Question

A rigid uniform circular disc rolls without slipping on a horizontal surface with uniform speed along the positive x-direction as shown.
$AC$ and $BD$ are the two diameters and at the instant shown in diagram $AC$ is horizontal and $BD$ is vertical then choose the correct options at the instant shown. ( $O$: centre of the disc)

• A. Sector BOC has greater kinetic energy than sector COD with respect to ground
• B. Sector ABC has greater kinetic energy than sector ADC with respect to ground
• C. Sector BOC has same kinetic energy as sector AOB with respect to ground
• D. Sector BOC has same kinetic energy as sector AOD with respect to the centre of mass frame of disc

Answer 1

Answer: (A), (B), (C)

Sector BOC has greater kinetic energy than sector COD with respect to ground
Sector ABC has greater kinetic energy than sector ADC with respect to ground
Sector BOC has same kinetic energy as sector AOB with respect to ground

Let 'dm' be the small particles

K.E. = $\frac{1}{2}\left(dm\right)v^2$

$\therefore v_{net}=\sqrt{v^2+w^2{R}^2+2vwR\cos \theta }$

In section ABC we have $(\theta <{90}^o)$

$\therefore v_{net}$ will greater $w\left(\cos \theta \right)$ have positive value.

In section ADC we have $(\theta >{90}^o)$

$\therefore v_{net}$ will decrease $w\left(\cos \theta \right)$ have negative value

$\Rightarrow$ We can say that,

$(K.E.{)}_{ABC}>(K.E.{)}_{ADC}$

As in sector BOC $\theta <{90}^o$ and in sector $COD\theta >{90}^o$

$\therefore (K.E.{)}_{BOC}>(K.E.{)}_{COD}$

As in sector AOB and BOC, $\theta <{90}^o$

$\therefore (K.E.{)}_{AOB}=(K.E{)}_{BOC}$

$\therefore$ the correct option is,

A) Sector BOC has greater kinetic energy than sector COD with respect to ground.

B) Sector ABC has greater kinetic energy than sector ADC.

C) Sector BOC has same kinetic energy as sector AOB.