mastering physics, Starting from rest, a constant force = 130 is applied to the

Question

mastering physics, Starting from rest, a constant force = 130 is applied to the free end of a 60- cable wrapped around the outer rim of a uniform solid cylinder, similar to the situation shown in Fig.10.9. The cylinder has mass 6.00 and diameter 40.0 and is free to turn about a fixed, frictionless axle through its center.

a. How long does it take to unwrap all the cable?

b. How fast is the cable moving just as the last bit comes off?

c. Now suppose that the cylinder is replaced by a uniform hoop, with all other quantities remaining unchanged. In this case, would the answers in the previous parts be larger or smaller? Explain.


HELP PLEASE!!!

Explanation / Answer

Given that the applied force is F = 130 N Length of cable is L = 60 m Mass of cylinder is M = 6.00 kg Diameter is D = 40.0 cm = 0.40 m ---------------------------------------------------------------- Now the torque is               = I             RF = I(a /R )                         (Since a = R: = R x F)             RF = (1/2)MR2(a /R )                  ------------ (1)             RF = (1/2)MR2(a /R )                  ------------ (1)                F = (1/2)Ma                a = 2F / M                a = 2*130 N / 6.00 kg                a = 43.33 m/s2 now from equation of motion             S = ut + (1/2)at2            L =  0 + (1/2)at2             t = (2L / a )             t = 1.66 s (b)    From equation of motion final velcoity is             v = u + at                = 0 + (43.33 m/s2) (1.66s)                = 72.108 m/s (c)            L =  0 + (1/2)at2             t = (2L / a )             t = 1.66 s (b)    From equation of motion final velcoity is             v = u + at                = 0 + (43.33 m/s2) (1.66s)                = 72.108 m/s (c)             t = (2L / a )             t = 1.66 s (b)    From equation of motion final velcoity is             v = u + at                = 0 + (43.33 m/s2) (1.66s)                = 72.108 m/s (c) If the cylinder replaced by hoop, the moment of inertia of the hoop is I = MR2                Then the acceleration of the hoop becomes                         a = F / M             (Since from equation (1) )                         a = 21.66 m/s2 Now the time t = (2L / a)                         =  2.353 s   Then final speed is v = at                                  = (21.66 m/s2)(2.353s)                                  = 50.98 m/s Now the time t = (2L / a)                         =  2.353 s   Then final speed is v = at                                  = (21.66 m/s2)(2.353s)                                  = 50.98 m/s               

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