Question

Two point masses having equal mass$\text{'}m\text{'}$are clamped on the disc at a distance$2m$and$4m$from the axis of rotation, which has an angular speed$\text{'}\omega \text{'}.$
Ratio of ${(K.E)}_A:{(K.E)}_B$is.

• A. 1:4
• B. 2:1
• C. 1:2
• D. 4:1

Answer 1

The correct option is A 1:4

Kinetic energy = $\frac{1}{2}m{v}^2=\frac{1}{2}m{r}^2{\omega }^2$
for 'A', $(K.E{)}_A=\frac{1}{2}m(2{)}^2{\omega }^2=2m{\omega }^2$
for 'B', $(K.E{)}_B=\frac{1}{2}m(4{)}^2{\omega }^2=8m{\omega }^2$

Note: As discussed in the videos, angular velocity of both the particles will be same.
$\frac{(K.E{)}_A}{(K.E{)}_B}=\frac{2m{\omega }^2}{8m{\omega }^2}=\frac{1}{4}$