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[Physics] Circular motion problem


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#1 PawnageMonkeyz

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Posted 06 February 2010 - 11:54 PM

Syllabus used is South Aus Year 12.

Question:
Small ball tied to 24 cm string and suspended from a fixed point P to make a conical pendulum. Ball describes a horizontal circle about a center vertically under point P. The string makes an angle of 15 degrees with the vertical. Find the speed of the ball.

TLDR version: How do you find speed in a conical pendulum when only length of the string and angle is given?

Answer: 0.4 ms^-1

I'm stuck findin the equations :\ any help to this would be appreciated smile.gif

#2 TheSilentShadow

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Posted 07 February 2010 - 10:05 PM

IPB Image

I've used x instead of theta because its easier to represent

Tsin x=(mv^2)/r   - horizontal component of force
Tcos x= mg - vertical component of force

get T/m on one side and equate both equations to each other

g/cos x = v^2/rsin x

Now, the velocity of the bob is (2*pi*r)/t

substitute this for v in the previous equation and get

g/cos x= (((2*pi)^2)*r)/((t^2)*sin x)

Solve this equation for t and you get

t=2*pi*(r/gtan x)^1/2

substitute r=Lsin x (using trigonometry)

t=2*pi*(Lcos x/g)*1/2

Substitute L=24*10^-2, x=15, g=9.8 and you will get t=0.966

Put this value into v=(2*pi*Lsin x)/t and the answer comes out to 0.404 m/s

Edited by TheSilentShadow, 07 February 2010 - 10:06 PM.


#3 hfswjyr

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Posted 08 February 2010 - 12:25 AM

I have totally forgotten all of this since coming out of high school, even though I am doing engineering. In fact, I feel dumber than 3 years ago.

Well done for remembering it all TSS! Good revision for me.

#4 愛 | Sieg

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Posted 10 February 2010 - 03:07 PM

It's all in the equations and understanding tongue.gif

I took the easy way out and went for a private school course, got 93 and didn't learn shit.

#5 EnErGiE

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Posted 12 February 2010 - 04:49 PM

QUOTE(TheSilentShadow @ Feb 8 2010, 12:05 AM) View Post

http://wpcontent.answers.com/wikipedia/com...endulum.svg.png

I've used x instead of theta because its easier to represent

Tsin x=(mv^2)/r   - horizontal component of force
Tcos x= mg - vertical component of force

get T/m on one side and equate both equations to each other

g/cos x = v^2/rsin x

Now, the velocity of the bob is (2*pi*r)/t

substitute this for v in the previous equation and get

g/cos x= (((2*pi)^2)*r)/((t^2)*sin x)

Solve this equation for t and you get

t=2*pi*(r/gtan x)^1/2

substitute r=Lsin x (using trigonometry)

t=2*pi*(Lcos x/g)*1/2

Substitute L=24*10^-2, x=15, g=9.8 and you will get t=0.966

Put this value into v=(2*pi*Lsin x)/t and the answer comes out to 0.404 m/s


You also do calculus?  wub.gif

#6 TheSilentShadow

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Posted 13 February 2010 - 10:43 PM

QUOTE(EnErGiE @ Feb 13 2010, 06:19 AM) View Post

You also do calculus?  wub.gif


Sure  smile.gif

#7 PawnageMonkeyz

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Posted 23 February 2010 - 06:11 AM

Whoa, I guess I do owe you thanks there, though it's ages ago.

Thank ya kindly smile.gif




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