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10 Nov 2019

The problem is as follows:

I have the notes in my notebook but cannot figure out what to dowith r

this is where I am right now:

Ie(ωe)+(Im)(ωm)=(Ie+Im)(ωfinal)

I know I need to find the final angular velocity of the earthbut I have no clue about what to do with the radiuses that pop upthroughout the equation since

I=mr2

ω=r/time

and

vt= rω

(vt in this equation would be 10^5 m/s for the meteor)

Please help!

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Keith Leannon
Keith LeannonLv2
25 Apr 2019

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

Given the radius vector and the velocity vector of an orbiting bodyidentify the trajectory as either circular, elliptical, hyperbolicor parabolic.

r = J + 2K
v = .9I + .123K

Alright, so I guess my question is if I'm doing thiscorrectly?

I am trying to find the eccentricity of the orbital path which willtell me what type of trajectory it is on.
So I start by finding the angular momentum which is h. h = r crossv which gives me .123I + .18J - .9K

Then I use the equation for eccentricity, mag(e) = (v cross h)/mu -r/magnitude(r).
I need to find mu which is a gravitational constant and since itsgeneric units of DU (distance units) and TU (time units) the muconstant has units of DU^3/TU^2. To get these units I usemag(r)*mag(v)^2.
I do not know if this is the correct way of getting mu, thegravitational constant usually equals G*M (gravitational constant *mass of central body) but all I am given is the r and vvectors.

Anyway continuing on with the problem I get the magnitude of e =.02655 which is between 0 and 1 so the orbit is an ellipse.

Again, I just am wondering if I am doing this right.

Thanks.

In order to solve the problem you must refer to the first partwhich i already solved and I am 200% sure of the answers, however Iam stuck on the second part and I need help asap. Please do not getscared when you look at this problem and thank you soo much inadvance.

A wheel 75 meters in diameter is mounted vertically on an infinitefrictionless plane at the origin. The infinite frictionless planehas a grid of Cartesian coordinates on it. The moment of inertia ofthe wheel is mr^2. The mass of the wheel is 2000kg. There is anengine connected to the wheel that can produce a torque of 2.34x10^6 Nm. There is a 50 kg object attached to the circumference ofthe wheel that is attached to the wheel with a remote releasemechanism. The wheel is placed on the origin so that the plane ofthe wheel contains the point x=50 meters and y=86.8 m. The wheel ispositioned so that the object is just above the plane at the 6o'clock position. The engine is engaged and begins to apply thetorque to the wheel at 12:00 noon on March 1, 2011. The wheelbegins to rotate in the counterclockwise direction. The torque isapplied for 4 x 10^4 revolutions. At that point the object isreleased from the wheel and a 75 Newton motor accelerates theobject in the direction it is released. The motor is fired for 2.34x 10^4 seconds. (the following questions i have already answered)What is the angular velocity and rotational KE of the object justbefore it is released? (answer for that is 6.39 x 10^2 rad/sec and5.89 x 10^11 J). What angle does the path of the object make withthe x axis? (answer is 60.05 degrees). Where is the object on theinfinite frictionless plane when the motor stops (what are itscoordinates....answer is x=4.85x10^8m y=8.40x10^8m). What is thevelocity and linear KE of the object when the motor stops?(5.91x10^4 m/sec and KE= 8.73 x 10^10J). How much work was done onthe object up until the time the motor stopped? (8.73 x 10^10J).(Hint: the mass of the wheel is 2050 kg while the torque is appliedand the object is attached).

YEA I know what you are thinking at this moment. Difficult right?Please give it a chance and have faith in yourself. Here is thepart I had difficulty with:

After the engine stops firing the object is turned around so thatthe engine can be used to stop the object. The motor is fired atits rated thrust to stop the object. Unfortunately it malfunctionsand shuts down after an hour and the object coasts from then on. Atthe end of an hour of coasting it encounters a ramp that is 2000meters high and slopes up at an angle of 20 degrees. The objectclimbs the ramp and flies off the ramp into the air when it reachesthe end of the ramp. It strikes the plane and digs a crater 5meters deep in the plane and comes to a stop. How far from the endof the ramp does it land? Where does it land on the plane (as incoordinates)? What is the resultant velocity of the object and whatis the acute angle of the resultant velocity with the plane when itstrikes the plane? What is the KE of the object when it hits theplane? What date and time is it when the object lands on the plane?How much work was done on the object to stop it? How long did ittake to stop? What was the deceleration of the object in m/sec^2?What was the force on the object in Newtons to stop it?

If you are from MIT I have faith that you can do this. I am only ajunior in high school taking first year physics that is not evenAP. My teacher is from MIT and said he did this kind of stuff whenhe was there. Please please please help me! I'm desperate.


You are given a circular conducting wire loop with one
end cut out and connected to a resistor, R, as shown in
the figure to the right. A metal rod of length l is set at an
initial angle ?0 from the resistor, and then rotates
clockwise about the center of the wire loop with angular
frequency through a uniform magnetic field B
directed into the page. The metal rod stays in contact
with the wire loop at all points of its motion. Unless
otherwise requested, all answers should be in simplified
algebraic form using the above given variables.

(a) Given that the initial angle ?0 is in radians, what is theinitial area of the closed circuit “wedge”defined by ?0? (2pts.)
(b) What is the magnetic flux through this initial “wedge” areadefined by ?0? (2 pts.)
Starting at time t=0 at position ?0, the rod is rotated clockwisewith angular velocity as shown in
the figure. Given this configuration, answer questions (c)-(g)below:
(c) What is the direction of the induced magnetic field inside the“wedge”? (2 pts)
(d) What is the direction of the induced current around the closedcircuit? (2 pts)
(e) What is the magnitude of the induced emf, E in the circuit? (4pts)
(f) What is the magnitude of the induced current in the circuit? (4pts)
(g) What is the power used by the resistor? (4 pts)
(h) The metal rod is now rotated in the opposite direction, i.e. inthe counter-clockwise direction. Given this information, completethe chart below as to what quantities you just calculated change,and what ones do not change. You do not have to show work, justgive the result. No partial credit will be given here. (5pts)

Direction of induced magnetic field inside the “wedge”.
Direction of the induced current around the closed circuit.
Magnitude of the induced emf, E in the circuit.
Magnitude of the induced current in the circuit.
Power used by the resistor.


The setup is now modified slightly so that the circular conductingwire loop no longer has a small
section removed as shown in both figures below. This modificationwould allow the metal rod to
continuously rotate on the circular wire loop as clearlyillustrated in the 3D side view. The metal rod
is rotated clockwise, as in the initial part of the problem.


(i) Given this information, complete the chart below. You do nothave to show work, just give the
result. No partial credit will be given here. (5 pts)

Direction of induced magnetic
field inside the “wedge”.
Direction of the induced current
around the closed circuit.
Magnitude of the induced emf, E in
the circuit.
Magnitude of the induced current
in the circuit.
Power used by the resistor.

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