Physics Homework/Practice 9 aar a solid cylinder (M-1.91 kg, R 0.115 m) pivots o
ID: 1782152 • Letter: P
Question
Physics Homework/Practice 9 aar a solid cylinder (M-1.91 kg, R 0.115 m) pivots on a thin, fixed, frictionless beari equals the weight of a o acceleration of the cylinder ng. A string wrapped around the cylinder pulls downward with a force F which 790 kg mass, i.e., F = 7.750 N. Calculate the angular Ir instead of the force F an actual mass m - 0.790 kg is hung from the string, find the angular acceleration of the cylinder. How far does m travel downward between 0.610 s and 0.810 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 0.684 m in a time of 0.570 s Find Iom of the new cylinder.Explanation / Answer
9. Net torque = I alpha
R F = (M R^2 / 2) alpha
alpha = 2 F / M R = 2 x 7.750 / (1.91 x 0.115)
= 70.6 rad/s^2
-----------------------------------
0.790 g - T = 0.790 a
T R = ( M R^2 / 2 )(a / R )
T = 1.91 a / 2
putting in previous equation,
7.750 - 1.91a / 2 = 0.790 a
a = 4.44 m/s^2
alpha = a / R = 38.6 rad/s^2
------------------
d = a (t2^2 - t1^2 ) /2
d = 4.44 (0.810^2 - 0.610^2) / 2
d = 0.63 m
----------------
0.684 = a (0.570^2 / 2)
a = 4.21 m/s^2
7.750 - T = 0.790 x 4.21
T = 4.424 N
R T = I (a/R)
I = 0.115^2 x 4.424 / 4.21 = 0.0139 kg m^2