Work/ energy on a rigid body (mechanics)

AI Thread Summary
The discussion revolves around calculating work and energy for a rigid body in mechanics, specifically focusing on a bar rotating about a pivot. Key equations include the kinetic energy (T) expressed as T = 1/2 mv² + 1/2 I (w1² - w2²), with initial velocity being zero since the bar starts from rest. Participants discuss how to determine the extension of a spring when a moment is applied, suggesting geometric methods to find initial and final stretches. The importance of considering gravitational potential energy (Vg = mgh) is highlighted, particularly in relation to the height of the center of mass during rotation. Ultimately, one participant successfully resolves their confusion with the calculations, indicating a collaborative problem-solving effort.
smruthi92
Messages
15
Reaction score
0
pls see attached file for the question.

so W = change in T + change in Ve + change in g

T = 1/2 mv2 + 1/2 I (w1^2 - w2^2), but how do u find the velocity, shouldn't it be 0 since it starts/stops at v=0.

for Ve = 1/2 kx2, how do u know how far it stretches once the moment is applied? :O :(

and for Vg = mgh, is it mg (1.458)?

im so sorry, i just have no idea about this one.
 

Attachments

  • physics.jpg
    physics.jpg
    16.8 KB · Views: 497
Physics news on Phys.org
smruthi92 said:
T = 1/2 mv2 + 1/2 I (w1^2 - w2^2), but how do u find the velocity, shouldn't it be 0 since it starts/stops at v=0.
You can treat the bar as have purely rotational KE about its pivot; it's initial speed is zero since it's released from rest.

for Ve = 1/2 kx2, how do u know how far it stretches once the moment is applied? :O :(
I suspect that you are supposed to figure out its initial stretch and its final stretch (using a bit of geometry).

and for Vg = mgh, is it mg (1.458)?
How does the position of the center of mass change as the bar moves?
 
Doc Al said:
You can treat the bar as have purely rotational KE about its pivot; it's initial speed is zero since it's released from rest.


I suspect that you are supposed to figure out its initial stretch and its final stretch (using a bit of geometry).


How does the position of the center of mass change as the bar moves?

oh awesome! so T would essentially just = 1/2 I (w2^2).

so W = 0,
1/2 I (w2^2) + 1/2 k (x2^2-x1^2) = 0?

i guess if it comes back to rest, x1 = 0. but i still don't know how to figure out that extended length using geometry? coz when ur rotating the rod by 69 degrees, could u maybe do 69/360 * length of spring to find it? :S its a bit odd.
 
smruthi92 said:
oh awesome! so T would essentially just = 1/2 I (w2^2).

so W = 0,
1/2 I (w2^2) + 1/2 k (x2^2-x1^2) = 0?

i guess if it comes back to rest, x1 = 0. but i still don't know how to figure out that extended length using geometry? coz when ur rotating the rod by 69 degrees, could u maybe do 69/360 * length of spring to find it? :S its a bit odd.

sorry i made a mistake. um so what i have is. to find x I am creating an arc with 69 degrees + tan inverse (1.458/1.8), so then i would find the extension to be 2 x pie x 1.8 x 108/230. minus that from the original length and then use the eqn above. but this is all assuming, its rotated about the point where the rod connects to the surface?
 
smruthi92 said:
oh awesome! so T would essentially just = 1/2 I (w2^2).
Yes.

so W = 0,
1/2 I (w2^2) + 1/2 k (x2^2-x1^2) = 0?
What happened to gravity?

smruthi92 said:
sorry i made a mistake. um so what i have is. to find x I am creating an arc with 69 degrees + tan inverse (1.458/1.8), so then i would find the extension to be 2 x pie x 1.8 x 108/230. minus that from the original length and then use the eqn above. but this is all assuming, its rotated about the point where the rod connects to the surface?
Not sure what you're saying here. In any case, you need to compare how the stretch of the spring changes. Figure out the length of the spring in each position. (When the bar is horizontal, how much of the spring's 1.458 m length is the stretch from its unstretched position?)
 
Doc Al said:
Yes.


What happened to gravity?


Not sure what you're saying here. In any case, you need to compare how the stretch of the spring changes. Figure out the length of the spring in each position. (When the bar is horizontal, how much of the spring's 1.458 m length is the stretch from its unstretched position?)

ok as I am assuming that they are rotating it at the end of the bar connected to the spring, and not say, the centre of mass?

so if we include Vg, then its mg h, where h = the height you find forming a triangle?
 
Doc Al said:
Yes.


What happened to gravity?


Not sure what you're saying here. In any case, you need to compare how the stretch of the spring changes. Figure out the length of the spring in each position. (When the bar is horizontal, how much of the spring's 1.458 m length is the stretch from its unstretched position?)

smruthi92 said:
ok as I am assuming that they are rotating it at the end of the bar connected to the spring, and not say, the centre of mass?

so if we include Vg, then its mg h, where h = the height you find forming a triangle?

OOOOHHH I WORKED IT OUT! THANKYOU SO MUCH FOR UR HELP! :d:d:d:d
 
Back
Top