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Aneli [31]
2 years ago
8

A disk-shaped platform has a known rotational inertia ID. The platform is mounted on a fixed axle and rotates in a horizontal pl

ane with an initial angular velocity of ÏD in the counterclockwise direction as shown. After an unknown time interval, the disk comes to rest. A single point on the disk revolves around the center axle hundreds of times before the disk comes to rest. Frictional forces are considered to be constant.
In a different experiment, the original disk is replaced with a disk for which frictional forces are considered to be negligible. The disk is set into motion such that it rotates with a constant angular speed. As the disk spins, a small sphere of clay is dropped onto the disk, and the sphere sticks to the disk.


Required:

Write down the claims which are correct about the angular momentum and the total kinetic energy of the disk-sphere system immediately before and immediately after the collision.
Physics
1 answer:
Zarrin [17]2 years ago
5 0

Answer:

Explanation:

The angular momentum of that same disk-sphere remains unchanged the very same way before and after the impact of the collision when the clay sphere adheres to the disk.

 \mathbf{I_w} = constant.

The overall value of such moment of inertia is now altered when the clay spherical sticks. Due to the inclusion of the clay sphere, the moment of inertia will essentially rise. As a result of this increase, the angular speed w decreases in value.

Recall that:

The Kinetic energy is given by:

\mathbf{K = \dfrac{1}{2} Iw^2} \\ \\\mathbf{K = \dfrac{1}{2} lw*w}

where;

\mathbf{I_w} is constant and w reduces;

As a result, just after the collision, the system's total kinetic energy decreases.

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For a vertical spring-mass oscillator that is moving up and down, which of the following statements are true? (more than one sta
omeli [17]

Answer:

At the lowest point in the oscillation, the momentum is zero.

At the lowest point in the oscillation, mg < ks_s

Explanation:

Since spring block system is performing to and fro motion along straight line

So here we can say at the lowest position of its path the velocity will become zero.

So we can say that momentum of the spring block system is given as

P = mv

P = 0

Also we know that after reaching the lowest point the block will again go up towards its mean position

So at the lowest point of the spring block system the block will move upwards again

So this will accelerate upwards hence

F_{spring} > mg

Ks > mg

6 0
3 years ago
Our Solar System's planets formed when fragments in space joined together to form growing spheres in a process known as ____​
Nimfa-mama [501]

Answer: accretion......

8 0
3 years ago
The atomic mass of an atom is determined by which of the following?
Flura [38]
The answer is A.number of protons in the nucleus.
3 0
3 years ago
Which of Newton's laws of motion best illustrates the principle of inertia?
andrew-mc [135]
Newton's first law of motion best illustrates the principle of inertia<span />
5 0
3 years ago
A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo
Talja [164]

Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

5 0
3 years ago
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