Find out the Acceleration of the mass centre:
A wheel of mass 200 kg and a diameter of 700 mm rolls without slipping down a plane inclined at an angle of 20^{o} with the horizontal. Find out the friction force F and the acceleration of the mass centre.
Solution
It is rather difficult to indicate the direction of F (F can have any value between - μN and μN. In this case, the friction should act up the plane otherwise the wheel shall slip down the plane. Also, we should keep in mind that friction is the only force which may cause moment about mass centre and, thus, it causes angular acceleration. Further, we should keep in mind that F shall not be equal to μN.
Choosing x axis parallel to the plane with the positive direction downward, the y axis is positive up.
∴ I = (½) mr ^{2} = 1 × 200 × (0.35)^{2} = 12.25 kg m^{2}.
Now equilibrium equations may be written as below:
∑ F_{x} = 0, mg sin θ - F - ma_{x} = 0-------- (1)
∑ F_{y} = 0, - mg cos θ + N = 0---------- (2)
∑ M c = 0, F. r - I α = 0------ (3)
∴ 0.35 F = 12.25 α
And further, we may write a_{x} = 0.35 α.
Substituting this in Eq. (1), we get
200 × 9.81 × sin 20^{o} = F + 200 × 0.35 α
671.04 = F + 70 α
∴ F = 671.04 - 70 α
And F = (12.25 /0.35) α = 35 α
∴ 671.04 - 70 α = 35 α
∴ 105 α = 671.04
α = 6.39 rad / sec^{ 2}
and F = 223.68 N
and a_{x} = 0.35 × 6.39
= 2.23 m/sec^{2}