A square weightless frame ABCD with the side a is rotating around the side AB with the constant angular velocity ω1. A uniform d
isk of radius R and mass m is rotating around the frame’s diagonal AC with the constant angular velocity ω2. The center of the disk coincides with the center of the frame. Assuming ω1 = ω2 = ω find
a. The kinetic energy of the system.
b. The magnitude of the torque that needs to be applied to the axis AB to keep it (the axis AB) still.
a. The kinetic energy of the system.
b. The magnitude of the torque that needs to be applied to the axis AB to keep it (the axis AB) still.
1 answer:
Answer:
1) KE of the system = 1/2(Iw^2)
2) the torque T that must be applied to keep it still must be equal to the torque of rotation. This is equal to
T = (Iw^2)/2©
Where;
I = moment of inertia of the body
W = angular velocity of the body
© = angle of rotation in rad
Explanation:
Detailed explanation and calculation is shown in the image below
You might be interested in
Answer:
The ball's initial kinetic energy
The ball comes to a stop at B. At this point its initial kinetic energy is converted into potential energy
Explanation:
A ball is fixed to the end of a string, which is attached to the ceiling at point P. As the drawing shows, the ball is projected downward at A with the launch speed v0. Traveling on a circular path, the ball comes to a halt at point B. What enables the ball to reach point B, which is above point A? Ignore friction and air resistance.
From conservation of energy which states that energy can neither be created nor be destroyed, but can be transformed from one form to another.
Ki+Ui=Kf+Uf
Ki=initial kinetic energy
Ui=initial potential energy
Kf=final kinetic energy
Uf=final potential energy
we know that
m=mass of the ball
ha=downward height a
hb=upward height b
u=initial velocity u
v=final velocity v, which is 0
g=acceleration due to gravity
v=0 at final velocity
1/2mu^2+mgha=0+1/2mv^2
ha=hb+Ki/mh
From the above equation, we can conclude that the ball's initial kinetic energy is responsible for making the ball reach point B.
Point B is higher than point A from the motion gained by the ball