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Two thin circular discs sof mass 2 kg an...

Two thin circular discs sof mass `2 kg` and radius `10 cm` each are joined by a rigid massless rod of length `20 cm`. the axis of the rod is perpendicular to the plane of the disc through their centres as shown in the figure. The object is kept in a truck in such as way that the axis of the objects horizontal and perpendicular to the directiion of motion of the truck. Its friction with the floor of the truck. Its friction with the floor of the truck is large enough to prevent slipping. If the tuck has an acceleration of `9 ms^(-2)`. Calculate

a. The force of friction on each disc.
b. The magnitude and direction of the frictional torque acting on each disc about the centre of mass `O` of the object. Take `x`-axis along the direction of the motion of the truck and `z`-axis along vertical upwards directio. Express the torque in the vector form in terms of unit vectors `hati,hatj and hatk` in the `x,y` and `z` direction respectively.
c. Find the minimum value of the coefficient of friction between the object and the floor of the truck which makes rolling of the object possible.

Text Solution

Verified by Experts

(i) 6, (ii) 0.6, (iii) 0.3
(i) FBD of the object with respect to truck.
In the reference from of truck it experiences a pseudo.
`vecF =-2mahati`

where a = acceleration of the truck.
Pseudo force does not provide torque about the centre ofthe disc. Because of this force, object has tendency to slidle along -ve x-axis. Hence, frictional force will act along +ve x-axis.
For no slipping, `alpha =(a^('))/(R)`
Hence, `f.R =2(mR^(2))/2(a^('))/R rArr a^(') = f/m` ...........(ii)

from eqs. (i) and (ii) we get `f=2/3 m.a`
Therefore, force of friction on each disc is `f/2 =(ma)/3 = 6N`
(ii) `vecf_(1) = 6hati , vecr_(1) =-0.1hatj - 0.3 hatk`
`rArr vectau(f_(1)) =vecr_(1) xx vecf_(1) =(-1)hatj -0.1hatk xx 6hati =0.6hatj + 0.6hatk, vecf_(2) =6hati`,
`vecr_(2) =0.1hatj + 0.1hatk`
`rArr vectau(f_(2)) = vecr_(2) xx vecf_(2) -(0.1 hatj +0.1 hatk) xx 6hati =-0.6 hatj -0.6 hatk`
(iii) Minimum value of frictionless force is `2mu mg`
`rArr 2/3 ma le 2 mu mg rArr mu ge a/(3g) ge 0.3`
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