Rotational Kinetic Energy Problems

In summary: So the U(final) + K(final) = U (initial) + K(initial) equation only applies to the disk... so the kinetic energy of the particle is only related to the disk's kinetic energy because it is attached to it?Yes, that's correct. The particle's kinetic energy is included in the disk's kinetic energy because it is attached to the disk and therefore moves with it. In summary, In the first problem, a uniform disk with a particle attached is given a gentle nudge and begins to rotate freely about a horizontal axis. The angular velocity of the disk when the particle is at its lowest point can be found by considering the conservation of energy and using the rotational inertia of the disk. The force required to
  • #1
yus310
81
1
Hello. I'm an intro physics student, who needs help with physics problem, because my graduate student refuses to help anybody in my class anyway... here is my first question on rotational kinetic energy

1) "A uniform disk of mass M and radius R can rotate freely about a horizontal axis through its center and perpendicular to the plane of the disk. A small particle of mass m is attached to the rim of the disk at the top, directly above the pivot. The system is given a gentle nudge, and the disk begins to rotate."

a) "What is the angular velocity of the disk when the particle is at its lowest point?"
b)"At this point, what force must be exerted by the disk on the particle to keep it stuck to the disk"

... First of all where should I start with part a)? And how does rotational kinetic energy apply to this problem. Second, how do I find the force for part b)

...

My second question is...

2) "A uniform ring 1.5 m in diameter is pivoted at one point on its perimeter so that it is free to rotate about a horizontal axis. Intially, the line joining the support point and the center is horizontal."

a) "If the ring is released from rest, what is its maximum angular velocity?"
b) "What minimum initial angular velocity must it be given if it is to rotate a full 360 degrees?"

... Once again any tips would be helpful... I appreciate it. Thanks.
 
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  • #2
My hint for both problems: What's conserved?
 
  • #3
Rotational Kinetic Energy is conserved... but how does the small particle apply in the first problem and how can I formulate into an equation to solve for the angular velocity?

Also how does one for problem #2, when the maximum angular velocity occurs?
 
  • #4
yus310 said:
Rotational Kinetic Energy is conserved...
No, total mechanic energy is conserved. What other form of energy is relevant here?

but how does the small particle apply in the first problem and how can I formulate into an equation to solve for the angular velocity?

Also how does one for problem #2, when the maximum angular velocity occurs?
Both involve conservation of energy. To express the rotational KE, you'll need to figure out the rotational inertia of both systems about the required axis.
 
  • #5
Doc Al said:
No, total mechanic energy is conserved. What other form of energy is relevant here?


Both involve conservation of energy. To express the rotational KE, you'll need to figure out the rotational inertia of both systems about the required axis.

... So it
U(final) + K(final) = U (initial) + K(initial)

... So, in the first problem, do you treat U(initial) as zero for both the disk and the particle, and the U(final) for the particle as mg(-2*pi*R)? Is this correct? And for the first problem, does the equation for solving for angular velocity like this...

mg(-2*pi*R)+((MR^2*angular velocity^2)/2)+((m*v^2)/2)=0

... Is this correct? for the second part, of this problem, is it referring to torque... because I'm a little confused to what force must be applied to the particle?

Thank you. I appreciate your help and knowledge greatly.
 
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  • #6
yus310 said:
... So it
U(final) + K(final) = U (initial) + K(initial)
Good.

... So, in the first problem, do you treat U(initial) as zero for both the disk and the particle, and the U(final) for the particle as mg(-2*pi*R)? Is this correct?
That would work fine.

And for the first problem, does the equation for solving for angular velocity like this...

mg(-2*pi*R)+((MR^2*angular velocity^2)/2)+((m*v^2)/2)=0
Almost. What's the rotational inertia of a disk? Also, express the KE of the particle in terms of angular velocity--since disk and particle rotate together.
 
  • #7
Doc Al said:
Good.


That would work fine.


Almost. What's the rotational inertia of a disk? Also, express the KE of the particle in terms of angular velocity--since disk and particle rotate together.

Sorry.. I really don't understand what you mean by this. I though that the
rotational inertia of a disk is "1/2*M*R^2" and wouldn't this correlate to final KE of the disk as "1/4*M*R^2*(angular velocity)^2" as the KE? Would the disk have an intitial KE?
And for the particle, would the KE= be 1/2*m*R^2*(angular velocity)^2? Is this right...
so would the equation be

Ui=0
Ki=0
Uf+Kf=0... So...

mg(-2*pi*R)+(1/4*M*R^2*(angular velocity)^2)+(1/2*m*R^2*(angular velocity)^2)=0?

Is this right?.. Once again thank you.
 
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  • #8
yus310 said:
I though that the
rotational inertia of a disk is "1/2*M*R^2" and wouldn't this correlate to final KE of the disk as "1/4*M*R^2*(angular velocity)^2" as the KE?
Right! (Check your equation and you'll see that you had it wrong before.)
Would the disk have an intitial KE?
Ignore the "gentle nudge".

And for the particle, would the KE= be 1/2*m*R^2*(angular velocity)^2? Is this right...
Good.

so would the equation be

Ui=0
Ki=0
Uf+Kf=0... So...

mg(-2*pi*R)+(1/4*M*R^2*(angular velocity)^2)+(1/2*m*R^2*(angular velocity)^2)=0?
Now you've got it.
 
  • #9
Is is meant by what force must be exerted... do they mean torque? And how do I go about it... hints?
 
  • #10
yus310 said:
Is is meant by what force must be exerted... do they mean torque? And how do I go about it... hints?
They mean force. Hint: Is the particle accelerating?
 
  • #11
Doc Al said:
They mean force. Hint: Is the particle accelerating?

... No, it is not accelerating, because it is held in place, but if you plug into Newton's second law.. F=ma, a=0, so what force are they referring to, meaning it just be the normal force, Fn=mg?..

Additionally, for the second problem, I had stated previously...I really don't how you can relate the rotational kinetic energy for the ring, if I don't have the stated mass... then how do I find the maximum angular velocity??
 
  • #12
yus310 said:
... No, it is not accelerating, because it is held in place, but if you plug into Newton's second law.. F=ma, a=0, so what force are they referring to, meaning it just be the normal force, Fn=mg?..
Hint: The particle is moving in a circle. (It's attached to the rotating disk.)

Additionally, for the second problem, I had stated previously...I really don't how you can relate the rotational kinetic energy for the ring, if I don't have the stated mass... then how do I find the maximum angular velocity??
Try it and see. Maybe you won't need the mass. :wink:
 
  • #13
Doc Al said:
Hint: The particle is moving in a circle. (It's attached to the rotating disk.)


Try it and see. Maybe you won't need the mass. :wink:

... Do you mean that I take my answer from part a) in the first problem and relate that the centripetal acceleration, a=r*(angular velocity^2), so I mutiply the angular velocity found in part a) and then take m*a=F, so ...

F= R*(angular velocity ^2)*(M + m), is this right...?
 
  • #14
yus310 said:
... Do you mean that I take my answer from part a) in the first problem and relate that the centripetal acceleration, a=r*(angular velocity^2),
Absolutely.
so I mutiply the angular velocity found in part a) and then take m*a=F, so ...
Good. Realize that here F is the net force. There may be more than one force acting on the particle. You want the force exerted by the disk on the particle.

F= R*(angular velocity ^2)*(M + m), is this right...?
Two problems: (1) The one I mentioned above and (2) what's M doing in there?
 
  • #15
Doc Al said:
Absolutely.

Good. Realize that here F is the net force. There may be more than one force acting on the particle. You want the force exerted by the disk on the particle.


Two problems: (1) The one I mentioned above and (2) what's M doing in there?

So.. Doc Al.. you're saying that it is only the centipetal acceleration that is keeping the particle on the disk??
 
  • #16
I'm saying that the particle is being centripetally accelerated, which requires a certain net force on the particle. (Given by F = ma.) Use that information to figure out how strongly the disk must pull on the particle. (Maybe its glued to the disk! But if it wasn't attached somehow, the particle would have slipped off long ago.) Don't neglect any other forces acting on the particle. (I only see two forces acting.)
 
  • #17
Doc Al said:
I'm saying that the particle is being centripetally accelerated, which requires a certain net force on the particle. (Given by F = ma.) Use that information to figure out how strongly the disk must pull on the particle. (Maybe its glued to the disk! But if it wasn't attached somehow, the particle would have slipped off long ago.) Don't neglect any other forces acting on the particle. (I only see two forces acting.)

Is the other force, the force of gravity meaning...

The disk must exert a force=F

F= m*R*(angular velocity^2)+(m*g)?

... Additionally I still don't quite understand the second part to the question I previously posted. The line connecting the pivot point and center, has me real confused...?
 
  • #18
yus310 said:
Is the other force, the force of gravity meaning...

The disk must exert a force=F

F= m*R*(angular velocity^2)+(m*g)?
Exactly!
... Additionally I still don't quite understand the second part to the question I previously posted. The line connecting the pivot point and center, has me real confused...?
You mean the hoop problem? (If you discuss multiple problems in a thread, be sure to be clear about which one you are referring to at any time.)

A picture would help, but I assume you have a hoop free to rotate about a pivot point somewhere on its perimeter. The center of the hoop starts out being at the same level (on a horizontal line) with the pivot point, then the thing is allowed to swing freely.
 
  • #19
Just wanted to say thank you, Doc Al, for your help. You really were a good teacher and helper. Thanks.
 

1. What is rotational kinetic energy?

Rotational kinetic energy is the energy possessed by a rotating object due to its motion. It is calculated by multiplying the moment of inertia (a measure of an object's resistance to rotational motion) by the square of its angular velocity.

2. How is rotational kinetic energy different from linear kinetic energy?

Rotational kinetic energy involves the rotational motion of an object, whereas linear kinetic energy involves the linear motion of an object. This means that rotational kinetic energy takes into account an object's shape and distribution of mass, while linear kinetic energy is only dependent on an object's mass and linear velocity.

3. What are some real-life examples of rotational kinetic energy?

Some examples of rotational kinetic energy include a spinning top, a moving Ferris wheel, a rotating fan, and a swinging pendulum. In each of these cases, the objects have rotational motion and therefore possess rotational kinetic energy.

4. How is rotational kinetic energy useful in solving problems?

Rotational kinetic energy is useful in solving problems because it allows us to analyze the motion and energy of rotating objects. By understanding the relationship between an object's moment of inertia, angular velocity, and rotational kinetic energy, we can make predictions and solve problems involving rotational motion.

5. How can rotational kinetic energy be changed?

Rotational kinetic energy can be changed by altering the moment of inertia or the angular velocity of an object. For example, increasing the moment of inertia (by changing the shape or mass distribution of an object) while keeping the angular velocity constant will increase the object's rotational kinetic energy. Similarly, increasing the angular velocity (by making an object spin faster) while keeping the moment of inertia constant will also increase the object's rotational kinetic energy.

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