Question

a solid disk rotates in the horizontal plane at an angular velocity of 4.9 x 10 rad/s with respect to an axis perpendicular to the disk at its center the moment of inertia of the disk is 0.14 kg from above sand is dropped straight down onto this rotating disk so that a thin unifrom ring of sand is formed at distance of 0.4 m from the axis the sand in the ring has mass of 0.5 kg after all the sand is in place what is the angular velocity of the disk

Answer 1

ωf = 3.1*10 rad/sec

Step-by-step explanation:

• Assuming no external torques acting while the sand is being dropped, total angular momentum must keep constant.
• So we can write the following equality:

`L_(o) = L_(f) (1)`

• For a rigid body rotating with respect to an axis, the angular momentum can be written as follows:

`L = I* \omega (2)`

• where I = moment of inertia
• ω = angular velocity

• Replacing (2) on both sides of (1) we get:

`I_(o)* \omega_(o) = I_(f)* \omega_(f) (3)`

• In (3) we know the values of I₀ and ω₀ (since they are givens), but we need to find the value of If first.
• The final moment of inertia, will be equal to the sum of the initial one, plus the one due to the ring of sand, that also rotates with respect to an axis perpendicular to the disk, as follows:
• 
  `I_(f) = I_(o) + I_ {ring} (4)`
• The moment of inertia of a circular ring is as follows:

`I_(ring) = m_(ring) *r^(2) (5)`

• Replacing by the givens in (5) we get:

`I_(ring) = m_(ring) *r^(2) = 0.5 kg * (0.4m)^(2) = 0.08 kg*m2 (6)`

• Replacing (6) in (4):
• 
  `I_(f) = I_(o) + I_ {ring} = 0.14kg*m2 + 0.08 kg*m2 = 0.22 kg*m2 (7)`

• Replacing I₀, ω₀ and If in (3), we can solve for ωf, as follows:
• 
  `\omega_(f) =(I_(o) *\omega_(o) )/(I_(f) ) = (0.14kg*m2*4.9*10rad/sec)/(0.22kg*m2) = 3.1*10 rad/sec (8)`