The 15 kg wheel has a radius of gyration about its center O of kO = 300 mm. When the wheel is subjected to the couple moment, it slips as it rolls. Determine the angular acceleration of the wheel and the acceleration of the wheel's center O. The coefficient of kinetic friction between the wheel and the plane is μk = 0.6

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porkchop6941

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The 15 kg wheel has a radius of gyration about its center O of kO = 300 mm. When the wheel is subjected to the couple moment, it slips as it rolls. Determine the angular acceleration of the wheel and the acceleration of the wheel's center O. The coefficient of kinetic friction between the wheel and the plane is μk = 0.6.

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kollybaba55

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Answer:
Angular acceleration of the wheel, $\alpha = 47.9 rad/s^2$
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:
$I_{0} = mk_{0} ^{2} \\I_{0} = 15 * 0.3^2\\I_{0} = 1.35 kg m^2$
Taking moment about point O in the diagram with the horizontal force being F.
$\sum M_{0} = I_{0} \alpha$
$M - F*r = 1.35 * \alpha\\100 - (\mu mg)*0.4 = 1.35 * \alpha\\100 - (0.6* 15*9.81)*0.4 = 1.35 * \alpha\\64.684 = 1.35 * \alpha\\\alpha = 64.684/1.35\\\alpha = 47.9 rad/s^2$
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