A ball on the end of a string is whirled around in a...

A ball on the end of a string is whirled around in a horizontal circle of radius $0.300 \mathrm{~m}$. The plane of the circle is $1.20 \mathrm{~m}$ above the ground. The string breaks and the ball lands $2.00 \mathrm{~m}$ (horizontally) away from the point on the ground directly beneath the ball's location when the string breaks. Find the radial acceleration of the ball during its circular motion.

Key Concepts

Projectile Motion

Projectile motion describes the motion of an object that is launched into the air and is subject only to the acceleration of gravity (assuming negligible air resistance). It involves both horizontal and vertical components, which are independent of each other. In many problems, the time of flight is determined from the vertical motion, and the horizontal motion is used to calculate the speed or displacement, making this concept essential for solving problems where an object transitions from circular motion into free-fall.

Uniform Circular Motion

Uniform circular motion refers to the motion of an object traveling at a constant speed along a circular path. Even though the speed is constant, the direction of the velocity vector continuously changes, which results in an acceleration directed toward the center of the circle. This concept is critical in analyzing scenarios where objects move along curved paths, as it explains how forces act on the object during its motion.

Centripetal Acceleration

Centripetal acceleration is the acceleration that keeps an object moving in a circular path and is always directed towards the center of rotation. It is given by the formula a_c = v²/r, where v is the tangential speed of the object and r is the radius of the circle. This concept helps in understanding how the speed of an object in circular motion constrains the magnitude of the inward acceleration required to maintain the circular path.

Transcript

00:01 In the given problem, here this is the horizontal circle whose height from the ground level is given as if we let it to be edge, this edge, this height is given as 1 .2 -0 meter.

00:26 Now suppose the stone, the ball is released here at this point, so it will start moving tangentially.

00:34 Then after getting released, it will cover a projectile path and finally we'll hit the ground here at this point.

00:45 And it is given that this distance over the horizontal ground, x is 2 .0 meter.

00:59 Now, this is the initial horizontal speed, which we have to find.

01:07 And here this is, this will be the initial vertical velocity, which is definitely zero.

01:15 Actually, we have to find the radial acceleration, but we know the expression for radial acceleration is ux square by r.

01:25 The linear velocity, the square of linear velocity divided by r.

01:30 And here, this linear velocity of the ball will actually serve as the initial horizontal velocity for the projectile motion of the ball.

01:39 So now, first of all, we will use vertical motion of the ball in its trajectory, in its projectile trajectory.

01:54 And using this vertical motion, we will apply second equation of motion, which says h equals to uy t plus half g t squared so it becomes for h this is 1 .20 meter for u y as u y is 0 so uy t will also become 0 then this is half into 9 .8 for g into t square so this t squared is 1 .20 into 2 divided by 9 .8 which comes out to be 0 .2449.

02:34 Hence this time comes out to be 0 .249 square root second, which becomes 0 .49 2 .495 second...

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