Finding Angle of Sphere Falling From Table Edge

In summary: As I said earlier, you assumed you could treat the ball as a point mass, and that was not justified.Have a look here:http://en.wikipedia.org/wiki/Rotating_reference_frame#Newton.27s_second_law_in_the_two_framesThe sum of the centrifugal force, Euler force, normal force, weight and friction must be zero (the Coriolis force is zero because the ball is stationary).In summary, a solid spherical ball placed on the edge of a table with a coefficient of static friction of 0.5 is given a slight push and begins to fall off. The question is to find the angle (in degrees with vertical) turned by the ball before it slips. Equations are set up using conservation
  • #1
Satvik Pandey
591
12

Homework Statement


A solid spherical ball is placed carefully on the edge of a table in the position shown in the figure. The coefficient of static friction between the ball and the edge of the table is 0.5 . It is then given a very slight push. It begins to fall off the table.

Find the angle (in degrees)(with vertical) turned by the ball before it slips.
2e19e72524.fbc259d1f7.UhpXQY.png

Homework Equations

3. The attempt at question

I have came up with some equations. Let ##\theta## be the angle(with vertical) at which the sphere begins to slip.

p3.png

By conservation of energy
##mgr-mgrcos\theta =\frac { 1 }{ 2 } { I }_{ 0 }{ \omega }^{ 2 }##

As ##v=r \omega##

So ##g(1-cos\theta )=\frac { 7 }{ 10 } \frac { { v }^{ 2 } }{ r } ##

By finding torque about the contact point

##mgsin\theta r=\frac { 7 }{ 10 } m{ r }^{ 2 }\alpha ##

As ##a=r \alpha##

So ##\frac { 5gsin\theta }{ 7 } =a##

Also from FBD of the block

##mgcos\theta -N=m\frac { { v }^{ 2 } }{ r } ##

and ##\\ mgsin\theta -\mu N=ma##

I don't know if these equations are right. Please help.
 
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  • #2
Satvik Pandey said:
##mgcos\theta -N=m\frac { { v }^{ 2 } }{ r } ##
I doubt that's right for the centripetal force on a sphere.
 
  • #3
haruspex said:
I doubt that's right for the centripetal force on a sphere.

Could please explain, why? How to proceed?
 
  • #4
Can this be solved by just considering the forces at the point of contact tangent to the sphere? That would give:
sin theta = 1/2 cos theta at the time of slipping.
 
  • #5
Jilang said:
Can this be solved by just considering the forces at the point of contact tangent to the sphere? That would give:
sin theta = 1/2 cos theta at the time of slipping.

Could you please show how did you find that.
 
  • #6
The frictional force at the point of contact would be mg cos theta x 1/2 at a tangent to the sphere and the gravitational force would be mg sine theta.
 
  • #7
Jilang said:
Can this be solved by just considering the forces at the point of contact tangent to the sphere?

If you can find the normal force at the contact, then it is simple. Your result follows from equating the normal force with the radial component of the weight, but how do you justify this?
 
  • #8
voko said:
If you can find the normal force at the contact, then it is simple. Your result follows from equating the normal force with the radial component of the weight, but how do you justify this?

voko ,what do you think about my equations?
 
  • #9
Satvik Pandey said:
voko ,what do you think about my equations?

I agree with haruspex.
 
  • #10
Jilang said:
Can this be solved by just considering the forces at the point of contact tangent to the sphere? That would give:
sin theta = 1/2 cos theta at the time of slipping.
Not in the way you write, as the ball is not moving in a straight line.

See haruspex.
 
  • #11
voko said:
I agree with haruspex.

Could you please explain why that is not correct?:p
 
  • #12
Satvik Pandey said:
Could you please explain why that is not correct?

Because that assumes that the entire mass of the ball is at its center.
 
  • #13
voko said:
If you can find the normal force at the contact, then it is simple. Your result follows from equating the normal force with the radial component of the weight, but how do you justify this?
Consider the element of the sphere just touching the corner, it is stationary.
 
  • #14
Then how should I proceed?
 
  • #15
Jilang said:
Consider the element of the sphere just touching the corner, it is stationary.

But the rest of the ball is not. How do you justify the insignificance of that?
 
  • #16
voko said:
But the rest of the ball is not. How do you justify the insignificance of that?
The sphere is rigid, you need to assume that the force on every single atom is equal.
 
  • #17
Jilang said:
The sphere is rigid, you need to assume at the force on every single atom is equal.

Why?
 
  • #18
Do you think that the force of gravity only acts on the centre of mass?
 
  • #19
Satvik Pandey said:
Then how should I proceed?

Personally, I would use a co-rotating frame. In that frame, the ball is in equilibrium. The sum of real and fictitious forces is zero.
 
  • #20
voko said:
Personally, I would use a co-rotating frame. In that frame, the ball is in equilibrium. The sum of real and fictitious forces is zero.
Does that give a different answer?
 
  • #21
Jilang said:
Do you think that the force of gravity only acts on the centre of mass?

If that is a question to me, the answer is no.

Regardless, I asked you to justify your statements. Please do so.
 
  • #22
voko said:
Personally, I would use a co-rotating frame. In that frame, the ball is in equilibrium. The sum of real and fictitious forces is zero.

What is 'co-rotating frame'?
Are my other equations correct?
 
  • #23
haruspex said:
No, it's ok - just didn't seem right, and I didn't have time to check it before.

voko has just said that it's not correct. What should I do?:(
 
  • #24
Satvik Pandey said:
Could please explain, why? How to proceed?

Calculate it from first principles by integration.
 
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  • #25
Satvik Pandey said:
voko has just said that it's not correct. What should I do?:(
No, I mean I hadn't checked ... I have now and it is wrong.
 
  • #26
haruspex said:
Consider a uniform bar with its centre as the point of contact. It would require a centripetal force to get around the corner even though its mass centre does not.
Calculate it from first principles by integration.
I don't understand what you want me to calculate. Please explain more.
 
  • #27
Hm, interesting. Unless I made a mistake in integration, total centrifugal force is ## m \omega^2 r ##, that of a point mass, which means Satvik's disputed normal force equation was correct :)
 
  • #28
voko said:
Hm, interesting. Unless I made a mistake in integration, total centrifugal force is ## m \omega^2 r ##, that of a point mass, which means Satvik's disputed normal force equation was correct :)

But was the way in which I found it incorrect?
 
  • #29
Satvik Pandey said:
But was the way in which I found it incorrect?

As I said earlier, you assumed you could treat the ball as a point mass, and that was not justified.

Have a look here:

http://en.wikipedia.org/wiki/Rotating_reference_frame#Newton.27s_second_law_in_the_two_frames

The sum of the centrifugal force, Euler force, normal force, weight and friction must be zero (the Coriolis force is zero because the ball is stationary with respect to itself). The centrifugal and the Euler force must be integrated to obtain the force balance.
 
  • #30
voko said:
Hm, interesting. Unless I made a mistake in integration, total centrifugal force is ## m \omega^2 r ##, that of a point mass, which means Satvik's disputed normal force equation was correct :)
I managed to convince myself it could not be right, but now I see a flaw. I'll trust your integration.
 
  • #31
Can this be solved without using rotating frame of reference because I haven't became familiar with it.
 
  • #32
Yes. And there is no large difference between a fixed and a rotating frame, the equations look similar.
Which net force is needed to keep the ball on its circular track?
 
  • #33
Yes, this can be solved in an inertial frame, but it is slightly more difficult in my opinion. Compose the equations of motion for each "small particle" within the ball. Every such particle moves in a circle with known velocity and acceleration, so the net force acting on every particle is also known. The integral of the net force over the entire ball must be equal to the sum of weight, normal force and friction.
 
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  • #34
mfb said:
Yes. And there is no large difference between a fixed and a rotating frame, the equations look similar.
Which net force is needed to keep the ball on its circular track?

Centripetal force.
 
  • #35
Right. Now you'll need its value.
 
<h2>1. How do you calculate the angle of a sphere falling from a table edge?</h2><p>To calculate the angle of a sphere falling from a table edge, you will need to know the height of the table, the distance from the edge of the table to where the sphere is released, and the time it takes for the sphere to fall. You can then use the formula <strong>θ = arctan(h/d)</strong> to find the angle, where h is the height and d is the distance.</p><h2>2. Why is it important to find the angle of a sphere falling from a table edge?</h2><p>Knowing the angle of a sphere falling from a table edge can help us understand the trajectory of the sphere and predict where it will land. This information can be useful in various fields such as physics, engineering, and sports.</p><h2>3. What factors can affect the angle of a sphere falling from a table edge?</h2><p>The angle of a sphere falling from a table edge can be affected by the height of the table, the distance from the edge of the table, the shape and weight of the sphere, air resistance, and any external forces acting on the sphere.</p><h2>4. How can you measure the angle of a sphere falling from a table edge?</h2><p>The angle of a sphere falling from a table edge can be measured using a protractor or a measuring tool with a built-in angle measurement feature. You can also use a high-speed camera to capture the trajectory of the sphere and analyze the footage to determine the angle.</p><h2>5. Are there any real-life applications of finding the angle of a sphere falling from a table edge?</h2><p>Yes, there are many real-life applications of finding the angle of a sphere falling from a table edge. For example, in sports such as basketball or golf, understanding the trajectory of a ball can help players make more accurate shots. In engineering, this information can be used to design structures that can withstand falling objects. In physics, it can be used to study the laws of motion and gravity.</p>

1. How do you calculate the angle of a sphere falling from a table edge?

To calculate the angle of a sphere falling from a table edge, you will need to know the height of the table, the distance from the edge of the table to where the sphere is released, and the time it takes for the sphere to fall. You can then use the formula θ = arctan(h/d) to find the angle, where h is the height and d is the distance.

2. Why is it important to find the angle of a sphere falling from a table edge?

Knowing the angle of a sphere falling from a table edge can help us understand the trajectory of the sphere and predict where it will land. This information can be useful in various fields such as physics, engineering, and sports.

3. What factors can affect the angle of a sphere falling from a table edge?

The angle of a sphere falling from a table edge can be affected by the height of the table, the distance from the edge of the table, the shape and weight of the sphere, air resistance, and any external forces acting on the sphere.

4. How can you measure the angle of a sphere falling from a table edge?

The angle of a sphere falling from a table edge can be measured using a protractor or a measuring tool with a built-in angle measurement feature. You can also use a high-speed camera to capture the trajectory of the sphere and analyze the footage to determine the angle.

5. Are there any real-life applications of finding the angle of a sphere falling from a table edge?

Yes, there are many real-life applications of finding the angle of a sphere falling from a table edge. For example, in sports such as basketball or golf, understanding the trajectory of a ball can help players make more accurate shots. In engineering, this information can be used to design structures that can withstand falling objects. In physics, it can be used to study the laws of motion and gravity.

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