Rotational Inertia

tiredphysicsstudent

Im struggling with this one question in AP Physics I
The students measure the mass of the contraption and the attached masses and determine that the value of m is so much greater than the mass of the contraption that the mass and rotational inertia of the contraption itself can be ignored. The students then perform several trials in which the distance r at which the masses are attached is increased with each trial, and the time t required for the object to fall at a constant distance D is measured. Should the students expect the value of t to increase, decrease or remain the same as the value of r increases? Explain your reasoning qualitatively using principals relating to forces and Newton's Second Law? Explain your reasoning qualitatively using principles related to energy conservation?

tiredphysicsstudent

I am pretty sure it increases but I don't understand how to connect this to Newtons Law or Energy

studiot

Perhaps if you were to tell us more, like describing 'the contraption' and specifying what m is we could help you.

topsquark

tiredphysicsstudent

Perhaps if you were to tell us more, like describing 'the contraption' and specifying what m is we could help you.
yea that would help wouldn't it lol sorry

tiredphysicsstudent

here's the contraption

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studiot

Well that's half an explanation.

It looks like the contraption is falling down a string, as with a yo-yo.

Is this the case or what?

Cervesa

If the contraption acts as stated by studiot, like a yo-yo, then in the vertical direction

$mg - T = ma \implies T = m(g-a)$ where $T$ is the string tension and $a$ is the magnitude of downward linear acceleration of the contraption's COM.

The string tension also exerts a torque

$R \times T = I \alpha \implies Rm(g-a) = mr^2 \cdot \dfrac{a}{R} \implies a = \dfrac{gR^2}{r^2+R^2}$

since $R$ and $g$ are constants, the value of $r$ has an inverse relationship with the contraption's linear acceleration.