Calculating Axle Force and Rotational Inertia of a Wheel with a Horizontal Spoke

[lehel]

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

 

[lehel]

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^26(10/32) lb = 10lb - V

V = 8lb which is the wrong answer :(

 

[Moetasim]

the values in the equation for rotational inertia at the axle, we get:

Rotational Inertia at Axle = 1/3(10/32 slugs)(1ft)^2 + (10/32 slugs)(1ft)^2

= 0.417 slug*ft^2

Now, using the equation for torque at the axle, we get:

Torque at axle = RI * alpha = (0.417 slug*ft^2)*12 rad/s^2 = 5 lbs*ft

Since the weight of the spoke is equally distributed between the attachment on the wheel and the axle, the vertical forces must balance out. This means that the vertical axle force must be equal to the weight of the wheel (10 lbs) plus the weight of the spoke (10 lbs), which is 20 lbs.

Using the equation for sum of horizontal forces, we get:

Sum of horizontal forces = m(a-radial) = Tension of spoke

(20/32 slugs)(a-radial) = 5 lbs

a-radial = 8 ft/s^2

Now, using the equation for vertical forces, we get:

(20/32 slugs)(a-tangential) = vertical axle force - 10 lbs

(20/32 slugs)(12 ft/s^2) = vertical axle force - 10 lbs

vertical axle force = 15 lbs

Therefore, the force of the axle on the wheel is 15 lbs, which is the same as the vertical axle force. The upward force of 18 lbs mentioned in the answer may be referring to the net force on the wheel, which includes the force of the axle as well as the force of the ground pushing up on the wheel. This would make sense since the wheel is in equilibrium when it is released, meaning that the net force on the wheel is zero. In this case, the upward force of 18 lbs would be the sum of the force of the axle (15 lbs) and the force of the ground (3 lbs).