Help in physics!!!!!!!

[shadow67]

hey i hav a question for u guys if u could please help me out. i am answering questions from a book for my board exam the question is: abodyof mass 1 kg is attached to a string 1m long and revolves in a horizontal circle of radius 0.4 m(40 cm) thus making a conical pendulum. find the tension in the string and the period of revolution of the pendulum. please try to visualize the situation its on the topic of 'circular motion'. the string is being whirled with a 1m string in a conical pendulum forms with the radius of the circle it form while it is being whirled is 0.4 m. the main question is written in bold. please help!

[CraigD]

To understand how to solve this problem, first try diagramming it. You should get something like the attached.

 

With what you’ve been given, the acceleration of gravity (9.8 m/s^2), the formula for centripetal force ([math]F=\frac{MV^2}r[/math]), the realization that the triangle formed by the string and the one formed by the 2 forces are similar, and the Pythagorean formula for the length of the sides of a right triangle ([math]c^2=a^2+b^2[/math]), you should be able to solve for the speed of the pendulum (v), the force (tension) on the string, and with the formula for the circumference of a circle ([math]C=2 \Pi r[/math]), its period of revolution.

[shadow67]

uh craig thanks for the explanantion but that was a bit hard for understanding could u please explain in some simple terms substituting those values please then i will understand thank you. by the way the answer for tension is 10.88N and and period of revolution is 1.93s. and isnt tension= mass * accelaration?

[CraigD]

Look at the 2 triangles in the diagram. Notice that in the upper one (which represents the string), we know the length of all its sides: the given 1 m (the string – the hypotenuse), the .4 m (radius of the circle – the “base”), from which we can calculate the “altitude” as [math]\sqrt{1^2-.4^2} \overset.= .9165[/math] m. ([math]\overset.=[/math] means “is approximately”)

 

From this, we can tell the ratio base/altitude: it’s [math].4/.9165 \overset.= .4364[/math].

 

We know that the same ratio is the lower triangle (which represents the force on the string) is the centrifugal force/force of gravity. Because we know (or can derive, or can look up) the formula for centrifugal force([math]F=\frac{Mv^2}r[/math]), and we know the force of gravity on a 1 kg mass is roughly 9.8 Newtons (kg-m/s^2), we can write the following equation:

 

[math]\frac{\left (\frac{1 kg \cdot v^2}{.4 m} \right )}{9.8 N} \overset.= .4364[/math]

 

Which can be algebraed into

 

[math]v = \sqrt{.4 \cdot 9.8 \cdot .4364} \overset.= 1.308[/math] m/s

 

Once you’ve got the velocity, you should be able to calculate the other things you want. Give it a try.

 

… isnt tension= mass * accelaration?

Yes. Tension is another word for force.

[shadow67]

thanks a lot craig

i understand it much better now. however jus two more questions from this topic if someone could please help.

 

1. given that the radius of the earth at the equator is 6.4 * 10 ^ 6m calculate the centripetal acceleration ( accelaration towards the centre ) of a point on the equator.

 

2. A mass of 5 kg is whirled in a horizontal circle at one end of a string 50 cm long and the othere end being fixed. if the string when hanging vertically will just support a load of 2000N without breaking find the maximum whirling speed in revolutions per second.

 

please help me in these questions since they are a bit different than the one answered previously. thanks for the help.

[Jay-qu]

We have no problem helping you with your physics homework, but this is a little less helping and more like actually doing the work for you. How do you expect to learn to apply these principles if you have not attempted the questions?

 

Have a go, or at least pose how you have tried to tackle the questions and then we can show you where you have gone wrong and help you answer the question.

 

J

[shadow67]

for any question i always try it myself at home and then if i cant answer i post it here. for the earth question only the radius is given and i was confused how to answer the question so i tried forming aright angle triangle like the previous one since earth is a sphere and has equal radius from its centre and substituted and found the accelaration but do not know if it is right. for the second one i did not understand the meaning of the term revolutions per second and what does the vertical string hav to do with spinning?;) please help.

[Jay-qu]

are you familar with the equation:

 

[math]a=\frac{v^2}{r}[/math]

 

r is going to be the radius of the earth and v is the velocity that one would be traveling on the surface of the earth. I will leave you to think about that one (hint: we complete on full revolution in approx. 24 hours)

 

With the second one, it is basically saying that the maximum tension force the string can handle is 2000N, so if it was in circular motion with the mass on the end, what speed would it be traveling to produce this 2000N tension force.

(hint: [math]F=\frac{mv^2}{r}[/math] )

 

hope this helps!

[shadow67]

thanks a lot jay-qu i understand a lot better now. i realise that information doesnt hav to be given everytime u hav to get out basic info like time for one revolution 24 hours.

[Tormod]

thanks a lot jay-qu i understand a lot better now. i realise that information doesnt hav to be given everytime u hav to get out basic info like time for one revolution 24 hours.

 

Just to add to what Jay-qu said - it is often fruitful to show what you have done and how you are trying to solve a problem, rather than simply ask for the solution. Try to explain what you have done and what steps you have taken, etc. That will make others more likely to help out, IMHO, because they don't think they're doing yuor homework but just helping you understand.

 

Glad you got it sorted though! :)