# Thread: Physics Problem: stationary bead relative to a revolving wire

1. ## Physics Problem: stationary bead relative to a revolving wire

I have been working on this problem for hours. I have been working on it on Google Docs with another user on this site for a while: https://docs.google.com/drawings/d/1...vre0Z6feA/edit

The problem: A small bead with a mass of 100g slides without friction along a semicircular wire with a radius of 10cm that is rotating about a vertical axis at a rate of 2.0 revolutions per second. Find the value of angle for which the bead will remain stationary relative to the rotating wire.

The answer: 52 degrees

Method: 1.588cos(theta) - the weight of the bead = 0 solve for theta

I'm not sure why that method works (gets the answer) and that's my biggest problem. I want to understand how this problem works.

This problem is difficult to understand without looking at the picture in the textbook. I drew the picture on the Google Docs link you may view. Tell me if you need any more information.

2. ## Re: Physics Problem: stationary bead relative to a revolving wire

Originally Posted by over9000
The problem: A small bead with a mass of 100g slides without friction along a semicircular wire with a radius of 10cm that is rotating about a vertical axis at a rate of 2.0 revolutions per second. Find the value of angle for which the bead will remain stationary relative to the rotating wire.

The answer: 52 degrees
I will use $\displaystyle \theta$ as the angle between the normal force of the wire and the horizontal as shown in the diagram ... it will be complementary to your angle.

$\displaystyle r = R\cos{\theta}$ , where $\displaystyle R$ is the radius of the semicircle and $\displaystyle r$ is the radius of the bead's rotation.

$\displaystyle N_y = N\sin{\theta} = mg$

$\displaystyle N = \frac{mg}{\sin{\theta}}$

$\displaystyle N_x = F_c$

$\displaystyle N\cos{\theta} = mr \omega^2$

$\displaystyle \frac{mg}{\sin{\theta}} \cdot \cos{\theta} = m R\cos{\theta} \cdot \omega^2$

$\displaystyle \frac{g}{\sin{\theta}} = R\omega^2$

$\displaystyle \frac{g}{R \omega^2} = \sin{\theta}$

$\displaystyle \theta = \arcsin\left(\frac{g}{R \omega^2}\right) \approx 38^\circ$

3. ## Re: Physics Problem: stationary bead relative to a revolving wire

what are does w stand for?

4. ## Re: Physics Problem: stationary bead relative to a revolving wire

someone else explained this to me: It's the equation that omega is squared.
angular acceleration (a_c) is velocity-squared divided by R, which is equal to omega-squared times R.
He put the radius at the point of the bead in terms of R, the max radius of the circle with r = R*cos(theta)
Ny = N*sin(theta) = mg, because the normal force on the wire is equal to the only counter force in that direction: weight.
Thus, solving for N, we have N = mg/sin(theta)
Nx = Fc = m*a_c = m*r*(omega-squared) = m*R*cos(theta)*(omega-squared)

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# a small block with a mass of 150gm slides without friction along a semicircular wire with a radius of 8cm that rotates about a vertical axis at a rate of 2 revolution per second.find the angle for which the block will remain stationary with respect to the

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