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## Homework Statement

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.

## Homework Equations

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

substituting