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Question: A, B, C and D are four masses each of mass m lying on the vertices of a square of side ‘a’. They alw...

A, B, C and D are four masses each of mass m lying on the vertices of a square of side ‘a’. They always move along a common circle with velocity v. Find v so that they always remain on the vertices of the square.

|F1| = |F2| F1+F2\left| \overrightarrow { \mathrm { F } } _ { 1 } + \overrightarrow { \mathrm { F } } _ { 2 } \right| = 2\sqrt { 2 } F1

A

GM(22+1)22a\sqrt { \frac { \operatorname { GM } ( 2 \sqrt { 2 } + 1 ) } { 2 \sqrt { 2 } a } }

B

GM(2+1)2a\sqrt { \frac { \mathrm { GM } ( \sqrt { 2 } + 1 ) } { 2 \mathrm { a } } }

C

D

None

Answer

GM(22+1)22a\sqrt { \frac { \operatorname { GM } ( 2 \sqrt { 2 } + 1 ) } { 2 \sqrt { 2 } a } }

Explanation

Solution

Fnet =√2 F1 + F2 =

2\sqrt { 2 } a = 2R or R = a2\frac { \mathrm { a } } { \sqrt { 2 } }

2\sqrt { 2 } Gm2a2+Gm22a2=mv2a2\frac { \mathrm { Gm } ^ { 2 } } { \mathrm { a } ^ { 2 } } + \frac { \mathrm { Gm } ^ { 2 } } { 2 \mathrm { a } ^ { 2 } } = \frac { \mathrm { mv } ^ { 2 } } { \mathrm { a } } \sqrt { 2 }

or v = GM(22+1)22a\sqrt { \frac { \operatorname { GM } ( 2 \sqrt { 2 } + 1 ) } { 2 \sqrt { 2 } a } }