M solid cylinder M-2.27 kg R-O 133 m ots en a thin fi ad fri tiale\" b aring A s
ID: 1784637 • Letter: M
Question
M solid cylinder M-2.27 kg R-O 133 m ots en a thin fi ad fri tiale" b aring A string wrapped around the cylinder pulls do n ward ith a farce F which equals the aight of a 0.570 kg mass, i e F 5 502 N. Calculata the angular acceleration of the c i der 3.70x10 rad/s 2 You are correct If instead of the kree F sn actual mass m-0.570 kg is hung from the string, find th..ngular acceleration cfthe cylinder 47.62 rodis Hint: The tension i, the st ng induces the to eue in both this part and the first part. T e tension is nat qual to m9, r t erw, the mass o noe accelerate do a d Determine a of the orces acti onthù mass, then app y Ne to s's second la and sa ve fo the tension, and apply it tù Nin ton second 320Previus Tries ow for does m trave cownward between 0.710 sand 0.910s after the mation begins? Tries 0/20 The cylinder is chanped to one with the same mass and radius, but o different moment af inertia. Starting frem ret, the maas now move a distance 0.338 m inatime of o.sso. Find iem of the new cylindr Tries a/20Explanation / Answer
Torque = F*R= I where I =MR^2/2 and is the angualar accceleration.
F*R= MR^2/2
= 2F/(MR ) = 2*5.502/(2.27*0.133) = 36.5 rad/s^2
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When the force mg = 5.502 is acting down, this force provides the linear
Acceleration' a 'of the mass and exerts a torque on M 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) = 5.502/ (1.135+0.57) = 3.23 m/s^2
= a/R = 3.23/0.133 = 24.3 rad/s^2
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How far does mass m travel downward between t = 0.71 s and t = 0.91 s (Assuming motion begins at time t = 0.0 s)?
s = 0.5*a(t2^2 -t1^2) = 0.5*3.23*(0.91^2 - 0.71^2) =0.523 m
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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 0.338 m in a time interval of 0.55 s. Find Icm of the new cylinder.
a = 2s/ (t^2) = 2*0.338/(0.55)^2 = 2.23 m/s^2
F = m g- ma = 5.502 - 0.57*2.23 = 4.2309 N
F = I /R =Ia/R^2 = 4.2309
I = 4.2309*0.133^2/2.23 = 0.0336 kg m^2
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