Question

figure shows four rotating disks that are sliding on a frictionless floor.three forces f ,2f or 3f act on each disk either at the rim, at the centre or halfway between rim and centre which disk is in equilibrium

Answer 1

Given:

Given figure contains four rotating discs which are sliding on a frictionless floor. Various forces are acting at various distances from the centre on the discs.

To find:

The discs which are in equilibrium.

Calculation:

In disc a)

$\rm{ \sum(Force) = F + 2F - 3F = 0}$

Hence, disc is in translational equilibrium.

$\rm{ \sum( \tau) = Fr - 2F( \frac{r}{2}) + 3F(0) }$

$= > \rm{ \sum( \tau) = Fr - Fr }$

$= > \rm{ \sum( \tau) = 0 }$

Hence, the disc is in Rotational Equilibrium too.

Similarly in disc c)

$\rm{ \sum(Force) = F + F - 2F = 0}$

Hence, disc is in translational equilibrium.

$\rm{ \sum( \tau) = Fr - Fr + 2F(0) }$

$= > \rm{ \sum( \tau) = Fr - Fr}$

$= > \rm{ \sum( \tau) = 0}$

Hence, disc is in rotational equilibrium too.

So, final answer is:

Disc a) and c) are in total equilibrium.