Question

8.
Particle A and B are moving in coplanar circular
paths centred at O. They are rotating in the same
sense. Time periods of rotation of A and B
around O are T, and T, respectively, with
T. > T. Time required for B to make one
rotation around O relative to A is :
(1) T - T
(2) T3 + TA
TBTA
(3) TB + TA
TBTA
(4) T8 - TA
1.
uniform sironlar motion the​

Answer 1

Answer:

$\frac{T_AT_B}{T_A-T_B}$

Explanation:

Time periods of rotation of A and B respectively are $T_A$ and  $T_B$

Let time required for B to make one rotation relative to A be $T_{BA}$

If the angular velocities of A and B are $\omega_A$ and $\omega_B$ respectively then

$\omega_A=\frac{2\pi}{T_A}$

And, $\omega_B=\frac{2\pi}{T_B}$

$\omega_{BA}=\omega_B-\omega_A$

$\implies \frac{2\pi}{T_{BA}}=\frac{2\pi}{T_B}-\frac{2\pi}{T_A}$

$\implies \frac{1}{T_{BA}}=\frac{1}{T_B}-\frac{1}{T_A}$

$\implies \frac{1}{T_{BA}}=\frac{T_A-T_B}{T_AT_B}$

$\implies T_{BA}=\frac{T_AT_B}{T_A-T_B}$

Hope this helps.

Answer 2

Answer:

Time periods of rotation of A and B respectively are T_AT

A

and T_BT

B

Let time required for B to make one rotation relative to A be T_{BA}T

BA

If the angular velocities of A and B are \omega_Aω

A

and \omega_Bω

B

respectively then

\omega_A=\frac{2\pi}{T_A}ω

A

=

T

A

2π

And, \omega_B=\frac{2\pi}{T_B}ω

B

=

T

B

2π

\omega_{BA}=\omega_B-\omega_Aω

BA

=ω

B

−ω

A

\implies \frac{2\pi}{T_{BA}}=\frac{2\pi}{T_B}-\frac{2\pi}{T_A}⟹

T

BA

2π

=

T

B

2π

−

T

A

2π

\implies \frac{1}{T_{BA}}=\frac{1}{T_B}-\frac{1}{T_A}⟹

T

BA

1

=

T

B

1

−

T

A

1

\implies \frac{1}{T_{BA}}=\frac{T_A-T_B}{T_AT_B}⟹

T

BA

1

=

T

A

T

B

T

A

−T

B

\implies T_{BA}=\frac{T_AT_B}{T_A-T_B}⟹T

BA

=

T

A

−T

B

T

A

T

B

Hope this helps.