Question

The 180-mm disk is at rest when it is placed in contact with a belt moving at a constant speed. Neglecting the weight of the link AB and knowing that the coefficient of kinetic friction between the disk and the belt is 0.54, determine the angular acceleration of the disk while slipping occurs.

Answer 1

Answer:

$\mathbf{\alpha = 25.88 \ rad/s^2 }$

Explanation:

Consider the force due to friction:

$F = \mu_kN -------- (1)$

where;

N = normal reaction

$\mu_k$ = coefficient of kinetic friction.

Via the horizontal direction, the forces of equilibrium are:

$\sum F_x = \sum(F_x) _{eff} \\ \\ N- F_{AB} \ cos \theta = 0 \\ \\ F_{AB} \ cos \theta = N ------ (2)$

where;

$\theta$ = angle of AB is associated with the horizontal;

$F_{AB}$ = force exerted by AB

Let's take a look at the equilibrium of forces along the vertical direction.

$\sum F_y = \sum (F_y)_{eff} \\ \\ F+F_{AB} sin \theta - W = 0\\ \\ \mu_k N + F_{AB} sin \theta -mg =0 \\ \\ F_{AB} sin \theta = mg - \mu_kN-------(3)$

By dividing (3) by (2), we get:

$\dfrac{F_{AB} sin \theta }{F_{AB} cos \theta} = \dfrac{mg - \mu_k N}{N} \\ \\ tan \theta = \dfrac{mg-\mu_kN}{N} \\ \\ Ntan \theta = mg - \mu_kN \\ \\ N(tan \theta + \mu_k ) = mg \\ \\ N = \dfrac{mg}{tan \theta + \mu_k}$

By replacing the obtained value of N into:

$F = \mu_k N$

we have:

$F = \mu_k (\dfrac{mg}{tan \theta + \mu_k}) \\ \\ \\ F = \dfrac{mg \mu_k}{tan \theta + \mu_k}$

Moment about center A can be expressed as:

$\sum M_A = \sum (M_A) _{eff} \\ \\ Fr = I \alpha\\ \\ \alpha = \dfrac{Fr}{I}$

where;

$\alpha$ = angular acceleration of the disc

I = mass moment of inertia

r = radius of disc

By replacing F and I; ∝ becomes:

$\alpha = \dfrac{\bigg( \dfrac{mg \mu_k}{tan \theta + \mu_k} \bigg) r}{\dfrac{1}{2}mr^2}$

$\alpha = \dfrac{2g}{r} \dfrac{\mu_k}{tan \ \theta + \mu_k}------- (4)$

where;

g = 9.8 m/s²

r = 0.18 m

$\mu_k = 0.54$

θ = 60°

$\alpha = \dfrac{2(9.8)}{0.18} (\dfrac{0.54}{tan \ 60+ 0.54})$

$\mathbf{\alpha = 25.88 \ rad/s^2 }$