A uniform circular disk of moment of inertia 8.0 kg m2 is rotating at 4.0 rad/s. A small lump of mass 1.0 kg is dropped on the disk and sticks to it at a distance of 1.0 m from the axis of rotation. What is the new rotational speed of the combined system?

The new rotational speed of the combined system is 32 rad/s

Explanation:

Given-

Moment of Inertia, I = 8 kg m²

speed of rotation, ω = 4 rad/s

Mass of the lump = 1 kg

distance = 1 m

New rotational speed = ?

We know,

When the lump is dropped on the rotating disc then the angular momentum remains conserved.

Angular Momentum of the disc, Lₐ = Iω

Angular momentum of the lump, L ₓ = mvR

Since the angular momentum remains conserved,

Lₐ = Lₓ

8 X 4 = 1 X v X 1

New rotational speed, v = 32 rad/s

Therefore, the new rotational speed of the combined system is 32 rad/s.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards