Question

anyone solve it... please please... xd... ☺☺☺☺☺​

Answer 1

SoluTion :

✴ Given :-

▪ Radius of outer ring = 3R

▪ Radius of inner disc = 2R

▪ Angular velocity of ring = ω(Clockwise)

▪ Angular velocity of disc = ω/2

(Anti-clocewise)

▪ Distance between centre and P point = R

▪ OP makes angle 30° with horizontal

✴ To Find :-

▪ Linear velocity of P point...

✴ Diagram :-

▪ Please see the attached image for better understanding.

✴ Calculation :-

✒ Since, outer ring is rolling without slipping, the point on the ring in contact with the ground (A) is at rest,

• i.e. $\sf\vec{V_A}$ = 0

✒ Thus, velocity of the point O is given by

$\sf\vec{V_O}=\vec{V_A}+(\vec{\omega_O}\times \vec{r_{AO}})\\ \\ \sf\vec{V_O}=0+(\omega\hat{j}\times 3R\hat{k})\\ \\ \sf\red{\vec{V_O}=3R\omega\hat{i}}$

✒ The point P lies on the inner disc having an angular velocity

• $\sf\vec{\omega_i}=-\dfrac{\omega}{2}\hat{j}$

✒ The position vector from O to P is

$\sf\vec{r_{OP}}=R\cos30\degree\hat{i}+R\sin30\degree\hat{k}\\ \\ \sf\vec{r_{OP}}=\dfrac{\sqrt{3}R}{2}\hat{i}+\dfrac{R}{2}\hat{k}$

✒ Thus, velocity of the point P is given by

$\sf\vec{V_P}=\vec{V_O}+(\vec{\omega_i}\times \vec{r_{OP}})\\ \\ \sf\vec{V_P}=3R\omega\hat{i}+[(-\dfrac{\omega}{2}\hat{j})\times (\dfrac{\sqrt{3}R}{2}\hat{i}+\dfrac{R}{2}\hat{k})]\\ \\ \bigstar\:\boxed{\tt{\purple{\vec{V_P}=\dfrac{11\omega{R}}{4}\hat{i}+\dfrac{\sqrt{3}\omega{R}}{4}\hat{k}}}}$

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✒Option-A) and B) both are correct

❄ Nice Question ...

Answer 2

Answer: answer is in attachment

Explanation: