Question

To measure the moment of inertia of a wheel-shaft system, a tape of negligible mass is wrapped around the shaft and pulled with a known constant force 'F' as shown in the figure. When a length '$l_0$' of the tape has unwound, the system has an angular speed '$\omega_0$'. If the moment of inertia 'I' of the wheel shaft system about an axis through 'C' is equal to $\frac{mR^2}{K}$. Then find the value of K. (take $l_0$ = 2R, r = R/2, F/$\omega_0^2$ = mR/9)

$\mu \neq 0$

Answer 1

K = 9/4

Answer 2

1. The tape unwinds through an angle $\theta$ given by

   $$\theta=\frac{l_0}{r}=\frac{2R}{R/2}=4 \, \text{radians}$$

2. In pure rotational motion, the work done by the force equals the rotational kinetic energy increase:

   $$\text{Work} = F\,l_0 = \frac{1}{2}I\,\omega_0^2.$$

   However, it is more direct to use the rotational kinematics relation for constant angular acceleration $\alpha$:

   $$\omega_0^2 = 2\,\alpha\,\theta.$$

3. The torque provided by the force acting at the shaft (of radius $r=R/2$) is

   $$\tau = F\,r = F\frac{R}{2}.$$

   And the relation between torque and angular acceleration is

   $$\tau = I\,\alpha \quad\Rightarrow\quad \alpha = \frac{F R}{2I}.$$

4. Substitute $\alpha$ into the kinematics equation:

   $$\omega_0^2 = 2\,\left(\frac{F R}{2I}\right)\,4 = \frac{4FR}{I},$$

   which rearranges to:

   $$I = \frac{4F R}{\omega_0^2}.$$

5. We are given:

   $$\frac{F}{\omega_0^2}=\frac{mR}{9}.$$

   Substitute into the expression for $I$:

   $$I = 4R\left(\frac{mR}{9}\right)=\frac{4mR^2}{9}.$$

6. Since the moment of inertia is also given by

   $$I=\frac{mR^2}{K},$$

   comparing the two expressions we get:

   $$\frac{mR^2}{K}=\frac{4mR^2}{9}\quad\Longrightarrow\quad K=\frac{9}{4}.$$