M, a solid cylinder (M=1.91 kg, R=0.113 m) pivots on a thin, fixed, frictionless
ID: 1502210 • Letter: M
Question
M, a solid cylinder (M=1.91 kg, R=0.113 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?.630 kg mass, i.e., F = 6.180 N.Calculate the angular acceleration of the cylinder. I solved this and the agular accelaration was 57..3 rad/s^2.
If instead of the force F an actual mass m = 0.630 kg is hung from the string, find the angular acceleration of the cylinder. This was 34.5 rad/s^2.
What I can't solve are these next two quesitons.
How far does m travel downward between 0.710 s and 0.910 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 0.433 m in a time of 0.550 s. Find Icm of the new cylinder.
Explanation / Answer
C) from the above data
tangental acceleration of the string, a = R*alfa
= 0.113*34.5
= 3.8985 m/s^2
at time t1 = 0.71 s
v1 = u + a*t1= 0 + 3.8985*0.71 = 2.768 m/s
at time t2 = 0.91 s
v2 = u + a*t2= 0 + 3.8985*0.91 = 3.548 m/s
distance travelled in the time gap, d = (v2^2 - v1^2)/(2*a)
= (3.548^2 - 2.768^2)/(2*3.8985)
= 0.632 m
d) let a is the acceleration of the hanging mass.
d = u*t + 0.5*a*t^2
d = 0 + 0.5*a*t^2
a = 2*d/t^2
= 2*0.433/0.55^2
= 2.863 m/s^2
Tension in the string, T = m*g - m*a
= 0.63*9.8 - 0.63*2.863
= 4.37 N
now Apply, Torque on the cyllinder = I*alfa
T*R = Icm*a/R
==> Icm = T*R^2/a
= 4.37*0.113^2/2.863
= 0.0195 kg.m^2
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