Aplayground merry-go-round has radius 2.80 m and moment of inertia 2400 kg⋅m2 about a vertical axle through its center, and it turns with negligible friction. a child applies an 20.0 n force tangentially to the edge of the merry-go-round for 25.0 s . if the merry-go-round is initially at rest, what is its angular speed after this 25.0 s interval?

Explanation:

given,

radius of merry- go- round = 2.80 m

moment of inertia = I = 2400 kg⋅m²

child apply force tangentially = 20 N

for time = 25 s

angular speed after 25 speed = ?

initial angular speed of the merry go round = 0 rad/s

we know,

torque = I α.............(1)

α is angular acceleration

and also

τ = F.r........................(2)

computing equation (1) and (2)

F . r = I α

the angular speed of merry-go-round after 25 second is equal to

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.

the answer is 100. i just did this lesson and got a 100% on it. : )

-fishtank