6 replies to this topic

[PawnageMonkeyz]

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

[TheSilentShadow]

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.

[hfswjyr]

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.

[愛 | Sieg]

It's all in the equations and understanding

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

[EnErGiE]

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?  

[TheSilentShadow]

You also do calculus?  

Sure  

[PawnageMonkeyz]

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

Thank ya kindly