Question

A U-shaped smooth wire has a semicircular bending between A and B as shown in the figure. A bead of mass m moving with uniform speed v through the wire enters the semicircular bend at A and leaves at B. The average force, exerted by the bead on the part AB of the wire is –

• A. 0
• B. $\frac { 4 \mathrm { mv } ^ { 2 } } { \pi \mathrm { d } }$
• C. $\frac { 2 \mathrm { mv } ^ { 2 } } { \pi \mathrm { d } }$
• D. None of these

Answer 1

Answer: (B)

$\frac { 4 \mathrm { mv } ^ { 2 } } { \pi \mathrm { d } }$

Answer 2

Q Change in momentum of bead,

Dp = 2mv and time to cover the half cycle,

Dt = $\frac { \text { Distance } } { \text { Speed } }$ = $\frac { \pi r } { v }$ = $\frac { \pi \mathrm { d } } { 2 \mathrm { v } }$

\ F = $\frac { \Delta \mathrm { p } } { \Delta \mathrm { t } }$ = $\frac { 2 \mathrm { mv } } { \pi \mathrm { d } / 2 \mathrm { v } }$ = $\frac { 4 \mathrm { mv } ^ { 2 } } { \pi \mathrm { d } }$