Two thin circular disks of mass 2kg and radius 10 cm each are joined by a rigid massless rod of length 20 cm. the axis of the rod is along the perpendicular to the planes of the disk through their centres. This object is kept on a truck in such a way that the axis of the object is horizontal and perpendicular to the direction of the motion of the truck. Its friction with the floor of the truck is large enough so that the object can roll on the truck without slipping. Take x axis as the direction of motion of the truck and z-axis as the vertically upwards direction. if the truck has an acceleration of `9m//s^2` Calculate:
(i) The force fo friction on each disk,
(ii) The magnitude and the direction of the frictional torque acting on each disk about the centre of mass O of the object. Express the torque in the vector form in terms of unit vectors `hati, hatj and hatk` in the x,y, and z directions.

Answer

A

Explanation

The observer, let us suppose is on the accelerated
frame. Therefore a pseudo force ma is applied individally
on each disc on the centre of mass. The firctional force is
acting in the + X direction which is producing an angular
acceleration `alpha.`

The torque acting on the
disc is
`tau = I alpah = fxxR`
`rArr f = (Ialpha)/(R ) ...(i)`
Let a' is the acceleration of
c.m. of the disc as seen by
the observer. Since the
case is of pure rolling and
from the perpective of the
observer
`a = alphaR ....(ii)`
`rArr (from (i) and (ii)`
`f = (Ia')/(R^2) ....(iii)`
Applying Newton's law for motion in X-direction
ma -f = ma'
`rArr a' = ((a - (f)/(m)) .....(iv)`
Also moment of inertia
`I= (1)/(2)mR^2 ....(v)`
From (iii), (iv) and (v)
`f=(1)/(2) (mR^2((a-(f)/(m))/(R^2) rArr 2f = ma-f`
`rArr 3f = ma rArr f= (ma)/(3) = (2xx9)/(3) = 6N (In + X direction)
`vecf = (6hati)N`
(ii) the positon vector fo point M, taking O as the origin
`vecr_m = -0.1vetj - 0.1hatk and position vector of point N`
`vecr_N = 0.1hatj - 0.1 hatk`
The torque due to friction on disc 1 about O
`vectau_1 = vecr_M xxvecf = (-0.1hatj - 0.1hatk)xx(6hati)`
`=0.6(hatk - hatj) N-m`
The torque due to friction on disc 2 about O
`vectau_2 = vecr_m xx vecf = (-0.1hatj - 0.1 hatk)xx(6hati)`
`=0.6(-hatj - hatk) N-m`
the magnitude of torque on each disc
`|tau_1| = |tau_2| = 0.6sqrt2 N_m`