100 points PLEASE ANSWER IN 3 PARAGRAPHS

Physics

1 answer:

Deffense [45]2 years ago

4 0

Answer:

In the previous section, we defined circular motion. The simplest case of circular motion is uniform circular motion, where an object travels a circular path at a constant speed. Note that, unlike speed, the linear velocity of an object in circular motion is constantly changing because it is always changing direction. We know from kinematics that acceleration is a change in velocity, either in magnitude or in direction or both. Therefore, an object undergoing uniform circular motion is always accelerating, even though the magnitude of its velocity is constant.

You experience this acceleration yourself every time you ride in a car while it turns a corner. If you hold the steering wheel steady during the turn and move at a constant speed, you are executing uniform circular motion. What you notice is a feeling of sliding (or being flung, depending on the speed) away from the center of the turn. This isn’t an actual force that is acting on you—it only happens because your body wants to continue moving in a straight line (as per Newton’s first law) whereas the car is turning off this straight-line path. Inside the car it appears as if you are forced away from the center of the turn. This fictitious force is known as the centrifugal force. The sharper the curve and the greater your speed, the more noticeable this effect becomes.

Figure 6.7 shows an object moving in a circular path at constant speed. The direction of the instantaneous tangential velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity; in this case it points roughly toward the center of rotation. (The center of rotation is at the center of the circular path). If we imagine Δs becoming smaller and smaller, then the acceleration would point exactly toward the center of rotation, but this case is hard to draw. We call the acceleration of an object moving in uniform circular motion the centripetal acceleration ac because centripetal means center seeking.

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suppose a child walks from the outer edge of a rotating Merry-Go-Round to the inside does angular velocity of the Merry-Go-Round

Akimi4 [234]

Okay here initially child walks on the outer edge of the disc and then started to move inside

So here as the child and Merry go round is an isolated system so there is no external Torque on this system from outside

As here we can see there is external force acting on this system by the hinge of the Merry go round as well as due to gravity so we can not use momentum conservation to solve such type of questions.

But as we can say that there is no external torque on this system about the hinge point so we will use conservation of angular momentum for this system

Here as we know that

$\tau = \frac{dL}{dt}$

where L = angular momentum

since here torque is ZERO

$0 = \frac{dL}{dt}$

L = constant

so here we can write initial angular momentum of the system as

$L = (I_1 + I_2)*\omega$

here we know that

$I_1$ = moment of inertia of merry go round

$I_2$ = moment of inertia of child

so here we can say

$(I_1 + I_2)* \omega_1 = (I_1 + I_2')\omega_2$

so here as the child moves from edge to inside the disc it moment of inertia will decrease because as we know that moment of inertia of child is given as

$I_2 = mr^2$