Posted by Ashley on Thursday, November 1, 2012 at 1:29pm.
A 190kg merrygoround in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merrygoround from rest to an angular speed of 0.700 rev/s in 2.00 s? (State the magnitude of the force.)

Physics  Jennifer, Thursday, November 1, 2012 at 3:25pm
F*r = I * a
Where F is the force on the merrygoround, r is the radius, I is the moment of inertia of the merrygoround, and a is the angular acceleration
for a solid, horizontal disk, the moment of inertia is: 1/2*m*r^2
F*r = 1/2*m*r^2*a
F = 1/2*m*r*a
w = w0 + a*t
where w is the angular speed, t is time, w0 is the initial angular speed
0.7 rev / s *(2*pi radians/rev) = 4.3 radians/s
4.3 = a*2
a = 2.2 rad/s^2
Plug this into the above equation to solve for F

Physics  yesi, Thursday, June 19, 2014 at 2:33am
wrong
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