# Movement of a point on the perimeter of a disc

• charbon
In summary, the problem involves a spinning disc of radius r moving clockwise on a horizontal line z = r of a vertical axis Oxz. The center of the disc, C, is in translation compared to the origin, O, and has an angle \theta with the vertical axis. The velocity and acceleration of a point A on the perimeter of the disc are expressed in R, and the question asks for the velocity needed for the lowest point of the disc to have a velocity of 0. The attempt at a solution involves using the velocity addition formula and the definition of velocity in an inertial frame.
charbon

## Homework Statement

a disc of radius r spins uniformly around its axis at an angular velocity $$\omega$$ in a clockwise motion. the center moves on the horizontal line z = r of a vertical axis Oxz of the referential Oxyz. We call R' = Cxyz, in translation compared to R= Oxyz, of origin C and we specify $$\theta$$ the angle made by a CA from the disc with Cz, A being a point of the perimeter of the disc.

a) express in R, the velocity and acceleration of A compared to R'
b) What velocity, in R, must we give to C in order for the velocity $$\vec{vb}$$/R of the lowest point of the disc be 0?

## Homework Equations

I am having a hard time finding b)

first off, I wrote the velocity-addition formula

$$\vec{vb}$$ = $$\vec{vb'}$$ + $$\vec{V}$$
thus
$$\vec{V}$$ = $$\vec{vb}$$ - $$\vec{vb'}$$

but $$\vec{vb}$$ = 0
so
$$\vec{V}$$ = - $$\vec{vb'}$$

This is where I am stuck. How do I find this vector?

The Attempt at a SolutionI tried using the definition of velocity of a point A in an inertial frame, v_A = \frac{d\vec{r}}{dt}where \vec{r} is the position vector of A relative to Oso I wrote\vec{vb'} = \frac{d\vec{r}}{dt}where \vec{r} = rcos\theta \hat{i} + rsin\theta \hat{j}but I am not sure if this is the correct approach and if so how do I continue?

## 1. What is the definition of "movement of a point on the perimeter of a disc"?

The movement of a point on the perimeter of a disc refers to the motion of a single point along the circular edge of a disc. It can include both rotational and translational motion.

## 2. What causes the movement of a point on the perimeter of a disc?

The movement of a point on the perimeter of a disc is caused by external forces acting on the disc, such as friction or applied torque.

## 3. How does the radius of the disc affect the movement of a point on the perimeter?

The radius of the disc plays a significant role in the movement of a point on the perimeter. A larger radius will result in a greater distance traveled for each rotation, while a smaller radius will result in a smaller distance traveled for each rotation.

## 4. What is the difference between uniform and non-uniform movement of a point on the perimeter of a disc?

Uniform movement refers to a constant speed and direction of the point on the perimeter, while non-uniform movement refers to a variable speed and/or direction. This can be caused by changes in external forces or the shape of the disc.

## 5. How is the movement of a point on the perimeter of a disc related to angular velocity?

The movement of a point on the perimeter of a disc is directly related to angular velocity, which is the rate of change of the angle of rotation. The higher the angular velocity, the faster the point will move along the perimeter.

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