# Wheel and spoke

## 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).

## 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

vertical-axle-force = 12.5 lbs.

thats as far as i got...

dunno how to get the 18lbs (upward) which is the answer

substituting

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actually, i think i realized that solving the problem involves just the free body diagram of the spoke.

the axle applies a force with unknown vertical component V and horizontal H

the net force on the vertical for the spoke becomes something like:

(10/32 slugs)(a-cm) = 10lb - V

where a-cm is the acceleration at the center of mass

horizontally:

by already knowing alpha, i can solve for a-cm:

a-cm = (1/2ft)alpha

plugging in, i get a-cm = 6 ft/s^2

6(10/32) lb = 10lb - V

V = 8lb which is the wrong answer :(