Example using Mass Center Moment Form (continued)

Ex. 1 (continued)

For the hub: I_zz^C = m_hub ( r_o² - r_i² ) = (2/32.2) [ (3/12)² – (2/12)² ] slug ft²

For the entire wheel sum each part to get total: I_zz^C = 0.755 slug ft² (result)

Now the equations of motion in scalar form are as follows:

( P – F) = m a_CX (1) (from Euler’s 1^st law)
( N – W) ) = m a_CY = 0 (2) (from Euler’s 1^st
- FR = I_zz^C α (3) (from Euler’s 2^nd law)

Check for the number of unknowns. F, a_CX, N, α (4 unknowns only 3 equations).
Therefore you need more information for a solution which comes from the kinematics.
The wheel will either roll or roll and slide.

Assume that the wheel rolls without sliding. Check afterwards to see if this assumption
is correct by comparing the friction force with its maximum value.

From kinematics of acceleration between points C and D of the wheel:

a_C = a_D + a_C/D = Rω² j – Rω² j + αk x R j = - R α i

From (3) α = ( - FR) / I_zz^C So a_CX = R( - FR) / I_zz^C Put into (1).

P – F = m [ FR² / I_zz^C ] so P = F [ 1 + mR² / I_zz^C ]

And F = P / [ 1 + mR² / I_zz^C ] = 5 / [ 1 + (13/32.2) (2²/0.755) ] = 1.59 lb (result)

Check on assumption: F_max = μ N = μW = 0.2(13) = 2.6 lb so F < F_max so wheel rolls
and assumption is correct.

Now from Eq. (1),

m a_CX = P – F so a_CX = ( P – F ) / m = ( 5 – 1.59 ) / (13/32.2) = 8.44 ft/sec²

a_C = 8.44 i ft/sec² (result)

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