Ball on a Turntable Simple Harmonic Motion

In summary, a spherical ball of mass "m", moment of inertia "I" and radius "a", rolls without slipping and without dissipation on a horizontal turntable with unknown radius and mass, while the turntable rotates about the vertical z-axis at an unspecified angular velocity. To solve this problem, the angular momentum at the center of mass and at point P were determined and equated to find the constant values C1, C2, and C3. Then, using the fact that the velocity of point P on the turntable and the ball are equal, the equations for x and y positions were derived. However, the final answer was incorrect according to the professor.
  • #1
anon11
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Homework Statement


ball turntable 2.png

A spherical ball of mass “m”, moment of inertia “I” about any axis through its center, and radius “a”, rolls without slipping and without dissipation on a horizontal turntable (of radius “r”) describe the balls motion in terms of (x,y) for a function of time.

**The turntable is rotating about the vertical z-axis at a constant unspecified angular velocity.

**Radius and mass of the turntable and the ball are unspecified.

Can someone please show me step by step how to solve this problem?

Homework Equations


(angular momentum at Center of Mass)=(moment of inertia of a ball)*(angular velocity) or
hcm=IW

The Attempt at a Solution


page 1.png

page 2.png
[/B]
This is my attempt at solving the problem. Can you please show me step by step how you would solve it and what answer you get. Thank you :)
 
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  • #2
Thread closed for Moderation... Thread re-opened so the OP can show more work. :smile:
 
Last edited:
  • #3
Here's my work in text format:

In part A
of my work I determined that angular momentum is conserved about the moving point P. (Point P being the point on the ball in contact with the turntable.) If I'm not mistaken that would mean that the angular momentum at moving point P is coupled to that of the turntable.

In part B I broke the angular momentum into its components to examine how the ball spins:
(angular momentum at Center of Mass)=(moment of inertia of a ball)*(angular velocity) or
hcm=IW

ν= (2/5)ma^2*<wxi + wyj + wzk> (angular momentum at Point P)=(angular momentum at Center of Mass)+(Radius “a”) cross product (linear momentum)

hp= hcm+ a x p

= hcm + a x mv

= hcm + a x m<vxi + vyj>

= hcm + <-mavyi + mavxj>Therefore,

(2/5)ma^2*wx - mavy = C1 C=constant

(2/5)ma^2*wy + mavx = C2

Wz=C3In part C I acknowledged that because the ball doesn’t slip “The velocity of point P on the turntable” and the “velocity of point P on the ball” will be equal to each other. I used this knowledge to break up the equation and eliminate the angular velocity components so that we can solve for x and y positions

Ωk x r = v + w (cross product) (-ak) Ωk= constant rotation about the z axis

My final answer was:
x=xo + RCos((2/7)Ωt)
y=yo + RSin((2/7)Ωt)

This is my attempt at solving the problem, but my professor has told me it's wrong. Can you please show me how you would solve it and what answer you get. Thank you :)
 
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Related to Ball on a Turntable Simple Harmonic Motion

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which an object oscillates back and forth about a central equilibrium point. This motion is characterized by a constant amplitude and frequency, and is caused by a restoring force that is directly proportional to the displacement of the object.

2. How does a ball on a turntable exhibit simple harmonic motion?

When a ball is placed on a turntable and the turntable is rotated at a constant speed, the ball will experience a centripetal force that pulls it towards the center of the turntable. This force acts as a restoring force, causing the ball to oscillate back and forth in a circular motion. This motion is considered simple harmonic because the force is directly proportional to the displacement of the ball from its equilibrium position.

3. What factors affect the period of a ball on a turntable's simple harmonic motion?

The period of a ball on a turntable's simple harmonic motion is affected by the mass of the ball, the radius of the turntable, and the speed at which the turntable is rotating. A larger mass or radius will result in a longer period, while a faster rotation speed will result in a shorter period.

4. What is the equation for the period of a ball on a turntable's simple harmonic motion?

The equation for the period of a ball on a turntable's simple harmonic motion is T = 2π√(m/r), where T is the period in seconds, m is the mass of the ball in kilograms, and r is the radius of the turntable in meters. This equation assumes that the turntable is rotating at a constant speed.

5. How is simple harmonic motion used in real-life applications?

Simple harmonic motion is used in a variety of real-life applications, such as pendulum clocks, musical instruments, and shock absorbers in vehicles. It is also used in scientific experiments to study the behavior of systems under periodic forces. Additionally, many mechanical and electrical systems can be modeled using simple harmonic motion equations to predict their behavior and make improvements.

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