Answer :
Answer:
Angular acceleration of the wheel, [tex]\alpha = 47.9 rad/s^2[/tex]
Acceleration of the wheel's center O, a = 5.886 m/s²
Explanation:
The mass of the wheel, m = 15 kg
Let N = Normal reaction
μ = 0.6
Let the force acting horizontally be = F ( i.e the frictional force between the wheel and the plane)
Taking vertical equilibrium of forces: N - mg = 0
N - (15*9.81) = 0
N = 15 * 9.81
N = 147.15 N
F = ma ( where a is the horizontal acceleration)
F = 15a...........(1)
Taking horizontal equilibrium of forces: F -μN = 0
15a - 147.15(0.6) = 0
15a =88.29
a = 88.29/15
a = 5.886 m/s²
b) From the diagram, M = 100 Nm
Radius of the wheel, r = 0.4 m
Radius of gyration, k₀ = 300 mm = 0.3 m
The moment of inertia passing through the point O is given by:
[tex]I_{0} = mk_{0} ^{2} \\I_{0} = 15 * 0.3^2\\I_{0} = 1.35 kg m^2[/tex]
Taking moment about point O in the diagram with the horizontal force being F.
[tex]\sum M_{0} = I_{0} \alpha[/tex]
[tex]M - F*r = 1.35 * \alpha\\100 - (\mu mg)*0.4 = 1.35 * \alpha\\100 - (0.6* 15*9.81)*0.4 = 1.35 * \alpha\\64.684 = 1.35 * \alpha\\\alpha = 64.684/1.35\\\alpha = 47.9 rad/s^2[/tex]
