Question 190: Centre of Mass and Rotational Motion

Question

A thin circular ring of mass M and radius r is rotating about its axis with an angular speed $$\omega$$. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become :

Accepted Answer

$$	au$$
net
= 0, so angular momentum is conserved
By angular momentum conservation
I
i
$$\omega$$
i
= I
f
$$\omega$$
f
(MR
2
)$$\omega$$ = (MR
2
+ 2mR
2
)$$\omega$$
f
$$\omega$$
f
= $${{(M{R^2})\omega } \over {M{R^2} + 2m{R^2}}} = {{M\omega } \over {M + 2m}}$$
$${\omega _f} = {{M\omega } \over {M + 2m}}$$

Answer

$$\omega {M \over {M + m}}$$

Answer

$$\omega {{M + 2m} \over M}$$

Answer

$$\omega {M \over {M + 2m}}$$

Answer

$$\omega {{M - 2m} \over {M + 2m}}$$

Centre of Mass and Rotational Motion

A thin circular ring of mass M and radius r is rotating about its axis with an angular speed $\omega$. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become :

Choose the correct answer: