How far does m travel downward between 0.450 s and 0.650 s after the motion begi

Question

How far does m travel downward between 0.450 s and 0.650 s after the motion begins?

The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.509 m in a time of 0.530 s. Find Icm of the new cylinder.

Hint: The tension in the string induces the torque in both this part and the first part. The tension is not equal to mg! If it were, the mass would not accelerate downward. Determine all of the forces acting on the mass, then apply Newton's second law and solve for the tension, and apply it to Newton's second law of rotational motion.

Explanation / Answer

Part A : T*r = I*

so, Tr = 0.5*m*r^2 *

so, = 2T/mr

also T = F = 5.592 N

so = 2*5.592 / (1.39 *0.129)

        = 62.4 rad/s2

Part B :-

if an actual mass is hung then by drawing the free body diagram of tthe mass, we get,

mg - T = ma

also T*r = I* = 0.5*M*r^2 *a/r

so, T = 0.5Ma

so, substituting the force equation,we get

mg - Ma/2 = ma

so , (m+M/2)a = mg

so, a = mg/(m+M/2)

= 0.57*9.8 / (0.57 + 1.39 /2 )        

        = 4.416 m/s2

so, angular acceleration = a/r

                                    = 4.416 / 0.129

                                    = 34.23 rad/s2


Part C:-
a = 4.416 m/s2 (obtained from above)

using the equation, s = ut +0.5*at^2

for start from rest, u =0

so, s= 0.5*a*t^2

so, s1 = 0.5*a*t1^2

and , s2 = 0.5*a*t2^2

so, s2-s1 = 0.5*a*(t2^2-t1^2)

                = 0.5*4.416*(0.65^2 - 0.45^2)

                = 0.486 m

Part D:-
using the same equation

0.509 = 0.5*a*0.533^2
so, a = 3.584 m/s2



solving backwards, we get
Tr = Ia/r
so, T = Ia/r^2

substituting in the main force equation, we get,
mg - Ia/r^2 = ma
so, Ia/r^2 = m(g-a)

so, I = m(g-a)*r^2/a

        = 0.57(9.8 - 3.584)*0.129^2/ 3.584

        = 0.0164 kg.m2

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