# Circular Motion of a horizontal disc Problem

• Peter G.
In summary: Ok, thanks!In summary, the smaller mass (the cart) must have a speed v such that the bigger mass (the seat of the cart) remains at rest.
Peter G.
Hi

A horizontal disc has a hole through its center. A string passes through the hole and connects a mass m on top of the disc to a bigger mass M that hangs below the disc. Initially, the smaller mass is rotating on the disc in a circle of radius r. What must the speed of m be such that the big mass stands still.

So, I managed to get to the answer with not too many problems:

Mg = ma
Mg = mv2 / r

and rearranged to: v = √ (Mgr / m)

My problem is, I don't understand what is happening. So, why/when would the big mass stop moving? Is it when the tension in the string due to the centripetal force equals the big mass Mg, so that the forces on the big Mass balance?

Thanks,
Peter G.

The rotation of the smaller mass creates a force on the string due to the centripetal acceleration. When the force due to centripetal acceleration, F=m1*(v^2/r), equals the force of gravity on the larger mass, F=9.81*m2, the two forces will cancel, and the two masses will be in equilibrium.

Ok, thanks

Hey, would you mind helping me out with this other question?

In an amusement park ride a cart of mas 300 kg and carrying four passengers, each of mass 60 kg is dropped from a vertical height of 120 m along a frictionless path that leads into a loop-the-loop machine of radius 30 m. The cart then enters a straight-line stretch from point A to C where friction brings it to rest: (Diagram on page 126)
(a) Find the velocity of the cart at A
(b) Find the reaction force from the seat of the cart onto passenger B
(c) What is the acceleration experience by the cart from A to C (assumed constant)

I did part (a) and got the correct answer: 48.99

I got problems with b:

The force they would feel would be the resultant force. In that position, the top of a loop, it would be: Centripetal force + mg, correct?

This was my working:

The energy at the bottom of the loop = (0.5 x 540 x 48.992)
The energy at 60 m from the ground = (540 x 10 x 60) + K.E

Therefore:

(540 x 10 x 60) + (0.5 x 540 x v2) = 648005.427
v = 34.64 m/s

Then:

F = mv2 / r
F = 21600.3618 N

F + mg = 27000.3618 N

But I must have gone very wrong somewhere! For the book says the correct answer is 1800N!

I think I see where you went wrong; it's a simple sign error. Recall that centripetal acceleration is always toward the center-point of the circle. Therefore, the resulting centrifugal force is toward the outside of the circle*.

*(On a technical level, what's actually happening is the body that is moving in a circle is trying to continue moving in a straight line, but is held in a circular motion by something else exerting a force on it. The force that is exerted on the body is called "centripetal force.")

I would like to emphasize the difference between "centripetal force" and "centrifugal force" here. The use of the term "centrifugal force" is frequently misused, as is generally discouraged in physics, but it is still useful in explaining forces involved in circular motion.

When you take a sharp turn in a vehicle, the only "true" force that is acting is the force of the door or seatbelt on your body that keeps you moving in a circular motion. This is the "centripetal force." However, from your perspective, you are pressed up against the door. This "false" force is centriFUGAL force, and is only a reactive force, similar to the difference between the force of gravity, and normal force. The normal force is occurring a result of the force of gravity.

Now, back to the problem. Reexamine the forces at the top of the loop. Gravity is always pulling down, as you well know. But (and this is key), which direction is the centripetal acceleration at that point, and which direction is the centripetal force? At the top of the loop, which direction is the car's inertia? Draw a free-body diagram if necessary.

## 1. What is circular motion?

Circular motion is the movement of an object along a circular path, where the object's distance from a fixed point remains constant while its direction changes.

## 2. What is the equation for circular motion?

The equation for circular motion is v = rω, where v is the linear velocity, r is the radius of the circle, and ω is the angular velocity.

## 3. How does the horizontal disc problem relate to circular motion?

The horizontal disc problem is a specific example of circular motion, where a disc is rotating horizontally around a fixed axis. The principles and equations of circular motion can be applied to analyze the movement of the disc.

## 4. What factors affect the circular motion of a horizontal disc?

The factors that affect the circular motion of a horizontal disc include the disc's mass, radius, and angular velocity, as well as any external forces acting on the disc, such as friction. The disc's moment of inertia, which depends on its shape and mass distribution, also plays a role in its circular motion.

## 5. How can the circular motion of a horizontal disc be used in real-world applications?

The circular motion of a horizontal disc can be seen in various real-world applications, such as car tires rotating while driving, the spinning of a record player, and the movement of a Frisbee in flight. Understanding the principles of circular motion can also be applied in engineering and design, such as in the construction of amusement park rides or the design of gears and pulleys.

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