Question

For a particle of a rotating rigid body, `v=r\omega.` So (1) `\omega \propto (1/r)` (2) `\omega \propto v` (3) `v \propto r` (4) `\omega` is independent of `r` Which options are correct?

Answer 1

1, 2, and 3

Explanation:

let's start by looking at a linear equation `y=mx`

if an equation has this format `y` `proportional` `x` or y is proportional to x

(1) rearrange the equation to solve for omega by dividing both sides by r
`v=r*omega`

`v/r=(r*omega)/r`

`v/r=omega`

here we see that `omega` is inversely proportional to r because `omega` is in the numerator and r is in the denominator on the other side of the equation.

(2) use the original equation `v=r*omega` to see how `v` is related to `omega`
they are both in the numerator, so `v proportional omega`

(3)use the original equation `v=r*omega` to see how `v` is related to `r`
they are both in the numerator, so `v proportional r`

(4) in the equation, `v` `r` and `omega` are all variables related to one another, so if (1) is true and `omega` is inversely proportional to r, then it is not independent of it too.

Answer 2

(3) & (4)

Explanation:

For a rotating rigid body, rotating about an axis with angular velocity `bbomega`, every particle making up the rigid body will have angular velocity `bbomega`.

Consider the statements in terms of 2 particles at `bbr_1` and `bbr_2`, with `bbr_1 ne bbr_2`.

(1)

As both particles have same angular velocity `bbomega`, then this is clearly not true

(2)

As both particles have same angular velocity `bbomega`, then:

`{(bbv_1 = bbomega bbr_1),(bbv_2 = bbomega bbr_2):} qquad square`

This statement is not true

(3)

On the other hand, because of the relation in
`square`, the statement `v propto r` is true

(4)

`bb omega` is the same for all particles irrespective of the value of `bbr`, so this statement is also true