Ch 5.2 Help

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Ch 5.2 Help

Post  gguest on Fri Sep 26, 2008 10:35 am

Question 1: (1 point)

The cabin of a merry-go-round moves in a vertical circle of 5.874 m radius making a full turn every 12.45 seconds. What is the apparent weight of a 31.48 kg passenger at the top of the circle (see sheet 12, Ch4 sheet 16)? Indicate with a negative (positive) sign whether the apparent weight at the top is smaller (larger) than the apparent weight at the bottom.

(How to solve):


Question 2: (1 point)

You swing an 0.3579 kg mass held by a wire in a vertical circle of 0.3211 m radius with a constant speed of 3.853 m/s. What is the magnitude of the tension in the wire at the bottom of the circle (see sheet 12 and Ch4 sheet 18" (the tension in the string is the apparent weight); replace the cabin with the 0.3579 kg mass; the centripetal force is due to gravity and tension in the wire)? Indicate with a positive (negative) sign whether the tension at the top of the circle is larger (smaller) than the tension at the bottom.

(How to solve):


Question 3: (1 point)

Simulate artificial gravity in a tube shaped space station with a 130.1 m radius far from earth (earth's gravitational pull is negligible) . Choose the tangential velocity of a point on the tube such that artificial gravity of 1.0894 g results ( g is the gravitational acceleration close to the surface of the earth) (see sheet 14,14'). Indicate with a negative (positive) sign whether the occupants of the station walk on the wall at the inner (outer) radius.

(How to solve):

Question 4: (1 point)

The coefficient of friction for the tire-road interface of a car is 0.2789 . What is the maximum speed in km/hr the car can have in a curve with a 58.74 m radius of curvature on a level road (not banked) before the car slides off the road (see sheet 16-18)? Indicate with a negative (positive) sign whether this velocity is larger (smaller) with everything the same except the road being banked (see sheet 16,19).

(How to solve):

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Re: Ch 5.2 Help

Post  thb1026 on Sat Sep 27, 2008 6:12 pm

I have no idea how to do question 3 on this 5-2 quiz! I am looking at sheet 14,14' and it's not helping at all! help!

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Re: Ch 5.2 Help

Post  guesttt on Sat Sep 27, 2008 9:29 pm

can you show how to do 1 and 2 if you got it

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5_2

Post  Jessica on Sun Sep 28, 2008 1:32 am

Number 1 and 2 are given to you straight on the sheets. For apparent weight you use Wapparent=m*(g+or -centripetal acceleration). I am not sure about 3. But for number for you need to consider that they want the final units in km/hour. Convert m/s you get using the formula given on sheet 18, then convert to km/hour (how many m are there in a km, how many s in an hour). HOPE THIS HELPS!

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5_2

Post  GUESTT on Sun Sep 28, 2008 1:37 am

NEED HELP WITH NUMBER 3 Simulate artificial gravity in a tube shaped space station with a 299.1 m radius far from earth (earth's gravitational pull is negligible) . Choose the tangential velocity of a point on the tube such that artificial gravity of 1.0466 g results ( g is the gravitational acceleration close to the surface of the earth) (see sheet 14,14'). Indicate with a negative (positive) sign whether the occupants of the station walk on the wall at the inner (outer) radius.

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Question 4

Post  Nejibana on Sun Sep 28, 2008 7:00 pm

The coefficient of friction for the tire-road interface of a car is 0.2789 . What is the maximum speed in km/hr the car can have in a curve with a 58.74 m radius of curvature on a level road (not banked) before the car slides off the road (see sheet 16-1Cool? Indicate with a negative (positive) sign whether this velocity is larger (smaller) with everything the same except the road being banked (see sheet 16,19).


F=uN
m(v^2/r)=u*m*g mass cancels out
(v^2)/r = ug
v^2=r*u*g
v^2 = .2789*58.74*9.81
v^2 = 160.713
square root to find v
v=12.6773
Then since we need km/hr we have to multiply by 60 for minutes and another 60 for hours and finally divide by 1000 for km
v=((12.6773*60)*60)/1000
v=45.64km/hr
The answer you get needs the negative sign

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Questions 1 and 3

Post  GuEsT001 on Sun Sep 28, 2008 9:33 pm

still confused on 1 and 3 any ideas?

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number three

Post  super Mo on Sun Sep 28, 2008 9:55 pm

Simulate artificial gravity in a tube shaped space station with a 259.3 m radius far from earth (earth's gravitational pull is negligible). Choose the tangential velocity of a point on the tube such that artificial gravity of 1.049 g results. (g is the gravitational acceleration close to the surface of the earth) (see sheet 14,14'). ...
I am not sure of the logic but...

Multiply the normal g by the artificial gravity given
9.81m/s^2 x 1.049m/s^2 = 10.2907
Think about the formula: a_c = v^2/r and solve for v
square root (259.3m x 10.2907m/s^2) = 51.656

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Questions 1 and 2

Post  Guests on Sun Sep 28, 2008 10:02 pm

Can someone help I can't figure out 1 and 2?

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Question 1

Post  Kath on Sun Sep 28, 2008 11:31 pm

The cabin of a merry-go-round moves in a vertical circle of 5.874 m radius making a full turn every 12.45 seconds. What is the apparent weight of a 31.48 kg passenger at the top of the circle (see sheet 12, Ch4 sheet 16)? Indicate with a negative (positive) sign whether the apparent weight at the top is smaller (larger) than the apparent weight at the bottom.


~~~~~~~~~~~~~~~~~~~~

For question one, you used Wapparent=m*(g-ac)

First, you have to find the velocity using (2*pi*r)/(t)
once you have velocity, use the equation ac=v^2/r to find centripetal acceleration

You can now plug those numbers into the apparent weight equation

Make sure to put a negative in front of your answer!

So, to do a walk-through for this equation, you would do
v= (2*pi*5.874)/(12.45) = 2.9644522
then to find ac do ac=(2.9644522)^2/(5.874) = 1.49608

then Wapparent = 31.48(9.81-1.49608) = 261.722

And the final answer, with a negative, is -261.722.

hope this is right and helps!

Kath
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Question 2

Post  hwilson on Sun Sep 28, 2008 11:43 pm

I have been working on number 2 for a while I just cant seem to get it right. Is there anyone who knows how to do it?

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Quest #2

Post  guest101 on Sun Sep 28, 2008 11:49 pm

For #2 use the centripetal acceleration equation ....ac = v^2/r
For my problem/#s this is how i did it...

ac = (3.775m/s)^2/.4175m = 34.133m/s^2

Then i plugged that # into the equation for apparent weight
Fn= m x (g +/- ac)

Fn= .2859kg x (9.81m/s^2 + 34.133m/s^2) = 12.5633kg

*The answer is negative*

Hope that helps ya Smile

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Question 2

Post  Kath on Sun Sep 28, 2008 11:58 pm

First off, for question two the answer is negative
I used the equation T=mg + m(v^2/r)
you are given all the variables so you just have to solve for T
It worked for me on maple Ta

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Re: Ch 5.2 Help

Post  confused on Mon Sep 29, 2008 2:25 pm

is number three positive or negative?

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Re: Ch 5.2 Help

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