Question

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!!!

Answer 1

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

Answer 2

the first one I know, can you help me with the others?..

Answer 3

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

Answer 4

I am very old and slow.

Answer 5

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

Answer 6

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.

Answer 7

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