Do the students exert a net torque on the rotating disc?

AI Thread Summary
The discussion centers on the dynamics of a rotating plate with two students walking outward. It clarifies that while the students exert a net torque on the plate, the overall angular momentum of the system remains conserved due to the closed nature of the system. As the students walk outward, their moment of inertia increases, causing the rate of rotation to decrease, which is consistent with conservation laws. The confusion arises from the relationship between torque and angular momentum, but it is explained that the torque exerted by the students is balanced by an equal and opposite torque from the plate. Ultimately, the students do produce a net torque, but this does not violate the conservation of angular momentum.
brushman
Messages
112
Reaction score
1

Homework Statement


A large plate is balanced at its center and two students of equal mass stand at its center. The plate is
rotated on a frictionless pivot about an axis through its center and perpendicular to its face. The students
then begin to walk out towards opposite edges.
(Select T-True, F-False, I-Increases, D-Decreases, S-Stays the same. ).
A) The students do no work in walking outward. False: The students do work against the centripetal force
B) The students produce a net torque on the plate. True: The rate of rotation slows down, there has to be a
torque to do that.
C) The rate of rotation ... as the students walk outward. Decreases: angular momentum is conserved,
moment of inertia increases so angular velocity has to decrease
D) The total angular momentum of the system ... as the students walk outward. :Same: closed system,
angular momentum is conserved
E) When the students reach the outer edge and stop, the moment of inertia of the system (plate+students)
is the same as when the students started. F: it increases

The Attempt at a Solution



As you can see, the correct answers are already there. What I don't under stand is B and C. Isn't this a direct contradiction? If there's a torque, then momentum must not be conserved, right?

What's the explanation for this!? I guess because it's a closed system. But wouldn't that also mean that kinetic energy is conserved which isn't necessarily true since omega is squared in the rotational energy, 1/2 Iw^2.

For example, if the kids started at the edge and walked in the kinetic energy would increase right?

I also don't see how the students would produce a net torque. It seems like their torques would cancel each other out.
 
Last edited:
Physics news on Phys.org
hi brushman! :smile:

there is a torque, because there is a tangential acceleration …

the students not only have a https://www.physicsforums.com/library.php?do=view_item&itemid=27" they also are changing their tangential component of velocity, and so have a tangential acceleration, which must be provided by a tangential force

(in a frame of reference rotating with the plate, that would be a https://www.physicsforums.com/library.php?do=view_item&itemid=86" , 2mω x vrel, perpendicular to the relative velocity, which is radial)

… that tangential force provides a torque on the students, increasing their https://www.physicsforums.com/library.php?do=view_item&itemid=313" on the plate, decreasing its angular momentum, so there's no overall torque and no overall change in angular momentum) :wink:
 
Last edited by a moderator:
So the students do exert a torque, and the reason momentum is still conserved is because the plate exerts an equal and opposite torque on the students.

Thanks.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top