# Rotational Mechanics(The concept of spinning)

• FedEx
In summary, a sphere with mass m and radius r, initially falling from a height of H with angular velocity omega, and a coefficient of restitution of e, collides with the ground and moves in the forward direction due to the coefficient of friction mu. To find the range of the sphere, velocities can be found using energy conservation and angular conservation. The formula for finding the final velocity in the x direction is (Vi + Awr)/(1+A), where A represents the moment of inertia divided by the product of mass and radius squared.

#### FedEx

well this is a really good question which i just thought while playing cricket.

A sphere of mass m and radii r(a tennis ball) falls freely from a height of H and angular velocity(clockwise) about its axis omega.Coefficient of restitution is e

The coefficient of friction is mu.
After the sphere touches the ground it moves in the forward direction.
We have to find the range of the sphere.

Now we can find the velocities by energy conservation.

We we can apply angular conservation about the point which touches the ground.

initial translational + initial rotational = final translational + final rotational

Final velocity * e = Initial velocity.

Lets suppose that after collision the velocities of the sphere is vx and vy. so we have four variables final angular velocity,final velocity,Vx and Vy.
But we have just two equations. we can get one more by angular conservation which i am unable to get through and we are still one short.

Using energy to find velocities is a crock usually.

You can treat the vertical and horizontal comonents seperately.

You know how high i'll bounce from COR*inital h. so you can work out the vertical component of velocity.

the forward motion will depend on rotational speed, moment of inertia, radius and initial sideways velocity (which in this case is zero as its a verticl drop). I really can't be bothred deriving the equation you'll have to do that one yourself.

xxChrisxx said:
I really can't be bothred deriving the equation you'll have to do that one yourself.

I didnt tell you to derive any equation. I am waiting for an insight to the problem. That how can we apply angular conservation. You have just told me the things which i know already except the way in which we can find Vy. But you have failed to give any other insight so that i can develop an equation for angular momentum

FedEx said:
I didnt tell you to derive any equation. I am waiting for an insight to the problem. That how can we apply angular conservation. You have just told me the things which i know already except the way in which we can find Vy. But you have failed to give any other insight so that i can develop an equation for angular momentum

Oh don't get me wrong, I know you didn't ask me to derive it. I didnt meant to imply that you did or that you were being lazy or anything like that. Sorry, I didnt mean for the post to come across in a harsh manner.

It's just that to he honest I couldn't remember the below formula and couldn't derive it when I wrote that, but didnt want to admit it! :D

I've been looking through my notes on oblique impacts and found this.

Vxnew = (Vx + Awr) / (1+A)
where A = I / ma^2

Set the inital velocity in x to be zero due to the vertical drop and bobs your uncle. This is a model for impacts that involve rolling when contact occurs. Which is the only thing that can happen in this case due to no horizontal inital velocity.

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xxChrisxx said:
Oh don't get me wrong, I know you didn't ask me to derive it. I didnt meant to imply that you did or that you were being lazy or anything like that. Sorry, I didnt mean for the post to come across in a harsh manner.

It's just that to he honest I couldn't remember the below formula and couldn't derive it when I wrote that, but didnt want to admit it! :D

I've been looking through my notes on oblique impacts and found this.

Vxnew = (Vx + Awr) / (1+A)
where A = I / ma^2

Set the inital velocity in x to be zero due to the vertical drop and bobs your uncle. This is a model for impacts that involve rolling when contact occurs. Which is the only thing that can happen in this case due to no horizontal inital velocity.

Sorry chris... I was a bit harsh too.I didn't knew that you were such a cheery guy.

Now back to the problem.

I am not able to understand how we get the formula mentioned by you. Do you know its possible derivation?

i'll have a crack at it. It's goig to be diffucult to get across on here. it'll probably need a diagram to help explain. also can someone please check I haven't messed up the derivation.

m=mass
T=torque
F=force
I=moment of inertia
w=angular velocity
We start off with:

F=ma

that can be rewritten as the integral, so we can get the inital and final velocities.

Fdt=mdv
F=m(vi - vf) (1)

Now we know that as the ball is spinning and it hits the ground, a force in x will occur. This force acting at the contact patch acts as a torque on the ball.

T=Fr

Like F=ma this can be rewritten as the integral to get.

T=I(wi-wf)
Fr=I(wi-wf) (2)

We also know that after impact, converting angular velocity to linear velocity we have.

vf=wf*r (3)What we do is then is:

rearrange (2) for F and put this into the rearranged F=ma. equation 1

F=(I(wi-wf))/r

(I(wi-wf))/r = m(vi-vf)

This gives us a problem as we don't know either vf or wf. However we do have equation 3 so we can rwrite it as 1 unknown. As we want to know vf, we rewrite wf in terms of this.(I(wi-wf))/r = m(vi-vf)
wf=vf/r

(I(wi-(vf/r)) = M(vi-vf) (4)

4 can be expanded and rearranged to get

Vi - (Iw/rm) = Vf ((I/r^2m) + 1)

That is the derevation. The eqaution can be used in the form above. But to make it look tidy, we rewrite the left ahnd bracket multiplied by r/r.

Vi - (Iwr/r^2m)

Vi - (Iw/r^2m)r

The (Iw/r^2m) was given the label A to make it easier to write. Tidying up the eqaution gives.

Vf = (Vi + Awr)/(1+A)

As there is no inital velocity in x. we have.

Vx final = Awr/(1+A)

Okey... I got the concept. Let me derive the formula on my own from the points which you have made. And let's see if it matches yours

## 1. What is rotational motion?

Rotational motion is the movement of an object in a circular or curved path around a fixed point, also known as the axis of rotation. It involves the concept of spinning or rotating and is a key aspect of rotational mechanics.

## 2. What is the difference between rotational and linear motion?

The main difference between rotational and linear motion is the type of path an object takes. Linear motion is a straight line movement, while rotational motion is a circular or curved movement. Additionally, rotational motion involves a fixed point of rotation, whereas linear motion does not.

## 3. How is rotational motion related to torque?

Rotational motion is directly related to torque, which is the force that causes an object to rotate around an axis. The greater the torque applied, the greater the rotational motion will be. This relationship is described by the equation: torque = force x distance from axis of rotation.

## 4. Can an object have both rotational and linear motion?

Yes, an object can have both rotational and linear motion at the same time. For example, a spinning top has both rotational motion around its axis and linear motion as it moves across the ground. These types of motion can occur simultaneously and are not mutually exclusive.

## 5. What are some real-life examples of rotational motion?

Some common examples of rotational motion include the spinning of a top, the rotation of a wheel on a car, the swinging of a pendulum, and the spinning of Earth on its axis. Other examples include the rotation of a ceiling fan, a spinning coin, and the rotation of a merry-go-round at a playground.