A child ties a rock to a string and whirls it around in a horizontal circle. (a) The string cann

A child ties a rock to a string and whirls it around in a...

Key Concepts

Period and Speed Relations

The period of circular motion is the time it takes for an object to complete one full cycle around the circle, and it is related to the speed and radius of the circle through the formula for centripetal acceleration. Understanding these relations is key to solving for unknown quantities like the speed of the object and the time period in circular motion problems.

Uniform Circular Motion

Uniform circular motion refers to motion in a circle at a constant speed. Even though the speed remains constant, the direction changes continuously, which requires a constant inward (centripetal) force. This concept ties together the ideas of velocity, acceleration, and force in a rotating system.

Force Decomposition

Force decomposition involves breaking a force into components along specific directions, typically along the horizontal and vertical axes. In circular motion problems with inclined strings, this concept is crucial because it allows the separation of the tension force into a horizontal component that provides the needed centripetal force, and a vertical component that counterbalances gravity.

Free-Body Diagram

A free-body diagram is a visual representation used to isolate an object and depict all forces acting upon it. In physics problems involving motion, such as circular motion, this diagram helps in decomposing forces into components, understanding how gravity, tension, and other forces interact, and setting up the necessary equations for analysis.

Centripetal Force

Centripetal force is the net force required to keep an object moving in a circular path. It is directed towards the center of the circle, and in the context of the problem, the horizontal component of the tension provides this force. Understanding centripetal force is essential for solving problems involving uniform circular motion.

Transcript

00:01 In this problem, it is given that a child is rotating a rock with the help of a string in horizontal circle.

00:10 So we have to, in first part, we have to explain that this string cannot be perfectly horizontal.

00:16 So to explain that, first of all, let us draw the force diagram for the stone or the rock.

00:25 So this rock is basically performing a circular motion in a horizontal circle.

00:32 So, what are, let us consider that this time the stone is performing or the string is perfectly horizontal.

00:39 So what are the forces that will be acting on the stone? so first one will be weight of the stone that will be acting in downward direction.

00:47 And there will be a tension force in the string which will be providing the centripetal acceleration to the stone because it is moving in a circular path.

00:56 So under this action, here we can see that the forces, this t, this, this, this, this, this tension force will be balancing the centrifugal force or this tension force will be responsible to provide the centripetal acceleration.

01:10 So forces in the horizontal directions will be balanced and the value of tension will adjust itself value.

01:17 But forces in the vertical direction is also unbalanced because here mg force is acting in downward direction and there is no other forces which is countering this mg.

01:27 So this mg will tend to lower the stone from its horizontal location.

01:32 And this stone will come through some distance in vertical direction.

01:36 Therefore, this string cannot be perfectly horizontal.

01:40 This is just because of the presence of mg component.

01:44 So what will you write? because of weight of stone, the string cannot be perfectly horizontal.

02:13 Now in part b, it is saying that the string, the length of the string is, let us say, say l is the length of the string which is 1 .22 meter and it is making an angle of 25 degree below the horizontal so theta is 25 degree so what will be the location so this is the string whose length is l and it is making an angle of theta below the horizontal so this this time the stone will be performing again it is performing a horizontal circular motion but radius of the circle will be r and here this angle is theta then this angle will also be theta so what will be the relation between r and l so r will be equals to l cost theta all right so we know the value of l which is 1 .22 and cost 25 degree so r is equals to 1 .1056 meter.

03:23 Now if we draw force diagram for the stone or for the rock then what are the forces acting on the rock? first one will be the tension force along the string.

03:34 So let us say this is the tension force...

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