M , a solid cylinder (mass = 1.590 kg , r = 0.424m) pivots on a thin, fixed, fri
ID: 1467137 • Letter: M
Question
M , a solid cylinder (mass = 1.590 kg , r = 0.424m) 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 0.810kg mass, i.e., F = 7.938N
b) If instead of the force F an actual mass m- 0.810kg is hung from the string (see figure below), what is the angular acceleration of the cylinder? Submit Answer Incorrect.Tries 4/6 Previous Tries Submit Answer Incorrect. Tries 4/6 Previous Tries c)How far does m travel downward between 0.390s and0.590s after the motion begins? Use equation of rotaional motion Submit Answer Incorrect. Tries 1/6 Previous TriesExplanation / Answer
B:-
When the force F = 7.938 is acting down, this force provides the linear
Acceleration,a of the mass and exerts a torque on cylinder giving an angualar
accceleration .
a = R
For the mass M,
F = I /R = [MR^2/2] /R= {Ma/2}
For the mass m, ma = mg -F = mg - Ma/2
a = mg/ (M/2 +m) = 7.938 / (0.795+0.810) = 4.95 m/sec^2
= a/R = 4.95 /0.424 = 11.67 rad/sec^2
So, this is the required angular accelaration.
C:-
Distance covered by block between given time period,d :-
d= 0.5*a*(t2^2 -t1^2) = 0.5*4.95*(0.59^2 - 0.39^2) =0.485m.
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