Two particles A and B are moving on two concentric circles of radii R1 and R2 with equal angular speed - At t-0- their positions and direction of motion are shown in the figure -The relative velocity vA-vB at t-2- is given by -

Using, θ=ω t we get, θ=ωπ2ω=π2 So, both have completed quarter circle Relative velocity, v_A v_B=ω R_1( î) ω R_2( î) ∴ nbsp;v_A v_B nbsp;=ω(R_2 R_1)î ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Physics

Conduction Law

Two particles...

Question

Two particles $A$ and $B$ are moving on two concentric circles of radii $R_{1} and R_{2}$ with equal angular speed $ω$ . At $t = 0$ , their positions and direction of motion are shown in the figure :

The relative velocity $_{A} -_{B} at t = \frac{π}{2 ω}$ is given by :

ω (R_{1} - R_{2})^i

No worries! We‘ve got your back. Try BYJU‘S free classes today!

ω (R_{1} + R_{2})^i

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- ω (R_{1} + R_{2})^i

No worries! We‘ve got your back. Try BYJU‘S free classes today!

ω (R_{2} - R_{1})^i

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $ω (R_{2} - R_{1})^i$
Using, $θ = ω t$ we get,

$θ = ω \frac{π}{2 ω} = \frac{π}{2}$

So, both have completed quarter circle

Relative velocity,

${\to v}_{A} - {\to v}_{B} = ω R_{1} (-^i) - ω R_{2} (-^i)$

$∴ {\to v}_{A} - {\to v}_{B} = ω (R_{2} - R_{1})^i$

Hence, option $(C)$ is the correct answer.

Suggest Corrections

Similar questions

Q. Two particles

A

and

B

are moving on two concentric circles of radii

R_{1}

and

R_{2}

with equal angular speed

ω

. At

t = 0

, their positions and direction of motion are shown in the figure. The relative velocity

_{A} -_{B}

t = \frac{π}{2 ω}

is given by:

Q. Two particles

A

and

B

are moving with equal angular speed

ω

as shown in figure. At

t = 0

, their positions are shown. Then, relative velocity

_{A} -_{B}

t = \frac{π}{2 ω}

Q. Two particles

A

and

B

are moving with equal angular speed

ω

as shown in figure. At

t = 0

, their positions are shown. Then, relative velocity

_{A} -_{B}

t = \frac{π}{2 ω}

Q. Two particles

A

and

B

are moving on two concentric circles of radii

R_{1}

and

R_{2}

respectively with equal angular speed

ω rad/s

. At

t = 0

, their positions and linear velocities are shown in the figure. Then relative velocity

(m/s

) of

A

with respect to

B

(i.e

_{A B})

t = \frac{π}{2 ω} s

is given by:

Q. Two particles

A

and

B

are moving in a horizontal place anticlockwise on two different concentric circles with different constant angular velocities

2 ω

and

ω

respectively. Find the relative velocity (in

m s

) of

B

w.r.t

A

after time

t = π / ω

. They both start at the position as shown in figure (Take

ω = 3 r a d s e c, r = 2 m

)

Join BYJU'S Learning Program

Two particles A and B are moving on two concentric circles of radii R1 and R2 with equal angular speed ω. At t=0, their positions and direction of motion are shown in the figure : The relative velocity −→vA−−→vB at t=π2ω is given by :

The correct option is D ω(R2−R1)^iUsing, θ=ωt we get, θ=ωπ2ω=π2 So, both have completed quarter circle Relative velocity, →vA−→vB=ωR1(−^i)−ωR2(−^i) ∴→vA−→vB=ω(R2−R1)^i Hence, option (C) is the correct answer.

Two particles $A$ and $B$ are moving on two concentric circles of radii $R_{1} and R_{2}$ with equal angular speed $ω$ . At $t = 0$ , their positions and direction of motion are shown in the figure :

The relative velocity $_{A} -_{B} at t = \frac{π}{2 ω}$ is given by :