M, a solid cylinder (M=2.23 kg, R=0.127 m) pivots on a thin, fixed, frictionless

Question

M, a solid cylinder (M=2.23 kg, R=0.127 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 0.870 kg mass, i.e., F = 8.535 N.
a)Calculate the angular acceleration of the cylinder.

b)If instead of the force F an actual mass m = 0.870 kg is hung from the string, find the angular acceleration of the cylinder.

c)How far does m travel downward between 0.730 s and 0.930 s after the motion begins?

d)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 0.450 m in a time of 0.470 s. Find Icm of the new cylinder

Explanation / Answer

Fr =I

I for solid cylinder is mr^2/2

8.535 *r =mr^2/2 *(a/r)

8.535 =1.115a

hence a =7.65 m/sec^2

angular acc ==a/r =7.65/0.127 =60.24 rad/sec^2

for mass of of 0.87 mass g =8.535/0.87 =9.81m/sec^2

hence 8.535-T =0.87a ............1

and Tr =mr^2/2*a/r

T=ma/2 =1.115a ............2

hence 1and 2

we get a= 4.3 m/sec^2

hence =a/r =4.3/0.127 =33.86 rad/s^2

s =1/2 at^2

S2-S1 =0.5a(t2^2 -t^1)

S2-S1 = 0.5 *4.3*(0.93^2-0.73^) =0.714 m

d)0.45 =1/2 at^2

for t=0.47 a =4.07

8.535 -T =0.87a

for a =4.07 ..T=4.99

and Tr=I

as =a/r

hence I =Tr^2/a = 0.01979 kgm^2

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