Question

Question 3 10 pts A horizontal disk with moment of inertia 0.36 kg-m2 is rotating with an angular speed of 6.5 rad/sec. A point mass of 0.52 kg is gently placed on the outer edge of the disk in a manner so that no torque is applied. The mass then rotates with the disk at an angular speed of 4.37 rad/sec. What is the radius of the disk in meters? 0.38 0.28 0.88 0.58

Answer 1

NO TORQUE IS ACTING, so by conservation of angular momentum;

I₁w₁=I₂w₂------------------(1)

I₁=0.36 kg m2

w₁=6.5 rad/s

w₂=4.37rad/s

putting the value in eq (1)

0.36*6.5=I₂*4.37

I₂=0.53 kg m²---------------------(2)

but I₂= M₂r² and I₁=M₁r²

M₂= (M₁+0.52) kg-----------------(3)

using I₁=M₁r²

0.36=M₁r²-------------------------(4)

I₂= M₂r²

0.53=(M₁+0.52)r²----------------(5)

dividing 5 by 4 we get,

$1.47=\frac{M_{1}+0.52}{M_{1}}$

solving this 1.47M₁=M₁+0.52

0.47M₁=0.52

M₁=1.10 Kg------------------------(6)

put in eq 4

r²=0.36/1.10
r=0.57 m