A conical pendulum is formed by attaching a 500 g ball to a...

A conical pendulum is formed by attaching a 500 g ball to a $1.0-\mathrm{m}-$ long string, then allowing the mass to move in a horizontal circle of radius $20 \mathrm{cm}$. Figure $\mathrm{P} 6.57$ shows that the string traces out the surface of a cone, hence the name. a. What is the tension in the string? b. What is the ball's speed? c. What is the period of the ball's orbit?

A conical pendulum is formed by attaching a 500 g ball to a $1.0-\mathrm{m}-$ long string, then allowing the mass to move in a horizontal circle of radius $20 \mathrm{cm}$. Figure $\mathrm{P} 6.57$ shows that the string traces out the surface of a cone, hence the name.
a. What is the tension in the string?
b. What is the ball's speed?
c. What is the period of the ball's orbit?

Key Concepts

Centripetal Force in Circular Motion

Centripetal force is the net inward force required to keep an object moving in a circular path. In the case of a conical pendulum, the horizontal component of the tension provides this force. Analyzing how centripetal force relates to the mass of the object, its speed, and the radius of its circular path is key to solving for unknown variables such as speed and period.

Relationships between Speed, Radius, and Period

The kinematics of uniform circular motion describe how the linear speed of an object moving in a circle is related to both its radius and the period of revolution. Specifically, the period (the time for one complete revolution) can be determined directly from the speed and radius. This relationship is crucial when deriving formulas to determine the speed of the object and the time taken for one complete orbit.

Conical Pendulum

A conical pendulum is a system in which an object (like a ball) is attached to a string or rod and moves in a horizontal circle while the string maintains a constant angle with the vertical. This creates a cone-shaped surface as the object moves, and analyzing this system helps bridge concepts from circular motion and force resolution.

Force Resolution and Tension

In a conical pendulum, the tension in the string is not aligned with the gravitational force. Instead, it is resolved into two components: a vertical component, which balances the gravitational force, and a horizontal component, which provides the required centripetal force for circular motion. Understanding this resolution is essential for analyzing the dynamics of the system.

Transcript

00:01 So in this problem, we have a conical pendulum that's formed by attaching a 500 gram ball to a 1 meter long string that allows the mass to move in a horizontal circle of radius 20 centimeters.

00:12 And we know that the string will chase out the surface of the cone, hence the name.

00:17 So we want to know the tension in the string, the ball speed, and the period of the ball's orbit.

00:23 So starting with the tension in the string, first let's just draw what this looks like.

00:35 Know that the ball is moving around in a circle like that.

00:41 This is o.

00:50 So we know that the tension and the string can be resolved into two components because we have an x and y direction.

00:58 So the vertical component bounces the weight mg of the ball.

01:03 So we have force cosine theta is equal to mg.

01:09 And the horizontal component provides the centrifugal force of f sine theta is equal to m v squared over r.

01:20 And this is both just coming from newton's second law, where we have g for a, and so the g will just be the gravitational acceleration.

01:31 Then here we have this intrapidial acceleration, which is just v squared over r.

01:36 So dividing equation 1 by equation 2, we'll give us tangent of theta is equal to v squared over rg, and then we want to square both equations 1 and 2, which will give us, and then we'll just take the square rate of this to solve for.

02:43 And now we have tangent of theta is equal to v squared over rg, and it's just equal to rh, which will give us that solving for v squared.

03:10 And then we're going to substitute in equation 5, this is equation 4.

03:48 We're just going to continue solving that down until we get to our final expression for the tension in the string, which is m .g.

03:58 It's a square rate of 1 plus r over h squared...

Please give Ace some feedback