Question

M, a solid cylinder (M=1.71 kg, R=0.133 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 0.690 kg mass, i.e., F = 6.769 N. Calculate the angular acceleration of the cylinder.

5.95×10¹ rad/s^2

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Previous Tries

If instead of the force F an actual mass m = 0.690 kg is hung from the string, find the angular acceleration of the cylinder.

Tries 0/20

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

Tries 0/20

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 0.379 m in a time of 0.470 s. Find I_cm of the new cylinder.

Answer 1

Torque τ = F*R= Iα where I =MR^2/2 and α is the angualar accceleration.
F*R= α MR^2/2
α = 2F/(MR ) =( 2*6.769)/1.71*0.133 =5.9526 *10 rad/s^2
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When the force mg =6.769 is acting down, this force provides the linear
Acceleration' a 'of the mass and exerts a torque on M giving an angualar accceleration α. a = R α
For the mass M,
F = I α/R = [MR^2/2] α/R= {Ma/2}
For the mass m, ma = mg -F = mg - Ma/2
a = mg/ (M/2+m) = 6.769 / (0.855+0.69) = 4.381 m/s^2
α = a/R = 4.381 /0.133= 32.93 rad/s^2
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How far does mass m travel downward between t = 0.73 s and t = 0.93 s (Assuming motion begins at time t = 0.0 s)?
s = 0.5*a(t2^2 -t1^2) = 0.5*4.381*(0.93^2 - 0.73^2) =0.727 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 0.379 m in a time interval of 0.47 s. Find Icm of the new cylinder.

a = 2s/ (t^2) = 2*0.379/0.47²=3.431 m/s²

F = m g- ma = 6.769- 0.69*3.431= 4.401 N

F = I α /R =Ia/R^2 = 4.401

I = 4.401 *0.133²/3.431 = 0.0226 kg m^2

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