Finding b for a Ball with Non-Uniform Density on a Circular Loop

In summary: In fact, it's actually quite common for people to not know how to solve for v in this case. The normal force is always directed perpendicular to the direction of motion. So in this case, the normal force would be directed towards the center of the circle (where the ball is located).
  • #1
bodensee9
178
0
[/B]

Homework Statement


Hello:

Can someone help with the following?

A ball of mass M and radius R rolls smoothly from rest down a ramp and onto a circular loop of radius 0.48m. The initial height of the ball is h = 0.36m. At the loop bottom, the magnitude of the normal force on the ball is 2.00 Mg. The ball consists of an outer spherical shell (of certain uniform density) that is glued to a central sphere(of different uniform density). The rotational inertia of the ball can be expressed in terms of the general form bMR^2, but b is not 0.4 as it is for a ball of uniform density. Find the b for this ball.

I thought to use conservation of energy? So this gives me:

M*g*0.36 = 1/2*I*w^2 + 1/2*M*v^2, where I is the rotational inertia and w is the angular velocity and v is the linear velocity.

But then I'm not sure what to do after like (for example, how do I get rid of the v's)? And is it true that the ball is not accelerating at the loop bottom with respect to the ground (since it is at the bottom) so that the normal force = force from gravity? But then what good is that since it only gives me M?

Thanks!
 
Physics news on Phys.org
  • #2
bodensee9 said:
[/B]

Homework Statement


Hello:

Can someone help with the following?

A ball of mass M and radius R rolls smoothly from rest down a ramp and onto a circular loop of radius 0.48m. The initial height of the ball is h = 0.36m. At the loop bottom, the magnitude of the normal force on the ball is 2.00 Mg. The ball consists of an outer spherical shell (of certain uniform density) that is glued to a central sphere(of different uniform density). The rotational inertia of the ball can be expressed in terms of the general form bMR^2, but b is not 0.4 as it is for a ball of uniform density. Find the b for this ball.

I thought to use conservation of energy? So this gives me:

M*g*0.36 = 1/2*I*w^2 + 1/2*M*v^2, where I is the rotational inertia and w is the angular velocity and v is the linear velocity.

But then I'm not sure what to do after like (for example, how do I get rid of the v's)?
Looks good :approve:. Now if we assume that the ball rolls without slipping, can you relate the angular velocity to the linear velocity?
bodensee9 said:
And is it true that the ball is not accelerating at the loop bottom with respect to the ground (since it is at the bottom) so that the normal force = force from gravity? But then what good is that since it only gives me M?
In fact it doesn't even give M, the M's cancel. If the ball weren't accelerating, then the normal reaction force would simply be Mg would it not?

HINT: The ball is undergoing circular motion so you should be able to relate the force acting on the ball to the linear velocity.
 
Last edited:
  • #3
Your equation,
[tex]0.36Mg=\frac{1}{2}I\omega^2+\frac{1}{2}Mv^2[/tex]
is correct.
Just replace I with its general form bMv^2, eliminate M obviously and then recognize that the linear velocity is [itex]v=r\omega[/itex]. Imagine the angular velocity is 2pi/s (one rotation per second) then you get [itex]v=2\pi r[/itex] which is what you expect if it linearly translates the distance of the circumference in one second.
 
  • #4
Hello,

Thanks. So I know that v = r*w (0.48*w) in this case. But I don't know how to find either w or v? If the ball is not slipping, then there is a frictional force that opposes the downward motion of the ball ... but I don't see where the normal force comes into use?
 
  • #5
bodensee9 said:
Hello,

Thanks. So I know that v = r*w (0.48*w) in this case. But I don't know how to find either w or v? ... but I don't see where the normal force comes into use?
Correct. However, you are not looking to solve for w or v you are looking to eliminate them in order to find b.

As I said previously, notice that the ball is moving in a circle. What is the requirement of the net force acting on any object undergoing circular motion?
 
  • #6
Um ... okay, there is a centripetal force, so then I know that the normal force, 2 = m*v^2/R (where R = 0.48). But then if I try to represent v using that equation I'm left with m's which I don't know the value of. Sorry.
 
  • #7
bodensee9 said:
2 = m*v^2/R (where R = 0.48)
Are you sure that that equation is correct?
bodensee9 said:
Sorry.
There's no need to apologise, not knowing how to answer a question straight away isn't anything to be ashamed of.
 

1. What is rolling motion?

Rolling motion is a type of motion where an object moves in a circular path while also rotating around its own axis.

2. How is rolling motion different from sliding motion?

Rolling motion is different from sliding motion because in rolling motion, the point of contact between the object and the surface it is rolling on is stationary, while in sliding motion, the point of contact is constantly changing.

3. What is the relationship between rolling motion and friction?

Friction plays a crucial role in rolling motion as it is the force that allows the object to roll without slipping. The direction of friction is opposite to the direction of motion, which helps to maintain the rolling motion.

4. How does the shape of an object affect its rolling motion?

The shape of an object can affect its rolling motion in several ways. Objects with a larger radius will roll more smoothly and efficiently, while objects with an irregular shape may experience more friction and have a harder time rolling. Additionally, objects with a lower center of mass will be more stable and easier to roll compared to objects with a higher center of mass.

5. How can rolling motion be used in real life applications?

Rolling motion has many practical applications in our daily lives. Some examples include the use of wheels on vehicles, gears in machinery, and rolling objects in sports such as bowling and baseball. It is also used in manufacturing processes, such as rolling metal sheets into different shapes.

Similar threads

Replies
10
Views
313
  • Introductory Physics Homework Help
Replies
32
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
2
Replies
35
Views
2K
  • Introductory Physics Homework Help
3
Replies
97
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
636
  • Introductory Physics Homework Help
Replies
2
Views
818
  • Introductory Physics Homework Help
Replies
1
Views
717
Back
Top