M, a solid cylinder (mass = 2.35 kg, r = 0.030 m) pivots on a thin, fixed, frict
ID: 2219073 • Letter: M
Question
M, a solid cylinder (mass = 2.35 kg, r = 0.030 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 0.650 kg mass, i.e., F = 6.370 N. Calculate the angular acceleration of the cylinder.(in rad/s^2)
How far does m travel downward between 0.490 s and 0.690 s after the motion begins?
Explanation / Answer
A) the cylinder has a inertia of I=1/2MR^2 ...with M=2.43 and R=0.131 applyinng newton 2nd law FR=I alpha....here F=5.592 N and R=0.131 It is possible to find alpha the angular acceleration B) in this problem we have 2 bodies , one in linear acceleration and the other in angular acceleration for the mass going down we write T-mg=-ma ......eq 1....here m=0.57kg , g=9.8m/sec^2 for the cylinder we have TR=I alpha....eq 2 ...we have 2 equations with 3 unknown quantities which are T, a and alpha a third equation involving alpha and... a.... exist we know that... alpha=a/R...eq 3... sooo a=R alpha from here it is easy to find alpha...( it's easier to find a first and find alpha after) C) with the linear aceleration found in B , uses kinematic equation to find the distances y=1/2 at^2 ....the distance travelled between the two specified times might be calculated by d=(1/2) a(0.63^2-0.43^2) D) with the distance and the time it is possible to find the new acceleration y=0.381= (1/2) a 0.51^2 where y=0.381m with this acceleration you can find the angular acceleration (eq 3) of the new cylinder and the tension in the string using eq..1 Finally with T and the angular acceleration it is a children play to caculate the new Inertia of the cylinder with eq...2 That's the most I can do I hope it helps but I have a little problem.. I don,t understand the expression ( I sub cm) You will have to ask someone else for that .
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