Using the angular momentum principle for 2 pucks

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The discussion revolves around two pucks on ice connected by a string, where one puck moves without rotating and the other both moves and rotates when a force is applied. It is established that the centers of both pucks travel the same distance due to the equal external force acting on their centers of mass. The calculation for the ratio of distances moved by the two pucks involves understanding the relationship between linear and angular motion, specifically using the angular momentum principle. The second puck's movement incorporates angular displacement, leading to the conclusion that the distance moved by the end of the string attached to it is twice that of the first puck. The conversation highlights the importance of distinguishing between the distances in the lab frame and the distances related to the pucks' motion.
shepherd882
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Homework Statement


Two pucks are lying on ice where they can slide and rotate with almost no friction. A string is tied to both pucks but it is tied to the middle of the first puck and wrapped around the second puck. You pull on both strings with the same force, F. The first puck moves without rotating while the second puck moves and rotates.
media%2F842%2F8428100b-d401-4c50-a9f5-ad1d4c933fff%2FphphHWdvo.png


a) The centres of both pucks will travel the same distance. Explain clearly why that is so.
b) Calculate the ratio of the total distance L that the end of the string attached to the second puck moved to the total distance d that the end of the string attached to the first puck moved. Beware that this is NOT a rolling without slipping problem and the relationship between angular speed and centre of mass speed may be different than in that case. There are at least two methods of solving this problem but they both require using the angular momentum principle. One method calculates the two types of kinetic energy and another method calculates the angle through which the second puck rotates.

Homework Equations


F=ma
Torque = I*w
I=0.5MR^2
Angular displacement = 0.5*a*t^2

The Attempt at a Solution


a) treat each puck as point-particle system. the external force (F) acting on the centre of mass for both pucks are the same. The centre of mass has a = F/m therefore distance traveled by both pucks is:
d=1/2*a*t^2
d= 1/2*F/M*t^2
b) for puck 2:
46502154_195563131368204_8563079905254834176_n.png?_nc_cat=107&_nc_ht=scontent.fyvr3-1.png

Divide angular displacement/distance:
R*angular displacement = 2D
(L=R*angular displacement)
Therefore, L=2D

not sure if this is right and for puck 1, how do you do it for slipping?

thanks for your help
 

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shepherd882 said:
R*angular displacement
In terms of the set-up, what distance does this calculation represent?
 
haruspex said:
In terms of the set-up, what distance does this calculation represent?

the distance that the string is pulled on puck 2?
 
shepherd882 said:
the distance that the string is pulled on puck 2?
It depends what you mean by that. It is not the distance that the pulled end travels in the lab frame.
 
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?

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