Ferris Wheel & Ice Cream (circular and projectile motion?)

In summary, a child riding a Ferris wheel with a constant angular velocity drops her ice cream cone at an angular position \varthetas and it comes flying back into the cone as she reaches the opposite point on the circle. The angle \varthetas is unknown, the centripetal acceleration of the child is being asked for, and the expression for the relative speed between the ice cream cone and ice cream blob is also unknown. This is a combination problem of circular and projectile motion.
  • #1
flavoric3
1
0

Homework Statement


A Ferris wheel with radius R turns with a constant angular velocity [tex]\omega[/tex]. A child rides the Ferris wheel with a fresh ice cream cone. When the child reaches an angular position [tex]\vartheta[/tex]s [see diagram below], she is so distracted that she tips the cone and drops the nice big blob of ice cream (with sprinkles!) from the top of the cone. Since her arm is hanging over the side, the ice cream falls away. Upon discovering the loss of her ice cream, she begins to cry.

But luck is with this little girl. Just as the little girl reaches the point on the circle exactly opposite to the point where she lost the ice cream, the blob of ice-cream comes sailing right back into the ice cream cone. Now, it’s true that ice cream cone was totally destroyed by the impact. But the whole experience was so weird that the little girl became convinced that she was the luckiest little girl in the whole world.

a) Find the (numerical value) of the angle [tex]\vartheta[/tex]s.
b) Find the centripetal acceleration of the little girl while she was on the Ferris wheel. Express your answer as a numeric multiple of g.
c) Find an expression for the relative speed between the ice cream cone and the ice cream blob as the blob hits the cone.

Homework Equations


I have no idea?

The Attempt at a Solution


I'm actually completely confused. I know this is a combination problem of circular and projectile motion, but I'm not sure how to combine them.
 
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  • #2
Hi Flavoric3, welcome to PF!:smile:

Would you mind posting the diagram so that we can visualize the problem properly?
 

Related to Ferris Wheel & Ice Cream (circular and projectile motion?)

1. What is the physics behind the motion of a Ferris wheel?

The motion of a Ferris wheel involves both circular and projectile motion. As the wheel rotates, the riders experience centripetal acceleration towards the center of the wheel, which keeps them moving in a circular path. At the same time, the riders also experience a constant velocity in the horizontal direction due to the rotation of the wheel, which is an example of projectile motion.

2. How does the speed of a Ferris wheel affect the forces experienced by the riders?

The speed of a Ferris wheel directly affects the centripetal acceleration experienced by the riders. As the wheel rotates faster, the centripetal acceleration increases, causing the riders to feel a stronger outward force. This is why riders typically feel more force at the top of the wheel where it is moving faster, compared to the bottom where it is moving slower.

3. What factors determine the height and distance traveled by a projectile launched from a Ferris wheel?

The height and distance traveled by a projectile launched from a Ferris wheel depend on the initial velocity and angle of launch. The higher the initial velocity, the higher the projectile will go and the farther it will travel. Similarly, the angle of launch also plays a role, with a higher angle resulting in a higher trajectory and longer distance traveled.

4. How does the shape of a Ferris wheel affect the forces experienced by the riders?

The shape of a Ferris wheel can affect the forces experienced by the riders in two ways. First, the size and curvature of the wheel can impact the speed and acceleration of the riders as the wheel rotates. Second, the shape can also affect the wind resistance experienced by the riders, which can create additional forces on the riders.

5. How does the type of ice cream impact its motion when thrown from a Ferris wheel?

The type of ice cream can impact its motion when thrown from a Ferris wheel in several ways. First, the weight and shape of the ice cream can affect its trajectory and distance traveled. Second, the melting rate of the ice cream can impact its aerodynamics and therefore its motion. Third, the texture and stickiness of the ice cream can also affect its motion, as stickier ice cream may adhere to the surface it lands on and have a shorter distance traveled.

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