Rotational kinetic energy of a disk

Click For Summary
To find the kinetic energy of a hanging mass released from a disk, one must consider whether the disk and mass fall together or if the mass moves relative to the disk. If they fall together, both experience the same gravitational acceleration, and the kinetic energy of the mass can be calculated using the formula 1/2mv², requiring the velocity as a function of time or distance. If the mass moves relative to the disk, the difference in velocity must be determined, which relates to the disk's angular velocity. The moment of inertia of the disk is also necessary for calculating rotational kinetic energy. Understanding these dynamics is crucial for accurately determining the kinetic energy of the mass at any given point after its release.
physgirl
Messages
99
Reaction score
0
ok, so a thread with hanging mass at the end is wrapped around a disk. the disk is dropped and mass is released. how do you find the kinetic energy of the weight at a given time/distance fallen from the disk? I'm not sure to whether use rotational kinetic energy or linear kinetic energy... all the basic values are given, except for the moment of inertia of the disk... so how would I find the kinetic energy of the mass at a given time after its release?
 
Physics news on Phys.org
It depends on the what happens when "the disk is dropped and mass is released."

If the disk and mass are dropped together, they are both subject to the same acceleration, g, unless there is more air resistance (force) on the disc.

Obviously, the kinetic energy of the mass if given by 1/2mv2, so one must determine the velocity of the mass as a function of time or distance traveled.

If the mass is moving relative to the disk, then one would need to determine the difference in velocity (speed) of the mass with respect to the rotational axis of the disk, and this speed difference would be manifest in the rotation of the disk, and the differential speed would be related to the angular velocity of the disk.

Note - the acceleration of any mass in 'freefall' is limited by the local acceleration due to gravity.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
26
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 21 ·
Replies
21
Views
5K
Replies
3
Views
1K
  • · Replies 33 ·
2
Replies
33
Views
3K
Replies
9
Views
3K
Replies
30
Views
3K
Replies
18
Views
5K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 15 ·
Replies
15
Views
3K