M, a solid cylinder (M=1.83 kg, R=0.129 m) pivots on a thin, fixed, frictionless

Question

M, a solid cylinder (M=1.83 kg, R=0.129 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.710 kg mass, i.e., F = 6.965 N. Calculate the angular acceleration of the cylinder. If instead of the force F an actual mass m = 0.710 kg is hung from the string, find the angular acceleration of the cylinder. How far does m travel downward between 0.470 s and 0.670 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 0.435 m in a time of 0.510 s. Find 1cm of the new cylinder.

Explanation / Answer

here

tension = force pulling the string down

T = I * / R

I = 0.5 * m * R^2 = 0.5 * 1.83 * 0.129^2 = 0.0152 kg m^2

The pulling force is still T the tension, but T = mg - ma, so mg is not equal to the tension!

T = mg - ma

= TR

I = TR

I*(a/R) = TR

T = I*a/R^2

mg-ma = Ia/R^2

a = mg/(I/R^2 + m)

a = 6.695 / (0.0152 / 0.129^2 + 1.83)

a = 2.44 m/s^2 (= tangential acceleration, or acceleration of mass m)

= a/R = 2.44/0.129

= 18.9 rad/s^2 = angular acceleration by mass 0.71 kg

then

a = 2.44 m/s^2

s = 1/2 at^2

s1 = 1/2*2.44*0.47^2

s1 = 0.269 m

s2 = 1/2*2.44*0.67^2

s2 = 0.547 m

difference
d = 0.278 m = distance moved between 0.47 and 0.67 seconds

then

s = 1/2 at^2

0.435 = 1/2 a*0.51^2

a = 3.34 m/s^2

a = mg/(I/R^2 + m)

3.34 = 6.965/(I/0.129^2 + 1.83)

I = 0,00424894 kgm^2

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