SATHEE: Rotation 3 Question 24

Rotation 3 Question 24

30. Two thin circular discs of mass $2 k g$ and radius $10 c m$ each are joined by a rigid massless rod of length $20 c m$ . 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 $9 m / s^{2}$ , calculate

$(1997, 5$ M) (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.

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Answer:

Correct Answer: 30. (a) $6 \hat{i}$

(b) $0.6 (\hat{k} - \hat{j}), 0.6 (- \hat{j} - \hat{k}), 0.85 N - m$

Solution:

Given, mass of disc $m = 2 k g$ and radius $R = 0.1 m$

(a) FBD of any one disc is

Frictional force on the disc should be in forward direction.

Let $a_{0}$ be the linear acceleration of $C M$ of disc and $α$ the angular acceleration about its CM. Then,

$\begin{aligned} a_{0} = \frac{f}{m} = \frac{f}{2} \\ α = \frac{τ}{I} = \frac{f R}{\frac{1}{2} m R^{2}} = \frac{2 f}{m R} = \frac{2 f}{2 \times 0.1} = 10 f \end{aligned}$

Since, there is no slipping between disc and truck. Therefore,

$\begin{aligned} ∴ α_{0} + R α = a or \frac{f}{2} + (0.1) (10 f) = a \\ or \frac{3}{2} f = a \Rightarrow f = \frac{2 a}{3} = \frac{2 \times 9.0}{3} N \\ ∴ f = 6 N \end{aligned}$

Since, this force is acting in positive $x$ -direction.

Therefore, in vector form $f = (6 \hat{i}) N$

(b) $τ = r \times f$

Here, $f = (6 \hat{i}) N$ ( for both the discs)

$and \begin{aligned} r_{P} & = r_{1} = - 0.1 \hat{j} - 0.1 \hat{k} \\ r_{Q} & = r_{2} = 0.1 \hat{j} - 0.1 \hat{k} \end{aligned}$

Therefore, frictional torque on disk 1 about point $O$ (centre of mass).

$\begin{aligned} τ_{1} = r_{1} \times f = (- 0.1 \hat{j} - 0.1 \hat{k}) \times (6 \hat{i}) N - m \\ = (0.6 \hat{k} - 0.6 \hat{j}) \\ or \\ τ_{1} = 0.6 (\hat{k} - \hat{j}) N - m \\ and | τ_{1} | = \sqrt{(0.6)^{2} + (0.6)^{2}} \\ = 0.85 N - m \end{aligned}$