force to turn or lift heavy things.

Equilibrium

Equilibrium means zero acceleration

Static Equilibrium: equilibrium in which the velocity is zero. “static”, not moving

Translational equilibrium: if the forces cancel to produce a net force of 0.

Rotational equilibrium: if the torques cancel to produce a net torque of 0.

Rotational Inertia

Rotational Inertia tells us how resistant an object is to rotational acceleration