Calculating the Release Point for Circular Motion

[newcool]

Hi, I received this problem today. If you put a small marble of mass m on top of a large round object that has a radius r, and then release the marble, when will it come off of the side so that the large round object will not be in contact with the marble?

The answer is arccosine of 2/3. I have tried making free body diagrams of the object when the normal force is 0 and got this equation: a * cosin(theta) = g.

I am stuck as to how to get the right answer.

Thanks for any help

 

[Doc Al]

As the marble rolls down the sphere, it will maintain contact as long as there is sufficient force to produce the required centripetal acceleration. Ask yourself: What provides the centripetal force? How does the required centripetal acceleration depend on the the angle that the marble makes with the vertical?

 

[newcool]

I made a free body diagram of when the sphere has a normal force of 0. I got one force,F_c pointing in towards the center and another force mg pointing down. Taking the y value of F_c, i got that:
<br /> F_c*cos(\theta) + mg = ma<br />
<br /> mv^2/r * cos(\theta) + mg = mv^2/r<br />
<br /> v^2/r * cos(\theta) + g = v^2/r<br />

I tried a different approach using the fact that the sum of all energies is equal to 0.
<br /> 0 = \Delta K + \Delta U_g<br />
<br /> 0 = 1/2 *mv^2 + -mgh<br />
<br /> mgh = 1/2 *mv^2<br />
<br /> gh = 1/2 *v^2<br />
<br /> h = r - r*cos(\theta)<br />


so gr(1 - cos(\theta)) = 1/2 * v^2<br />

However, I have gotten nowhere with these equations. Any input on what I did wrong would be appreciated.

Thanks

 

[Pyrrhus]

Ok the answer is acrosinus of 2/3 = 48.2 degrees.

Force analysis

n - mg \cos \theta = -m \frac{v^2}{R}

The object will fall when n = 0 so

v^2 = Rg \cos \theta

when it loses contact with the surface

Assuming isolated system

I can use Conservation of mechanical energy

K + \Omega = K_{0} + \Omega_{0}

\frac{1}{2}mv^2 + mgR \cos \theta = 0 + mgR

Plugging our speed when it loses contact

\frac{1}{2}mRg \cos \theta + mgR \cos \theta = mgR

which gives:

\cos \theta = \frac{2}{3}

\theta = 48.2^{o}

which is the angle when it will lose contact with the surface.

That's of course assuming an isolated system.

 

[Doc Al]

I made a free body diagram of when the sphere has a normal force of 0. I got one force,F_c pointing in towards the center and another force mg pointing down. Taking the y value of F_c, i got that:
<br /> F_c*cos(\theta) + mg = ma<br />
<br /> mv^2/r * cos(\theta) + mg = mv^2/r<br />
<br /> v^2/r * cos(\theta) + g = v^2/r<br />

One big problem here: You seem to be treating the "centripetal force" as though it were a separate force (like friction or weight). Not so! "Centripetal" just means "towards the center": The centripetal force is just those real forces that act towards the center. The only forces acting on the mass m are: (1) mg, acting down, and (2) N, the normal force, acting normal to the surface.

So what's the centripetal force? Just the components of those forces acting towards the center of the circular motion, thus producing the centripetal acceleration:
mg cos\theta - N = F_c = mv^2/r
Of course, you'll set the normal force to zero, so:
mg cos\theta = mv^2/r

I tried a different approach using the fact that the sum of all energies is equal to 0.
<br /> 0 = \Delta K + \Delta U_g<br />
<br /> 0 = 1/2 *mv^2 + -mgh<br />
<br /> mgh = 1/2 *mv^2<br />

<br /> gh = 1/2 *v^2<br />
<br /> h = r - r*cos(\theta)<br />


so gr(1 - cos(\theta)) = 1/2 * v^2<br />

Exactly correct. But that's not a "different approach"--it's a necessary part of solving this problem!

However, I have gotten nowhere with these equations. Any input on what I did wrong would be appreciated.

Now just combine the two equations and solve for \theta.

 

[Pyrrhus]

Too bad, i just did the work

Well, also read Doc Al's explanation, it's quite good

 

[newcool]

Thanks for the help everyone. Just one question, Doc, how did you get that
mg cos\theta - N = F_c = mv^2/r

Isn't mg pointing straight down so the component that points towards the center is
mg/cos\theta

 

[Pyrrhus]

We got a vector

\vec{R}

with y component

R \cos \theta

and x component

R \sin \theta

where do you get \frac{mg}{\cos \theta} ?

 

[Pyrrhus]

I think you got a misconception with centripetal force, Centripetal force is a role asigned to forces, because they act towards the center. The forces acting on the body are normal and the weight, and the components acting towards the center are equal to m \frac{v^2}{r}.

 

[Doc Al]

Thanks for the help everyone. Just one question, Doc, how did you get that
mg cos\theta - N = F_c = mv^2/r

Isn't mg pointing straight down so the component that points towards the center is
mg/cos\theta

Yes, mg points straight down. Therefore its component towards the center will be mg cos\theta, not mg/cos\theta. (Draw yourself a picture.)

I believe you are thinking like this: That mg is the vertical component of the "centripetal force" (F_c), so F_c cos\theta = mg ==> F_c = mg/cos\theta. This is incorrect thinking.

One thing you must realize is that "centripetal" is just a decription of the direction that a force has: it just means "towards the center". (Another term used is "radial".) It is not a kind of force. It's just like describing a force as a horizontal force.

Think like this: The mass m must have an acceleration towards the center since it moves in a circle. So, let's apply Newton's 2nd Law in that radial (or centripetal) direction. As Cyclovenom and I have explained, the only forces acting on the mass are gravity and the normal force. We know that N points away from the center. What's the component of gravity (mg) towards the center? mg cos\theta. So:
F_{towards-center} = ma_{towards-center}
mg cos\theta - N = mv^2/r

Make sense?

 

[newcool]

Thanks for all the help, Doc and Cyclove, I understand it now