Question

The unbalanced wheel has a mass of 15 kg and rolls without slipping on the flat horizontal surface. The wheel has a radius of gyration about its mass center G of 64 mm, and its mass center is 40 mm from the center of the wheel. When the mass center G passes the horizontal line through O, the angular velocity of the wheel is 4 rad/s.
For this instant, compute:
a. The angular acceleration of the wheel.
b. The normal force N and the friction force F acting on the wheel at its point of contact with the horizontal surface.
c. The minimum coefficient of friction for which the wheel does not slip.

Answer 1

Answer:

a) $a_w=4.331m/s^2$

b) $F=53.7N$

c)$\mu=96.5N$

Step-by-step explanation:

From the question we are told that

mass of unbalanced wheel $M=15kg$

Radius of gyration $G=64m$

Mass center $M_c =40mm=$$4*10^{-2} m$

Angular velocity $\omega=4rads/s$

a)

Generally the equation for moment at point A is mathematically given as

$\sum M_a=I \alpha+\sum M \=ad$

Where $a_w=wheel\ accelaration$

$15*9.81*(0.04)=0.06144*\frac{a_w}{0.1}+15(0.04)(0.04)\frac{a_w}{0.1}+(15a_w-15(0.04)(4^2)0.1)$

$5,87=(a_w*0.8544+15a_0-0.96)$

$a_w=4.331m/s^2$

b)

Generally the equation for equilibrium of system on the horizontal axis is mathematically given as

$\sum f_x=ma_w$

$F=ma_w-m \=r w^2$

$F=(25*4.221)-(15*0.04*4^2)$

$F=63.3-9.6$

$F=53.7N$

c)

Generally the equation for equilibrium of system on the vertical axis is mathematically given as

$\sum f_y=ma_w$

Where

$F=\mu-mg+m \=r \alpha$

$\alpha=\frac{a_0}{0.1}$

$\mu-mg+m \=r \alpha=0$

$\mu-15*9.81+15*0.04*\frac{4.221}{0.1}$

$\mu-96.5N=0$

$\mu=96.5$