M, a solid cylinder (M=2.19 kg, R=0.131 m) pivots on a thin, fixed, frictionless

Question

M, a solid cylinder (M=2.19 kg, R=0.131 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 0.610 kg mass, i.e., F = 5.984 N. Calculate the angular acceleration of the cylinder.

If instead of the force F an actual mass m 0.610 kg is hung from the string, find the angular acceleration of the cylinder. Im 2.68×101 rad/s^2 You are correct Your receipt no. is 158-8449 Tries How far does m travel downward between 0.690 s and 0.890 s after the motion begins? 5.55x10 1 m You are correct. Your receipt no.is 158-5031@Previous Tries Th2g linder is heanpod to one with the same mass nd eadius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.329 m in a time of 0.510 s. Find Icm of the new cylinder 04100 kg m 2 You know how to write an equation for the torque from the second part. You are given quantities to calculate the acceleration (review motion in one dimension if this is not obvious), and in this case you need to solve for the moment of inertia Submit Answer Incorrect. Tries 4/20 Previous Tries

Explanation / Answer

I thimk you did first two parts ,you need only last part

last part:

initial velocity of mass is Vo = 0 m/sec

distance moved is S = 0.329 m

time taken is t = 0.51 sec

then using kinematic equations

S= (Vo*t)+(0.5*a*t^2)

0.329 = (0*t)+(0.5*a*0.51^2)

0.329 = 0.5*a*0.51^2

a = 2.53 m/s^2

angular accelaration is a = r*alpha

2.53 = 0.131*alpha

alpha = 2.53/0.131 = 19.32 rad/s^2

using Torque T = Inew*alpha

R*F = Inew*alpha

0.131*5.984 = Inew*19.32

Inew = (0.131*5.984)/19.32

Inew = 0.0405 kg-m^2

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