Angular Acceleration and Projectile Motion

Click For Summary
SUMMARY

The discussion focuses on calculating the angular acceleration required for a marble launched from a spring-loaded wheel to land in a cup. The marble is released after 11/12 of a revolution, equating to 330 degrees, and is launched at a 30-degree angle. Key equations utilized include the projectile motion equations for horizontal and vertical positions, as well as the formula for angular acceleration, a = (omega(final)^2 - omega(initial)^2)/2(delta(theta)). The participants emphasize the need to derive the launch velocity based on the given parameters.

PREREQUISITES
  • Understanding of projectile motion equations
  • Familiarity with angular kinematics
  • Knowledge of trigonometric functions in physics
  • Ability to manipulate equations for solving physics problems
NEXT STEPS
  • Study the derivation of launch velocity in projectile motion
  • Learn about angular acceleration calculations in rotational dynamics
  • Explore the relationship between angular displacement and linear motion
  • Investigate the application of trigonometric functions in physics problems
USEFUL FOR

Students studying physics, particularly those focused on mechanics and projectile motion, as well as educators seeking to explain angular dynamics in practical scenarios.

MyNewPony
Messages
30
Reaction score
0

Homework Statement



An amusement park game, shown in the figure (see attachment), launches a marble toward a small cup. The marble is placed directly on top of a spring-loaded wheel and held with a clamp. When released, the wheel spins around clockwise at constant angular acceleration, opening the clamp and releasing the marble after making 11/12 revolution.

What angular acceleration is needed for the ball to land in the cup?

Homework Equations



x1 = vcos(theta)t
y1 = y(initial) + vsin(theta)t - 4.9t^2

omega = v/r

a = (omega(final)^2 - omega(initial)^2)/2(delta(theta))

The Attempt at a Solution



1) 11/12 of a revolution = 330 degrees, therefore it is launched at an angle of 30 degrees.
2) It is shot from an initial height of rcos30 = 0.2*sqrt3/2 m
3) It has a range of 1 + rsin30 = 1 + 0.2*1/2 = 1.1 m

I know I'm supposed to find launch velocity from this data but how do I go about doing so?
 

Attachments

  • Angular Acceleration.jpg
    Angular Acceleration.jpg
    7 KB · Views: 760
Physics news on Phys.org
MyNewPony said:

The Attempt at a Solution



1) 11/12 of a revolution = 330 degrees, therefore it is launched at an angle of 30 degrees.
2) It is shot from an initial height of rcos30 = 0.2*sqrt3/2 m
3) It has a range of 1 + rsin30 = 1 + 0.2*1/2 = 1.1 m
Sounds good. (I assume that the cup is at the same height as the center of the wheel? Hard to tell from the diagram.)

I know I'm supposed to find launch velocity from this data but how do I go about doing so?
Set up the usual projectile motion equations for vertical and horizontal position as a function of time.
 
I'm trying to do this same problem, but I don't know how. I was hoping I could figure it out with some online help, but I don't understand where you got all the equations and such.
 

Similar threads

Replies
67
Views
5K
Replies
26
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
9
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
3K
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 15 ·
Replies
15
Views
6K
  • · Replies 7 ·
Replies
7
Views
3K