How Does Mass Affect Angular Velocity in Rotational Systems?

[melissa_y]

1) A 5.4m diameter merry-go-round is rotating freely with an angular velocity of 0.730 rad/s. Its total moment of inertia is 1665kg*m2. Four people standing on the ground, each of mass 60.8kg, suddenly step onto the edge of the merry-go-round. What is the angular velocity of the merry-go-round now? Use units of "rad/s".

2)A solid sphere of radius 13.0 cm and mass 12.0 kg starts from rest and rolls without slipping a distance of L = 7.0 m down a house roof that is inclined at 32o. What is the angular speed about its center as it leaves the house roof? Use units of "rad/s".

 

[anathema]

I would like to clarify a few things before providing an answer to the questions.

In the first question, it is important to know the moment of inertia of the merry-go-round with the four people standing on it. If we assume that the people are standing on the edge of the merry-go-round, then the moment of inertia would increase due to the added mass at a greater distance from the axis of rotation. Therefore, the total moment of inertia would change from the given value of 1665kg*m2. Additionally, the distribution of mass on the merry-go-round would also affect the moment of inertia. Without this information, it would be difficult to accurately determine the new angular velocity of the merry-go-round.

In the second question, it is important to know the moment of inertia of the solid sphere. The moment of inertia of a solid sphere is given by (2/5)*m*r^2, where m is the mass and r is the radius. Without knowing the moment of inertia, it would be difficult to calculate the angular speed of the sphere.

Assuming that we have all the necessary information, I will provide an answer to the questions.

1) To calculate the new angular velocity of the merry-go-round, we can use the conservation of angular momentum. Initially, the angular momentum of the merry-go-round is given by L = I*ω, where I is the moment of inertia and ω is the angular velocity. After the four people step onto the edge of the merry-go-round, the new moment of inertia would be greater than 1665kg*m2. Let's say the new moment of inertia is I'. Using the conservation of angular momentum, we can write:

L = I*ω = I'*ω'

Solving for ω', we get:

ω' = (I*ω)/I' = (1665kg*m2*0.730 rad/s)/I'

As mentioned earlier, the value of I' is unknown. Therefore, we cannot accurately calculate the new angular velocity of the merry-go-round without this information.

2) To calculate the angular speed of the sphere, we can use the conservation of energy. Initially, the sphere has only potential energy due to its position on the roof. As it rolls down the roof, this potential energy is converted into kinetic energy. At the bottom of the roof, all the potential energy is converted into kinetic energy. Therefore, we can write:

mgh =

 

[thepatientmental]

1) When the four people step onto the merry-go-round, the total moment of inertia increases to 1665kg*m2 + 4*(60.8kg)*(5.4m)^2 = 3547.68 kg*m2. By the conservation of angular momentum, the initial and final angular momenta are equal. Therefore, the final angular velocity is given by:
Iω = I'ω'
1665kg*m2 * 0.730 rad/s = 3547.68 kg*m2 * ω'
ω' = (1665kg*m2 * 0.730 rad/s)/3547.68 kg*m2 = 0.343 rad/s

2) The potential energy of the sphere at the top of the roof is converted into rotational kinetic energy as it rolls down the roof, so the final angular speed is given by:
KErot = PE = mgh = (1/2)Iω^2
(1/2)(12.0 kg)(9.8 m/s^2)(7.0 m) = (1/2)(2/5)(12.0 kg)(0.13 m)^2ω^2
ω = (2*9.8*7)/(2/5*0.13^2)^(1/2) = 6.45 rad/s