Question

Two thin circular discs of 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 along the perpendicular to the planes of the disc 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 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 $9m/s^2,$ calculate
(a) the force of friction on each disc and
(b) the magnitude and direction of the frictional torque acting on each disc about the centre of mass O of the object. Express the torque in the vector form in terms of unit vectors $\hat{i},\hat{j}$ and $\hat{k}$ in x,y and z-directions.

Answer 1

Given, mass o f disc $m=2kg$ and radius $R=0.1m$
(a) FBD o f any one disc is
Frictional force on the disc should be in forward direction.
Let a0 be the linear a cceleration o f CM o f disc and a the angular acceleration about its CM. Then,
$a_0=\frac{f}{m}=\frac{f}{2}$....(i)
$\alpha =\frac{\tau }{I}=\frac{fr}{\frac{1}{2}m{R}^2}=\frac{2f}{mR}=\frac{2f}{2\times 0.1}=10f$....(ii)
Since, there is no slipping between disc and truck.
Therefore,
$\therefore {\alpha }_0+R_{\alpha }=a$ or $\left(\frac{f}{2}\right)+(0.1)(10f)=a$
or $\frac{3}{2}f=a\Rightarrow f=\frac{2a}{3}=\frac{2\times 9.0}{3}N$
$\therefore f=6N$
Since, this force is acting in positive x-direction.
Therefore, in vector form $f=(6\hat{i})N$
(b) $\tau =r\times f$
Here, $f=(6\hat{i})N$ ( for both the discs)
$r_P=r_1=-0.1\hat{j}-0.1\hat{k}$
and $r_Q=r_2=-0.1\hat{j}-0.1\hat{k}$
Therefore, frictional torque on disk 1 about point O (centre o f mass).
${\tau }_1=r_1\times f=(-0.1\hat{j}-0.1\hat{k})\times (6\hat{i})N-m$
$=(0.6\hat{k}-0.6\hat{j})$
or ${\tau }_1=0.6(\hat{k}-\hat{j})N-m$
and $|{\tau }_1|=\sqrt{(0.6{)}^2+(0.6{)}^2}$
$=0.85N-m$
Similarly, ${\tau }_2=r_2\times f=0.6(-\hat{j}-\hat{k})$
and $|{\tau }_1|=|{\tau }_2|=0.85N-m$