a wheel (hoop) of 10lbs has one 10lb spoke which is horizontal and rests on the ground (held). it is connected to an axle. the radius of the wheel (length of spoke) is 1ft. when the wheel is let go, what is the force of the axle on the wheel?
(the problem also asked for the rotational inertia at the axle).
the answer given is 18lbs.
Rotational Inertia at Axle = 1/3(10/32 slugs)(1ft)^2 + (10/32 slugs)(1ft)^2
Sum of horizontal forces = m(a-radial) = Tension of spoke
torque at axle = RI * alpha = radius * spoke-weight
The Attempt at a Solution
RI = 0.417slug*ft^2
i set up an fbd of the wheel.
torque at axle = (0.417slug*ft^2)(alpha) = (1ft)(5lbs) (i assume that the weight of the spoke is equally distributed between the attachment on the wheel and the axle)
i then get rotational accelerate (alpha) = 12 rad/s^2
for the horizontal force, i have (20/32 slugs)(a-radial) = Tension
vertical forces, I have (20/32slugs)(a-tangential) = vertical-Axle-force - 5lbs
a-tangential = radius*alpha = 12ft/s^2
vertical-axle-force = 12.5 lbs.
thats as far as i got...
dunno how to get the 18lbs (upward) which is the answer