Question

A particle (m = 2 kg) located at point (3m, 0,2m)
has an angular velocity (3i - 4j+ 2k) rad/s about
point P(1m, -1m, 3m). What will be the Kinetic
energy of the particle?​

Answer 1

since the body is rotating about point p....

so,p will be the centre of circle...

coordinates of p(1,-1,3)

point on circle=point on which it is located=(3,0,2)

now

take out r using distance formula

r={(1-3)^2+(-1)^2+(3-2)^2}^1/2

r=(4+1+1)^1/2

r=√6

we know

kinetic energy=(1/2)×I×W^2

here,I represent moment of inertia,W represent angular velocity

I=MR^2

I=2×(√6)^2

I=12

W=(9+16+5)^1/2

W=√30

put the value of I and W in KE.. formula

KE=1/2×12×(√30)^2

KE=6×30

KE=180

Answer 2

Answer:

K. E =180

Explanation:

I=mr^2=12. W=30. K. E =1/2mr^2=180