# Ch 5.2 Help

## Ch 5.2 Help

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):

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):

**gguest**- Guest

## Re: Ch 5.2 Help

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!

**thb1026**- Posts : 8

Join date : 2008-09-27

## 5_2

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!

**Jessica**- Guest

## 5_2

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.

**GUESTT**- Guest

## Question 4

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

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

**Nejibana**- Posts : 1

Join date : 2008-09-22

## number three

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

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

**super Mo**- Guest

## Question 1

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!

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

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**- Guest

## Question 2

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?

**hwilson**- Guest

## Quest #2

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

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

**guest101**- Guest

## Question 2

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

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

**Kath**- Guest

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