M, a solid cylinder (mass = 2.39 kg, r = 0.060 m) pivots on a thin, fixed, frict
ID: 3162539 • Letter: M
Question
M, a solid cylinder (mass = 2.39 kg, r = 0.060 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 0.690 kg mass, i.e., F = 6.762 N. Calculate the angular acceleration of the cylinder. ___________ If instead of the force F an actual mass m = 0.690 kg is hung from the string, what is the angular acceleration of the cylinder. __________________ How far does m travel downward between 0.530 s and 0.730 s after the motion begins? _________________Explanation / Answer
here,
mass , m = 2.39 kg
r = 0.06 m
when applied force ,F = 6.762 N
the accelration , a = net force /effective mass
a = F /( 0.5 *m * r^2 /r^2)
a = 6.762 /( 0.5 * 2.39)
a = 5.7 m/s^2
the angular accelration , alpha = r * a = 0.34 rad/s
when weight is hung on ,W = 6.762 N
m' = 0.69 kg
the accelration , a' = net force /effective mass
a' = F /( 0.5 *m * r^2 /r^2 + m')
a' = 6.762 /( 0.5 * 2.39 + 0.69)
a' = 3.54 m/s^2
the angular accelration , alpha' = r * a' = 0.21 rad/s
for time t1 = 0.53 s
the distance travelled , s1 = o + 0.5 * a' * t1^2
s1 = 0.5 * 3.54 * 0.53^2 = 0.527 m
for time , t2 = 0.73 s
the distance travelled , s2 = 0 + 0.5 * a' * t2^2
s2 = 0.5 * 3.54 * 0.73^2 = 0.943 m
the distance travelled between t1 and t2 , s = s2 - s1 = 0.42 m
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