Answer:
559.5 N at the bottom and 519.6 N at the top of the wheel
Explanation:
If it completes 1 revolution (or 2π rad) per 28s then its angular speed is
[tex]\omega = 2\pi/28 = 0.224 rad/s[/tex]
The centripetal acceleration would be:
[tex]a_c = \omega^2 R = 0.224^2*7.2 = 0.363 m/s^2[/tex]
Let gravitational acceleration g = 9.81 m/s2.
At the bottom of the wheel the net acceleration would be g plus the centripetal acceleration:
[tex]a_b = a_c + g = 9.81 + 0.363 = 10.17 m/s^2[/tex]
So the weight at the bottom of the wheel would be
[tex]W_b = a_b*m = 10.17*55 = 559.5 N[/tex]
Similarly at the top of the wheel the net acceleration is g subtracted by the centripetal acceleration:
[tex]a_t = g - _a_c = 9.81 - 0363 = 9.45 m/s^2[/tex]
And the weight at the top is
[tex]W_t = a_t*m = 9.45*55 = 519.6 N[/tex]
