Question

A uniform disc of mass 6 kg and radius 20 cm is kept as shown. Find tension in the string just after the is released. String is attached at point A of disc. Such that line OA is horizontal and having value 10

Answer 1

39.2 N

Answer 2

We are given a disc of mass

$$m=6\text{ kg}$$

and radius

$$R=0.2\text{ m}.$$

A light inextensible string is attached to the disc at a point $A$ so that the line $OA$ (from the centre to the point of attachment) is horizontal and has length

$$OA=0.1\text{ m}.$$

(The diagram shows the string leaving the disc vertically upward from $A$.) When the disc is released the constraint that the string is inextensible requires that the vertical acceleration of $A$ vanishes.

Let

• $a$ be the vertical acceleration of the centre $O$ (positive upward),
• $\alpha$ be the angular acceleration (with the sign chosen so that the rotation produces an upward acceleration of $A$ due to the rotation),
• $T$ be the tension in the string, and
• $g=9.8\text{ m/s}^2$.

Step 1. Translation (Newton’s 2nd law for the disc):

The only forces acting on the disc are the weight (acting at $O$) and the tension $T$ (applied at $A$). Writing the equation in the vertical direction we have

$$T - mg = m\,a\quad \Longrightarrow\quad a = \frac{T-mg}{m}\,. \tag{1}$$

Step 2. Rotation (Torque about the centre $O$):

The weight $mg$ acts through $O$ and produces no torque about $O$. The tension $T$ acts vertically at $A$ whose position vector from $O$ is $\vec{r}_A=(0.1,0)$ (in meters). Thus the lever arm (the perpendicular distance from $O$ to the line of action of $T$) is

$$0.1\text{ m}.$$

Thus the torque is

$$\tau = T\times 0.1.$$

For a uniform disc about its centre,

$$I=\frac{1}{2}mR^2 = \frac{1}{2}\times 6\times (0.2)^2 = 0.12\text{ kg\,m}^2.$$

Thus Newton’s 2nd law for rotation gives

$$T(0.1) = I\,\alpha\quad \Longrightarrow\quad \alpha = \frac{T\,(0.1)}{0.12}\,. \tag{2}$$

Step 3. Constraint: Acceleration of point $A$ is zero

Point $A$ is attached to the inextensible string so that its vertical acceleration must vanish. The acceleration of $A$ is the sum of the translational acceleration $a$ (of the centre $O$) and the contribution from the angular acceleration:

$$a_A = a + a_{\rm rot}.$$

The rotational contribution is obtained by differentiating the position of $A$ relative to $O$. With

$\vec{r}_A = (0.1,\,0)$ and angular acceleration $\alpha$ (about $O$), we have

$$a_{\rm rot} = \alpha \times (0.1) \,,$$

where the sign is determined by geometry. (Since $OA$ is horizontal and the string is vertical, a positive $\alpha$ gives an upward contribution at $A$.)

Thus the vertical constraint is

$$a + 0.1\,\alpha = 0\,. \tag{3}$$

Step 4. Solve the equations

Substitute $a$ from (1) and $\alpha$ from (2) into (3):

$$\frac{T-mg}{m} + 0.1\left(\frac{T\,(0.1)}{0.12}\right)=0 \,.$$

Plug in $m=6$ kg and $mg=6\times 9.8 = 58.8\text{ N}$:

$$\frac{T - 58.8}{6} + 0.1\left(\frac{0.1\,T}{0.12}\right)=0\,.$$

Simplify the second term:

$$0.1\left(\frac{0.1\,T}{0.12}\right) = \frac{0.01\,T}{0.12} = \frac{T}{12}\,.$$

Thus the equation becomes:

$$\frac{T - 58.8}{6} + \frac{T}{12} = 0\,.$$

Multiply the entire equation by 12 to eliminate the denominators:

$$2(T - 58.8) + T = 0\,.$$

That is,

$$2T - 117.6 + T = 0 \quad \Longrightarrow \quad 3T = 117.6\,.$$

So,

$$T = \frac{117.6}{3} = 39.2\text{ N}\,.$$