A stone at the end of a sling is whirled in a vertical...

A stone at the end of a sling is whirled in a vertical circle of radius $1.20 \mathrm{~m}$ at a constant speed $v_{i}=1.50 \mathrm{~m} / \mathrm{s}$ as in Figure $\mathrm{P} 4.57$. The center of the string is $1.50 \mathrm{~m}$ above the ground. What is the range of the stone if it is released when the sling is inclined at $80.0^{\circ}$ with the horizontal (a) at $A ?$ (b) at $B ?$ What is the acceleration of the stone (c) just before it is released at $A$ ? (d) just after it is released at $A ?$

Key Concepts

Uniform Circular Motion

This concept refers to the motion of an object traveling along a circular path at a constant speed. Even though the speed remains constant, the direction of motion is continually changing, which means the object is experiencing an acceleration directed toward the center of the circle, known as centripetal acceleration. This idea is fundamental when analyzing objects moving in circular paths, such as a stone whirled in a sling.

Centripetal Acceleration

Centripetal acceleration is the acceleration that is required for an object to follow a circular path and is directed towards the center of the circle. It is given by the square of the object's speed divided by the radius of the circle. This acceleration is not due to any change in speed, but rather to the continuous change in the direction of the velocity vector.

Projectile Motion

Projectile motion describes the motion of an object that is launched into the air and moves under the influence of gravity alone, once any external forces (like tension in a string) are no longer acting on it. The object follows a parabolic trajectory determined by the initial velocity and the acceleration due to gravity. Concepts such as time of flight, range, and maximum height are key features in analyzing projectile motion.

Vector Decomposition of Velocity

This involves breaking a velocity vector into its horizontal and vertical components. When an object is launched at an angle, its initial velocity can be separated into two perpendicular components that determine its behavior along the horizontal (constant velocity) and vertical (accelerated by gravity) axes. This decomposition is essential for solving problems in projectile motion and for analyzing the motion immediately after an object is released from circular motion.

Free-Fall Acceleration

Once an object is released from a constraint, such as a string in circular motion, it is subject only to the acceleration due to gravity (assuming air resistance is negligible). This gravitational acceleration is constant near the Earth's surface and acts downward, determining the object's vertical motion until it impacts the ground.

Transcript

0:00 Hello.

00:01 In this problem we have a stone that is whirling in this circle with radius 1 .2 meters and it's moving at speed 1 .5 meters per second.

00:12 The center of the circle is 1 .5 meters high above the ground and we are asking about the range of the stone if it is released from any of these two positions.

00:28 Position a here with 80 degrees with the horizon and position b with also 80 degrees from the horizon but on the other side of the loop.

00:40 The stone is moving in this direction which is counterclockwise and we know that the velocity is always tangential.

00:51 So at point a it's as the red arrow is droid pointing like this and at the point b, it's pointing like this.

01:02 So we're asking about the range at both a and b.

01:05 So let's start by position a.

01:12 At position a, first we need to get the y coordinates or how high is the stone above the ground.

01:20 So we know that from the center to the ground, this is already 1 .5 meters.

01:25 So we have 1 .5 meters already.

01:28 And then we need to know how high is the ball or the stone where it is right now so from the drawing here you can see that this is just 1 .2 multiply by sign of the angle so this is 1 .2 sign 80 so we have 2 .68 meters and that's the y coordinate or the y position at 0 .8 meters.

02:00 Now let's get the speed or the velocity of the stone at this point.

02:07 The velocity is 1 .5 and from here by simple geometrical trick we can know that this angle is 10 degrees.

02:21 So the velocity in the x direction or the horizontal direction is just 1 .5 cosine 10 where the velocity in the direction is 1 .5, sign 10.

02:40 Doing both on the calculator, we'll find that the velocity in the x direction is 1 .48 meters per second, and in the y direction, it's 0 .2 .6 meters per second.

02:54 Now, from the equations of kinematics, we are ready to get the time of the flight itself, first of all.

03:01 So now we know that y equals to y dot the initial position plus v y multiplied by t minus 4 .9 t squared.

03:16 Y is where the stone will reach the ground...

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