What is the angular speed of the cylinder when hit by a clay ball?

In summary, the problem involves a wad of sticky clay fired at a solid cylinder that is initially at rest and rotates on an axis through the center of mass. The line of motion of the projectile is perpendicular to the axis of rotation and at a distance d < R from the center. The solution utilizes the conservation of angular momentum equation and takes into consideration the change in initial velocity vector due to the clay sticking to the cylinder. The resulting formula for the angular speed of the cylinder just as the clay hits the surface is \omega = \frac{2 d m v_i}{(2 m + M) r^2}.
  • #1
lizzyb
168
0

Homework Statement



A wad of sticky clay (mass m, velocity v), is fired at a solid cylinder (mass M, radius R). The cylinder is initially at rest and rotates on an axis through the center of mass. The line of motion of the projectile is perpendicular to the axis of rotation and at a distance d < R from the center.

Find the angular speed of the cylinder just as the clay hits the surface.

Homework Equations



Conservation of Angular Momentum
See: https://www.physicsforums.com/showthread.php?t=63404

The Attempt at a Solution



I did:

[tex]L_i = d m v_i[/tex]
[tex]L_f = d m v_f + \frac{1}{2} M r^2 \omega = d m r \omega + \frac{1}{2} M r^2[/tex]
[tex]L_i = L_f [/tex]
[tex]d m v_i = d m r \omega + \frac{1}{2} M r^2 \omega[/tex]
[tex] \omega = \frac{d m v_i}{(d m r + \frac{M r^2}{2})}[/tex]

Which isn't correct; it may have something to do with the momentum of the wad of sticky clay; the reason I used L = m v d is because d would be the moment arm but perhaps I'm confusing this with torque? What value of r should I use when the force isn't perpindicular?
 
Physics news on Phys.org
  • #2
lizzyb said:

Homework Statement



A wad of sticky clay (mass m, velocity v), is fired at a solid cylinder (mass M, radius R). The cylinder is initially at rest and rotates on an axis through the center of mass. The line of motion of the projectile is perpendicular to the axis of rotation and at a distance d < R from the center.

Find the angular speed of the cylinder just as the clay hits the surface.

Homework Equations



Conservation of Angular Momentum
See: https://www.physicsforums.com/showthread.php?t=63404

The Attempt at a Solution



I did:

[tex]L_i = d m v_i[/tex]
[tex]L_f = d m v_f + \frac{1}{2} M r^2 \omega = d m r \omega + \frac{1}{2} M r^2 \omega[/tex]
[tex]L_i = L_f [/tex]
[tex]d m v_i = d m r \omega + \frac{1}{2} M r^2 \omega[/tex]
[tex] \omega = \frac{d m v_i}{(d m r + \frac{M r^2}{2})}[/tex]

Which isn't correct; it may have something to do with the momentum of the wad of sticky clay; the reason I used L = m v d is because d would be the moment arm but perhaps I'm confusing this with torque? What value of r should I use when the force isn't perpindicular?

You lost an ω in your second equation, so I added it in the quote. Why do you think it is wrong? Could it be that changing r to R is causing a problem?
 
Last edited:
  • #3
The solution in the back of the book is:
[tex]\omega = \frac{2 m v_i d}{(M + 2m) R^2}[/tex]
so using what I did:
[tex]\omega = \frac{d m v_i}{(dmr + \frac{M r^2}{2})} = \frac{2 m v_i d}{ 2 d m r + M r^2 }[/tex]
which is close but not it.

I don't think changing R to r makes a difference (let r = R).

But there is a relationship between R and d, viz. [tex]\cos \theta = \frac{d}{r}[/tex] but while that will take care of the r^2 it adds the cos.
 
Last edited:
  • #4
lizzyb said:
The solution in the back of the book is:
[tex]\omega = \frac{2 m v_i d}{(M + 2m) R^2}[/tex]
so using what I did:
[tex]\omega = \frac{d m v_i}{(dmr + \frac{M r^2}{2})} = \frac{2 m v_i d}{ 2 d m r + M r^2 }[/tex]
which is close but not it.

OOPs.. I missed it. Sorry. They are correct. In your second equation on the right hand side, the distance between the line of the velocity and the axis for the clay is R, not d.

I better go get some coffee.:yuck:
 
Last edited:
  • #5
How do you figure?

[tex]L_f = d m v_f + \frac{1}{2} M r^2 \omega = d m r \omega + \frac{1}{2} M r^2 \omega[/tex]

The clay is coming in on the top part of the cylinder yet below the cylinder's radius, at a height d, which would be the distance between the velocity vector and the cylinder's axis.

oh wait, you mean because this is after the clay has stuck to the cylinder that the initial velocity vector has changed? In which case the distance between the line of velocity is r and not d as before?

[tex]d m v_i = m r^2 \omega + \frac{1}{2} M r^2 \omega[/tex]
[tex] \omega r^2(m + \frac{1}{2} M) = \omega \frac{r^2}{2}(2m + M) = d m v_i[/tex]
[tex] \omega = \frac{2 d m v_i}{(2 m + M) r^2}[/tex]
 
Last edited:
  • #6
lizzyb said:
How do you figure?

[tex]L_f = d m v_f + \frac{1}{2} M r^2 \omega = d m r \omega + \frac{1}{2} M r^2 \omega[/tex]

The clay is coming in on the top part of the cylinder yet below the cylinder's radius, at a height d, which would be the distance between the velocity vector and the cylinder's axis.

oh wait, you mean because this is after the clay has stuck to the cylinder that the initial velocity vector has changed? In which case the distance between the line of velocity is r and not d as before?

[tex]d m v_i = m r^2 \omega + \frac{1}{2} M r^2 \omega[/tex]
[tex] \omega (r^2(m + \frac{1}{2} M) = \omega (\frac{r^2}{2}(2m + M) = d m v_i[/tex]
[tex] \omega = \frac{2 d m v_i}{(2 m + M) r^2}[/tex]

That's it. The clay cannot penetrate the cylinder, so it has to abrubtly change direction at impact to move tangent to the cylinder.
 
  • #7
ok, that makes sense! thanks (again)!
 

1. What is Angular Momentum?

Angular momentum is a measure of an object's rotational motion. It is defined as the product of the moment of inertia and the angular velocity of an object.

2. How is Angular Momentum of a clay ball calculated?

The angular momentum of a clay ball can be calculated by multiplying its moment of inertia (determined by its mass and distribution of weight) by its angular velocity (determined by its rotational speed).

3. Why is Angular Momentum important to understand in relation to clay balls?

Angular momentum is important to understand when studying clay balls because it determines how they behave when rotating or spinning. This can have implications for their stability, motion, and interactions with other objects.

4. Can the Angular Momentum of a clay ball change?

Yes, the angular momentum of a clay ball can change if there is an external torque acting on it. This could be due to a change in its moment of inertia or angular velocity.

5. How does the shape of a clay ball affect its Angular Momentum?

The shape of a clay ball can affect its angular momentum by changing its moment of inertia. A more compact shape will have a smaller moment of inertia, resulting in a higher angular velocity and therefore a higher angular momentum.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
636
  • Introductory Physics Homework Help
10
Replies
335
Views
7K
  • Introductory Physics Homework Help
Replies
10
Views
803
Replies
13
Views
791
  • Introductory Physics Homework Help
Replies
17
Views
278
  • Introductory Physics Homework Help
Replies
12
Views
808
Replies
39
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
608
Back
Top