Mastering Circular Motion: Tips for Solving Conservation of Energy Problems

[steejk]

Not really sure where to start with this. I know it has something to do with conservation of energy but not really sure how to go about it.

 

[ehild]

Given that the bead moves along a circle. What is the resultant force acting on it? Draw the free-body diagram.

ehild

 

[ideasrule]

Start by drawing a free-body diagram on the bead, labeling all forces, and writing out Newton's second law for the radial direction. It should include "v" somewhere, which you can find using the conservation of energy.

EDIT: oops, edit conflict with ehild

 

[tiny-tim]

hi steejk!

start by writing the conservation of energy equation to find v as a function of θ …

what do you get? ​

 

[steejk]

hi steejk!

start by writing the conservation of energy equation to find v as a function of θ …

what do you get? ​

Ok so at bottom TE = KE = 0.5mu

And at point B TE = 0.5mv² + mgh = 0.5mu²

How can I work out h?

 

[ideasrule]

Ok so at bottom TE = KE = 0.5mu^2

And at point B TE = 0.5mu^2 - mgh

Use different symbols for the two velocities; the "u" in your second equation is different from the "u" in your first equation.

How can I work out h?

You can relate it to r and theta using some trigonometry.

 

[steejk]

Use different symbols for the two velocities; the "u" in your second equation is different from the "u" in your first equation.

Sorry

You can relate it to r and theta using some trigonometry.

Hmm. I know its probably something obvious but I'm not getting it.

What would acosθ equal?

 

[tiny-tim]

What would acosθ equal?

cos = adj/hyp

sin = opp/hyp

tan= opp/adj

 

[ehild]

Hmm. I know its probably something obvious but I'm not getting it.

What would acosθ equal?

Believe me: your life would be much easier with a drawing. From it, you would see at once how the initial speed is related with the one at angle theta, and how to get it. And you could find out the normal force from a free body diagram at the position labelled with theta.

ehild

 

[steejk]

Believe me: your life would be much easier with a drawing. From it, you would see at once how the initial speed is related with the one at angle theta, and how to get it. And you could find out the normal force from a free body diagram at the position labelled with theta.

ehild

Okay so from my diagram does N = (mu^2)/a + mgcosθ?

 

[steejk]

cos = adj/hyp

sin = opp/hyp

tan= opp/adj

But where is the right angle triangle?

 

[tiny-tim]

ah, sometimes you have to draw the right-angled triangle yourself!

draw a horizontal line from B, meeting OA at D …

then OBD is a right-angled triangle

the angle at O is θ

hyp (hypotenuse) is the long side, OB, = a

adj (adjacent) is the side next to θ, OD (the height of O above B)

so cosθ = adj/hyp = OD/a
​