Question

A ring of radius R is rotating with an angular speed $\omega _ { 0 }$ about a horizontal axis. It is placed on a rough horizontal table. The coefficient of kinetic friction is.The time after which it starts rolling is

• A. $\frac { \omega _ { 0 } \mu _ { \mathrm { k } } R } { 2 \mathrm {~g} }$
• B. $\frac { \omega _ { 0 } \mathrm {~g} } { 2 \mu _ { \mathrm { k } } R }$
• C.
• D.

Answer 1

Answer: (D)

Answer 2

Acceleration produced in the centre of mass due to friction

$\mathrm { a } = \frac { \mathrm { f } } { \mathrm { M } } = \frac { \mu _ { \mathrm { k } } \mathrm { Mg } } { \mathrm { M } } \mu _ { \mathrm { k } } \mathrm { g }$

Where M is the mass of the ring, …(i)

Angular retardation produced by the torque due to friction

$\alpha = \frac { \tau } { \mathrm { I } } = \frac { \mathrm { fR } } { \mathrm { I } } = \frac { \mu _ { \mathrm { k } } \mathrm { MgR } } { \mathrm { I } }$ …(ii)

As v = u + at

(Using (i))

As $\omega = \omega _ { 0 } + \alpha t$

(Using (ii))

For rolling without slipping

V = $\mathrm { R } \omega$

$\therefore \frac { \mathrm { v } } { \mathrm { R } } - \omega _ { 0 } - \frac { \mu _ { \mathrm { k } } \mathrm { MgR } } { \mathrm { I } } \mathrm { t }$

$\frac { \mu _ { \mathrm { k } } \mathrm { gt } } { \mathrm { R } } = \frac { \omega _ { 0 } } { 1 + \frac { \mathrm { MR } ^ { 2 } } { \mathrm { I } } } \Rightarrow \mathrm { t } = \frac { \mathrm { R } \omega _ { 0 } } { \mu _ { \mathrm { k } } \mathrm { g } \left( 1 + \frac { \mathrm { MR } ^ { 2 } } { \mathrm { I } } \right) }$

For ring, $\mathrm { I } = \mathrm { MR } ^ { 2 }$

$\therefore \mathrm { t } = \frac { \mathrm { R } \omega _ { 0 } } { \mu _ { \mathrm { k } } \mathrm { g } \left( 1 + \frac { \mathrm { MR } ^ { 2 } } { \mathrm { MR } ^ { 2 } } \right) } = \frac { \mathrm { R } \omega _ { 0 } } { 2 \mu _ { \mathrm { k } } \mathrm { g } }$