Class 091104

`q001.  The mass of the Earth is about 6 * 10^24 kg.  You should know your own mass (if you want to use your ideal mass, that's OK).  You are about 6400 km from the center of the Earth.

Using Newton's Law of Universal gravitation, find the force exerted by gravity on your mass.

Convert this force to pounds to see if it makes sense.

My mass is about 78 kg.  That's a midrange mass, so we'll use it for this example.

The force exerted by gravity on my mass is

• F = G m_earth * m_me / r^2,

where r is the distance from the center of the Earth and G is 6.67 * 10^-11 N m^2 / kg^2.

Thus

• F = 6.67 * 10^-11 N m^2 / kg^2 * (6 * 10^24 kg) * (78 kg) / (6400 km)^2.

You should immediately see that km is not a unit that matches the units of G, which is expressed in terms of kg, m and sec.  Since 6400 km = 6 400 000 m, we modify our calculation and get

• F = 6.67 * 10^-11 N m^2 / kg^2 * (6 * 10^24 kg) * (78 kg) / (6 400 000 m)^2.

Doing the arithmetic we get

• F = 760 N.

My weight in pounds is about 170. 

• 170 lbs = 170 lbs * (9.8 N / (2.2 lb) ) = 757 N,

which except for a little approximation error is equal to the calculated force exerted by gravity.

`q002.  If you were 10% further from the center of the Earth than you are right now, how far would you be?  What force would gravity be exerting on you at that distance?  If you were in free fall at this distance, what would be your acceleration?  What average force do you think gravity would exert on you if you climbed from the surface of the Earth to that distance?  How much would your gravitational PE change during the climb?  Where would you get the energy to manage this?

If I was 10% further then, since 10% of 6 400 000 m is 640 000 m, I would be 6 400 000 m + 640 000 m = 7 040 000 m from the center of the earth.

At this distance the force exerted by gravity is about

• F = 6.67 * 10^-11 N m^2 / kg^2 * (6 * 10^24 kg) * (78 kg) / (7 040 000 m)^2 = 600 N.

In free fall the gravitational force would be the net force, so the acceleration would be

• a = F_net / m = 600 N / (78 kg) = 6.2 m/s^2.

At the surface gravity exerts 760 N of force; at the given distance the force is 600 N.  The forces don't differ drastically so the average force is pretty close to the average of the two forces.

• F_ave = (760 N + 600 N) / 2 = 680 N.

Climbing 'straight up' to the increased distance, gravity oppose your efforts with an average force of about 680 N, so the work done by gravity would be

• `dW_grav = - 680 N * 7 040 000 m = -4 800 000 000 Joules, or in scientific notation
• `dW_grav = -6.8 * 10^2 N * 7.04 * 10^6 m = -4.8 * 10^9 Joules.

Your change in gravitational PE would be equal and opposite to this, so

`dPE_grav = 4 800 000 000 Joules or 4.8 * 10^9 Joules.

To get the energy for this climb you would have to eat about 10^10 Cheerios, or equivalent, above and beyond whatever amount you would need to eat to maintain your weight if you were sedentary.  Imagine 3000 rows, each with 3000 Cheerios, and 1000 layers of Cheerios.

`q003.  If you doubled your mass, what would happen to the gravitational force exerted on you by Earth's mass?  What would happen to this force if your mass remained unchanged, but Earth's mass doubled?  What would happen if both your mass and that of the Earth doubled?  What would happen if your mass, the Earth's mass and your distance from the center of the Earth doubled?

If my mass doubled, then F = G m_earth * m_me / r^2 would double.

If the Earth's mass doubled, then F = G m_earth * m_me / r^2 would double.

If both my mass and the Earth's mass doubled, then F = G m_earth * m_me / r^2 would quadruple.

If r doubles then r^2 quadruples.  r^2 is in the denominator so if nothing else changed F = G m_earth * m_me / r^2 would become 1/4 as great.  However if, as in the question, my mass and that of the Earth both doubled, then the numerator becomes 4 times as great, as does the denominator, and the force of gravity won't change.

`q004.  If you drive at 30 mph around a circle of radius 100 feet, what is your centripetal acceleration?  Using your mass, calculate the corresponding centripetal force.

Centripetal acceleration is a_cent = v^2 / r.

30 mph means 30 miles / hour.  100 ft is expressed in feet.  We need consistent distance units, and the standard unit of time is the second.

If we use ft as the distance unit, then 30 miles / hour becomes 30 ( 5280 ft) / (3600 sec) = 46 ft / sec, roughly.

Our centripetal acceleration is thus

• a_cent = v^2 / r = (46 ft / sec)^2 / (100 ft) = 21 ft / sec^2.

1 ft = .31 meters so if we want to use metric units this becomes

• 21 ( .31 m) / sec^2 = 6 m/s^2, again an approximate calculation.

Advice for Test 1

If an object is on an incline:

• Immediately sketch the appropriate force diagram, obtaining the components of its weight parallel and perpendicular to the incline (i.e., mg_x and mg_y, respectively). 
• Note where the parallel component is in the positive or negative x direction. 
• If no other y forces are present, the normal force is equal and opposite to the perpendicular component of the weight, and you should note this.
• If the object moves along the incline, the work done by gravity is done only by the component of the gravitational force which is parallel to the incline (i.e., by mg_x).  Be sure to consider whether the displacement is in the positive or negative x direction, so this result will have the correct sign.  The change in gravitational PE is equal and opposite to the work done on the object by gravity.

`q005.  A ball rolls with unchanging velocity down a constant-velocity ramp.  When given a velocity up the ramp its acceleration is -20 cm/s^2.  What is the coefficient of rolling friction between ball and ramp?  What would be the acceleration of the ball if the ramp was perfectly level?  (Apply the advice given previously to these situations). 

Let m stand for the mass of the ball.  Its acceleration when moving up the ramp is -20 cm/s^2, so the net force on the ball as it goes up the ramp (assuming up to be the positive direction, a choice made implicitly in the problem) is

F_net_up = -20 cm/s^2 * m.

The qualitative difference between the net force up the ramp and the net force down is that friction acts upward when the ball rolls down the ramp, and downward when the ball rolls up.  It follows that the difference in forces is double the frictional force. 

So

• f_friction = 1/2 F_net_up = -10 cm/s^2 * m.

This is much less than the acceleration of gravity; we conclude that the ramp is nearly, but not quite, level.  It follows that the magnitude of the normal force is very close to the magnitude of the weight of the ball.  The weight of the ball is the force exerted on it by gravity, so weight = m g, and it follows that the normal force has magnitude

• | F_normal | = m g.

Since | f_friction | = mu *   F_normal |, we have

• mu = | f_friction | / | F_normal | = 10 cm/s^2 * m / (m g) = 10 cm/s^2 / g = 10 cm/s^2 / (980 cm/s^2) = .01, approx.

`q006.  What would be `dPE, `dKE and `dW_NC_ON for a 100 gram ball as it rolls 40 cm down the constant-velocity ramp of the preceding?  Answer the same if the ball rolls 40 cm up that incline.  Answer the same if it rolls 40 cm along the same ramp, but with the ramp perfectly level.

A 100 gram ball has weight 4.9 N.  If it rolls along the ramp in the preceding, where the coefficient of rolling friction is .01, it experiences a frictional force of

• f_frict = .01 * 4.9 N = .049 N.

If it rolls 40 cm along the ramp, then friction is a nonconservative force directed against motion, and

• `dW_NC_ON = F `ds = -.049 N * .4 m = -.02 J, approx..
• `dW_NC_ON = `dKE + `dPE

On a constant-velocity ramp the velocity, and therefore the KE, does not change.  Thus `dKE = 0 and

• `dPE = `dW_NC_ON so that
• `dPE = -.02 J, approx..

If the ball rolls 40 cm up the incline, then `dPE = +.02 J, approx..  The frictional force is again in the direction opposite the displacement, so the work done by friction is still negative, and

• `dKE = `dW_NC_ON - `dPE = -..02 J - .02 J = -.04 J.

We conclude that the ball slows down, losing and equal amount of KE to both friction and PE.

If the ball rolls 40 cm along a level ramp, then `dPE = 0 and friction still does .02 J of work against the ball, so

• `dKE = `dW_NC_ON = -.02 J.

The ball again slows.

Homework:

Your label for this assignment: 

ic_class_091104

Copy and paste this label into the form.

Answer the questions posed above.

You should know everything in the first six problems of Set 5 in the Introductory Problem Set, which will give you a good, and not difficult, introduction to working with vectors.  A link that should get you there is at http://vhmthphy.vhcc.edu/ph1introsets/default.htm . 

You should also read Chapters 7 in your text, which can serve as a good coherent reference.