Question

Consider a large disk of radius \(R\) and two smaller disks, each of radius \[ r=\frac{R}{50} \] lying on its circumference, as shown in the figure. The smaller disks are initially in contact with each other, with an angular separation \(\Delta\theta\) between their centers. They are made to roll without slipping in opposite directions, with constant angular velocities \(\omega\) and \(2\omega\) while the large disk is held stationary. The time at which the smaller disks are again in contact is: \[ \text{[Use }\sin(\Delta\theta)=\Delta\theta\text{ and ignore gravity.]} \]

• A. \(\tau = 51\times\dfrac{\left(2\pi-\frac4{51}\right)}{\omega}\)
• B. \(\tau = 51\times\dfrac{\left(2\pi-\frac2{51}\right)}{3\omega}\)
• C. \(\tau = 51\times\dfrac{\left(2\pi-\frac4{51}\right)}{3\omega}\)
• D. \(\tau = 51\times\dfrac{\left(2\pi-\frac2{51}\right)}{\omega}\)

Hint:

For rolling without slipping: \[ v=r\omega \] Relative angular speed for opposite motions: \[ \Omega_{\mathrm{relative}} = \Omega_1+\Omega_2 \]

Answer 1

The Correct Option is B

Step 1: Find the initial angular separation.
Distance between centres of the small disks: \[ =2r \] Radius of motion of centres: \[ R+r = R+\frac{R}{50} = \frac{51R}{50} \] Using: \[ (R+r)\Delta\theta=2r \] \[ \frac{51R}{50}\Delta\theta = 2\cdot\frac{R}{50} \] \[ 51\Delta\theta=2 \] \[ \Delta\theta=\frac2{51} \]

Step 2: Find angular speeds of revolution of centres.
For rolling without slipping: \[ v=r\omega \] Angular speed of centre about large disk: \[ \Omega_1=\frac{r\omega}{R+r} \] Substituting: \[ \Omega_1= \frac{\frac{R}{50}\omega}{\frac{51R}{50}} = \frac{\omega}{51} \] Similarly for second disk: \[ \Omega_2= \frac{r(2\omega)}{R+r} = \frac{2\omega}{51} \] Since they move in opposite directions, relative angular speed: \[ \Omega=\Omega_1+\Omega_2 = \frac{3\omega}{51} \]

Step 3: Find time to meet again.
Initially separation: \[ \frac2{51} \] To come into contact again, relative angular displacement required: \[ 2\pi-\frac2{51} \] Thus: \[ \tau= \frac{2\pi-\frac2{51}} {\frac{3\omega}{51}} \] \[ = 51\times\frac{\left(2\pi-\frac2{51}\right)}{3\omega} \]

Step 4: Identify the correct option.
Therefore: \[ \boxed{ \tau= 51\times\frac{\left(2\pi-\frac2{51}\right)}{3\omega} } \] Hence correct option is: \[ \boxed{\mathrm{(B)}} \]