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Isaac Physics Help!!

Hi,
I'm stuck on a conical pendulum Q from Isaac physcis. I got sqrt(g/(l+r)) as my answer but this is wrong. Can anyone help pls??


A mechanical governor consists of a narrow central axle to which are hinged to two light rods of length ℓ, each attached to the centres of spherical masses of radius r.

At what angular velocity ω, in terms of g, (written as l in the editor) and r, will the spheres lose contact with the axle?

Reply 1

Original post by laura.sta
Hi,
I'm stuck on a conical pendulum Q from Isaac physcis. I got sqrt(g/(l+r)) as my answer but this is wrong. Can anyone help pls??
A mechanical governor consists of a narrow central axle to which are hinged to two light rods of length ℓ, each attached to the centres of spherical masses of radius r.
At what angular velocity ω, in terms of g, (written as l in the editor) and r, will the spheres lose contact with the axle?

consider one of the spheres
when it loses contact there is no contact force on the axel so the only two forces are the weight of the sphere and the tension in the rod
draw a free body diagram of the sphere showing these forces
vertically ----- its in equilibrium
horizontally ---- centripetal force = m a
remember sin/cos = tan
enough hints
Original post by laura.sta
Hi,
I'm stuck on a conical pendulum Q from Isaac physcis. I got sqrt(g/(l+r)) as my answer but this is wrong. Can anyone help pls??


A mechanical governor consists of a narrow central axle to which are hinged to two light rods of length ℓ, each attached to the centres of spherical masses of radius r.

At what angular velocity ω, in terms of g, (written as l in the editor) and r, will the spheres lose contact with the axle?

It would be better that you show your working by attaching a pic of your calculation.
It could be your physics or maths or both that contribute to the wrong answer.

Reply 3

Original post by Drummy
consider one of the spheres
when it loses contact there is no contact force on the axel so the only two forces are the weight of the sphere and the tension in the rod
draw a free body diagram of the sphere showing these forces
vertically ----- its in equilibrium
horizontally ---- centripetal force = m a
remember sin/cos = tan
enough hints

image.jpg
hi, thank you for your help!
i’m still stuck as i’m still getting the wrong answer, i’ve attached a photo of the Q and my working if you could give me any extra advice? Thanks :smile:image.jpg
Original post by laura.sta
image.jpg
hi, thank you for your help!
i’m still stuck as i’m still getting the wrong answer, i’ve attached a photo of the Q and my working if you could give me any extra advice? Thanks :smile:image.jpg

First, tan(theta) = sin(theta)/cos(theta)

Next, you need to apply Pythagoras Theorem to find the adjacent side length.

Reply 5

Original post by Eimmanuel
First, tan(theta) = sin(theta)/cos(theta)
Next, you need to apply Pythagoras Theorem to find the adjacent side length.

Hi, sorry i'm still a bit confused
Original post by laura.sta
Hi, sorry i'm still a bit confused

How did get tan θ = g/(rω2) ?

How did you get tan θ = r/(l-r)? Show how you draw the right-angled triangle.

Spoiler

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