M, a solid cylinder (M=2.35 kg, R=0.111 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 0.750 kg mass, i.e., F = 7.357 N.
1)Calculate the angular acceleration of the cylinder.
The answer here is: 5.64×101 rad/s^2
2)If instead of the force F an actual mass m = 0.750 kg is hung from the string, find the angular acceleration of the cylinder. ?
3)How far does m travel downward between 0.470 s and 0.670 s after the motion begins?
4)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 0.284 m in a time of 0.470 s. Find Icm of the new cylinder.
Thank You very much!!!
I am very old and slow.
tension up on weight T = mg-ma
= .75 (9.81 - a)
moment = T(r) = .75(9.81-a)(.111)
again Moment = I alpha but alpha = a/r
so
.75(9.81-a)(.111) = .0145 (a/.111)
.8167- .08325 a = .1306 a
a = 3.82 m/s^2
alpha = a/r = 34.4 rad/s^2
I = (1/2) m r^2 = .5*2.35*.111^2 = .0145
moment = I alpha
7.357 * .111 = .0145 alpha
alpha = 56.4 radians/s^2 yes
d = 0.284 m in a time of 0.470 s
.284 = (1/2) a (.47)^2
a = 2.57
again:
.75(9.81-a)(.111) = I (a/.111)
but now a = 2.57
.75(7.24)(.111) = I (2.57/.111)
I = .026 kg m^2
how far between 0.470 s and 0.670 s
well, I have a = 3.82 m/s^2
d = 0 + 0 + (1/2) a t^2
at .47sec d = (1/2)(3.82)(.47)^2
at .67sec d = (1/2)(3.82)(.67)^2
then subtract
I really appreciate all your help, if I said anything that made it sounds disrespectful I am deeply apologizing..
Thank you very much for all your help.