Problems and Questions 1. After the sys tem is released from rest, is the hangin

Question

Problems and Questions 1. After the sys tem is released from rest, is the hanging mass m in free fall during this experiment? Explain why or why not. 2. How was the inertia (mass) of the string accounted for in this lab? 3. Does the system's acceleration really stay constant as the hanging mass falls, or does it increase or decrease? Answer this question by thinking about the inertia of the string as the hanging mass falls, then inspect your formulas for apreficn. Do the numerators change as the hanging mass falls, or do the denominators change, or neither one? What assumption did you make in deriving these formulas? 17

Explanation / Answer

1. for the system in the experiment, the hanging mass is not in free fall, as the acceleration of the box is not g (9.8 m/s/s) but acceleration of the block = a

where a < g

2. theoretically

for hanging mass Mh, cart mass Mc, pulley mass Mp, pulley radius R, and string length l, mass m

at time t, when length y of the string is hanging

from force balance

Mh*g - T1 = Mh*a

T1 + m*y*g/l = T2

(T2 - T3)*R = 0.5*Mp*R^2*a/R

and

T3 = Mc*a ( assuming fricitonless surface, and no air drag)

hence

2(T1 + myg/l - Mc*a) = Mp*a

2(Mh(g - a) + my*g/l - Mc*a) = Mp*a

Mh*g - Mh*a + my*g/l - Mc*a = 0.5Mp*a

a = (Mh*g + my*g/l)/(0.5Mp + Mc + Mh)

hence

dv/dt = dv/dy * dy/dt = vdv/dy = (Mh*g + m*yg/l)/(0.5Mp + Mc + Mh)

vdv = (Mh*g + m*yg/l)dy/(0.5Mp + Mc + Mh)

integrating

let initial hanging length be 0

final hanging length be y

v^2/2 = (Mh*g*y + m*y^2g/2l)/(0.5Mp + Mc + Mh)

now in the experiment , we havent accounted for the mass of the string using this formula though we could

although mass of string is too low to cause apprfeciable change inthe result

3. from part 2, we can see accelration of the system increases as the hanginng part of the string increases in length with time

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