I am pushing with a contact force of 40 N on the handle of the winch shown in th

Question

I am pushing with a contact force of 40 N on the handle of the winch shown in the diagram.

The handle of the winch is L = 0.750 m long, the radius of the winch R = 0.300 m. The moment of inertia (rotational inertia) of the winch is 24 kg m2 and friction in the bearings exerts a torque of 14.0 N m, which resists the winch turning.

The winch starts from rest. After one quarter of a rotation (pi/2 radians), if I exert a constant torque (by maintaining the 30 degree angle to the handle and the 40 N of force) what is the speed of the rope (2 s.f.)?

0.36 rad/s isn't correct

Explanation / Answer

net torque on the winch, Tnet = F*L*sin(90+30) - 14

= 40*0.75*sin(120) - 14

= 12 N.m

let alfa is the angular acceleration.

now Apply, Tnet = I*alfa

==> alfa = Tnet/I

= 12/24

= 0.5 rad/s^2

after quarter cycle,

theta = pi/2 rad

now apply, w2^2 - w1^2 = 2*alfa*theta

w2^2 - 0^2 = 2*alfa*theta

w2 = sqrt(2*alfa*theta)

= sqrt(2*0.5*pi/2)

= 1.25 rad/s <<<<<<<<<<<<<<<-------------------Answer

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