# 11.1 Rolling motion  (Page 2/6)

 Page 2 / 6

From [link] (a), we see the force vectors involved in preventing the wheel from slipping. In (b), point P that touches the surface is at rest relative to the surface. Relative to the center of mass, point P has velocity $\text{−}R\omega \stackrel{^}{i}$ , where R is the radius of the wheel and $\omega$ is the wheel’s angular velocity about its axis. Since the wheel is rolling, the velocity of P with respect to the surface is its velocity with respect to the center of mass plus the velocity of the center of mass with respect to the surface:

${\stackrel{\to }{v}}_{P}=\text{−}R\omega \stackrel{^}{i}+{v}_{\text{CM}}\stackrel{^}{i}.$

Since the velocity of P relative to the surface is zero, ${v}_{P}=0$ , this says that

${v}_{\text{CM}}=R\omega .$

Thus, the velocity of the wheel’s center of mass is its radius times the angular velocity about its axis. We show the correspondence of the linear variable on the left side of the equation with the angular variable on the right side of the equation. This is done below for the linear acceleration.

If we differentiate [link] on the left side of the equation, we obtain an expression for the linear acceleration of the center of mass. On the right side of the equation, R is a constant and since $\alpha =\frac{d\omega }{dt},$ we have

${a}_{\text{CM}}=R\alpha .$

Furthermore, we can find the distance the wheel travels in terms of angular variables by referring to [link] . As the wheel rolls from point A to point B , its outer surface maps onto the ground by exactly the distance travelled, which is ${d}_{\text{CM}}.$ We see from [link] that the length of the outer surface that maps onto the ground is the arc length $R\theta \text{​}$ . Equating the two distances, we obtain

${d}_{\text{CM}}=R\theta .$

## Rolling down an inclined plane

A solid cylinder rolls down an inclined plane without slipping, starting from rest. It has mass m and radius r . (a) What is its acceleration? (b) What condition must the coefficient of static friction ${\mu }_{\text{S}}$ satisfy so the cylinder does not slip?

## Strategy

Draw a sketch and free-body diagram, and choose a coordinate system. We put x in the direction down the plane and y upward perpendicular to the plane. Identify the forces involved. These are the normal force, the force of gravity, and the force due to friction. Write down Newton’s laws in the x - and y -directions, and Newton’s law for rotation, and then solve for the acceleration and force due to friction.

## Solution

1. The free-body diagram and sketch are shown in [link] , including the normal force, components of the weight, and the static friction force. There is barely enough friction to keep the cylinder rolling without slipping. Since there is no slipping, the magnitude of the friction force is less than or equal to ${\mu }_{S}N$ . Writing down Newton’s laws in the x - and y -directions, we have
$\sum {F}_{x}=m{a}_{x};\phantom{\rule{0.5em}{0ex}}\sum {F}_{y}=m{a}_{y}.$

Substituting in from the free-body diagram,
$\begin{array}{ccc}\hfill mg\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta -{f}_{\text{S}}& =\hfill & m{\left({a}_{\text{CM}}\right)}_{x},\hfill \\ \hfill N-mg\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta & =\hfill & 0,\hfill \\ \hfill {f}_{\text{S}}& \le \hfill & {\mu }_{\text{S}}N,\hfill \end{array}$
we can then solve for the linear acceleration of the center of mass from these equations:
${\left({a}_{\text{CM}}\right)}_{x}=g\left(\text{sin}\phantom{\rule{0.2em}{0ex}}\theta -{\mu }_{S}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta \right).$

However, it is useful to express the linear acceleration in terms of the moment of inertia. For this, we write down Newton’s second law for rotation,
$\sum {\tau }_{\text{CM}}={I}_{\text{CM}}\alpha .$

The torques are calculated about the axis through the center of mass of the cylinder. The only nonzero torque is provided by the friction force. We have
${f}_{\text{S}}r={I}_{\text{CM}}\alpha .$

Finally, the linear acceleration is related to the angular acceleration by
${\left({a}_{\text{CM}}\right)}_{x}=r\alpha .$

These equations can be used to solve for ${a}_{\text{CM}},\alpha ,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{f}_{\text{S}}$ in terms of the moment of inertia, where we have dropped the x -subscript. We write ${a}_{\text{CM}}$ in terms of the vertical component of gravity and the friction force, and make the following substitutions.
${a}_{\text{CM}}=g\text{sin}\phantom{\rule{0.2em}{0ex}}\theta -\frac{{f}_{\text{S}}}{m}$

${f}_{\text{S}}=\frac{{I}_{\text{CM}}\alpha }{r}=\frac{{I}_{\text{CM}}{a}_{\text{CM}}}{{r}^{2}}$
From this we obtain
$\begin{array}{cc}\hfill {a}_{\text{CM}}& =g\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta -\frac{{I}_{\text{CM}}{a}_{\text{CM}}}{m{r}^{2}},\hfill \\ & =\frac{mg\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta }{m+\left({I}_{\text{CM}}\text{/}{r}^{2}\right)}.\hfill \end{array}$

Note that this result is independent of the coefficient of static friction, ${\mu }_{\text{S}}$ .
Since we have a solid cylinder, from [link] , we have ${I}_{\text{CM}}=m{r}^{2}\text{/}2$ and
${a}_{\text{CM}}=\frac{mg\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta }{m+\left(m{r}^{2}\text{/}2{r}^{2}\right)}=\frac{2}{3}g\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta .$

Therefore, we have
$\alpha =\frac{{a}_{\text{CM}}}{r}=\frac{2}{3r}g\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta .$
2. Because slipping does not occur, ${f}_{\text{S}}\le {\mu }_{\text{S}}N$ . Solving for the friction force,
${f}_{\text{S}}={I}_{\text{CM}}\frac{\alpha }{r}={I}_{\text{CM}}\frac{\left({a}_{\text{CM}}\right)}{{r}^{2}}=\frac{{I}_{\text{CM}}}{{r}^{2}}\left(\frac{mg\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta }{m+\left({I}_{\text{CM}}\text{/}{r}^{2}\right)}\right)=\frac{mg{I}_{\text{CM}}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta }{m{r}^{2}+{I}_{\text{CM}}}.$

Substituting this expression into the condition for no slipping, and noting that $N=mg\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta$ , we have
$\frac{mg{I}_{\text{CM}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta }{m{r}^{2}+{I}_{\text{CM}}}\le {\mu }_{\text{S}}mg\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta$

or
${\mu }_{\text{S}}\ge \frac{\text{tan}\phantom{\rule{0.2em}{0ex}}\theta }{1+\left(m{r}^{2}\text{/}{I}_{\text{CM}}\right)}.$

For the solid cylinder, this becomes
${\mu }_{\text{S}}\ge \frac{\text{tan}\phantom{\rule{0.2em}{0ex}}\theta }{1+\left(2m{r}^{2}\text{/}m{r}^{2}\right)}=\frac{1}{3}\text{tan}\phantom{\rule{0.2em}{0ex}}\theta .$

#### Questions & Answers

a non-uniform boom of a crane 15m long, weighs 2800nts, with its center of gravity at 40% of its lenght from the hingr support. the boom is attached to a hinge at the lower end. rhe boom, which mAKES A 60% ANGLE WITH THE HORIZONTAL IS SUPPORTED BY A HORIZONTAL GUY WIRE AT ITS UPPER END. IF A LOAD OF 5000Nts is hung at the upper end of the boom, find the tension in the guywire and the components of the reaction at the hinge.
what is the centripetal force
Of?
John
centripetal force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle. Fc = MV²/r
Sampson
centripetal force is the force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle. Fc = MV²/r
Sampson
I do believe the formula for centripetal force is F=MA or F=m(v^2/r)
John
I mean the formula is Fc= Mass multiplied by square of velocity all over the Radius of the circle
Sampson
Yes
John
The force is equal to the mass times the velocity squared divided by the radius
John
That's the current chapter I'm on in my engineering physics class
John
In Example, we calculated the final speed of a roller coaster that descended 20 m in height and had an initial speed of 5 m/s downhill. Suppose the roller coaster had had an initial speed of 5 m/s uphill instead, and it coasted uphill, stopped, and then rolled back down to a final point 20 m bel
A steel lift column in a service station is 4 meter long and .2 meter in diameter. Young's modulus for steel is 20 X 1010N/m2.  By how much does the column shrink when a 5000- kg truck is on it?
what exactly is a transverse wave
does newton's first law mean that we don't need gravity to be attracted
no, it just means that a brick isn't gonna move unless something makes it move. if in the air, moves down because of gravity. if on floor, doesn't move unless something has it move, like a hand pushing the brick. first law is that an object will stay at rest or motion unless another force acts upon
Grant
yeah but once gravity has already been exerted .. i am saying that it need not be constantly exerted now according to newtons first law
Dharmee
gravity is constantly being exerted. gravity is the force of attractiveness between two objects. you and another person exert a force on each other but the reason you two don't come together is because earth's effect on both of you is much greater
Grant
maybe the reason we dont come together is our inertia only and not gravity
Dharmee
this is the definition of inertia: a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.
Grant
the earth has a much higher affect on us force wise that me and you together on each other, that's why we don't attract, relatively speaking of course
Grant
quite clear explanation but i just want my mind to be open to any theory at all .. its possible that maybe gravity does not exist at all or even the opposite can be true .. i dont want a fixed state of mind thats all
Dharmee
why wouldn't gravity exist? gravity is just the attractive force between two objects, at least to my understanding.
Grant
earth moves in a circular motion so yes it does need a constant force for a circular motion but incase of objects on earth i feel maybe there is no force of attraction towards the centre and its our inertia forcing us to stay at a point as once gravity had acted on the object
Dharmee
why should it exist .. i mean its all an assumption and the evidences are empirical
Dharmee
We have equations to prove it and lies of evidence to support. we orbit because we have a velocity and the sun is pulling us. Gravity is a law, we know it exists.
Grant
yeah sure there are equations but they are based on observations and assumptions
Dharmee
g is obtained by a simple pendulum experiment ...
Dharmee
gravity is tested by dropping a rock...
Grant
and also there were so many newtonian laws proved wrong by einstein . jus saying that its a law doesnt mean it cant be wrong
Dharmee
pendulum is good for showing energy transfer, here is an article on the detection of gravitational waves: ***ligo.org/detections.php
Grant
yeah but g is calculated by pendulum oscillations ..
Dharmee
thats what .. einstein s fabric model explains that force of attraction by sun on earth but i am talking about force of attraction by earth on objects on earth
Dharmee
no... this is how gravity is calculated:F = G*((m sub 1*m sub 2)/r^2)
Grant
gravitational constant is obtained EXPERIMENTALLY
Dharmee
the G part
Dharmee
Calculate the time of one oscillation or the period (T) by dividing the total time by the number of oscillations you counted. Use your calculated (T) along with the exact length of the pendulum (L) in the above formula to find "g." This is your measured value for "g."
Dharmee
G is the universal gravitational constant. F is the gravity
Grant
search up the gravity equation
Grant
yeahh G is obtained experimentally
Dharmee
sure yes
Grant
thats what .. after all its EXPERIMENTALLY calculated so its empirical
Dharmee
yes... so where do we disagree?
Grant
its empirical whixh means it can be proved wrong
Dharmee
so cant just say why wouldnt gravity exists
Dharmee
the constant, sure but extremely unlikely it is wrong. gravity however exists, there are equations and loads of support surrounding the concept. unfortunately I don't have a high enough background in physics but have this discussion with a physicist
Grant
can u suggest a platform where i can?
Dharmee
stack overflow
Grant
stack exchange, physics section***
Grant
its an app?
Dharmee
there is! it is also a website as well
Grant
okayy
Dharmee
nice talking to you
Dharmee
***physics.stackexchange.com/
Grant
likewise :)
Grant
What is the percentage by massof oxygen in Al2(so4)3
A spring with 50g mass suspended from it,has its length extended by 7.8cm 1.1 determine the spring constant? 1.2 it is observed that the length of the spring decreases by 4.7cm,from its original length, when a toy is place on top of it. what is the mass of the toy?
solution mass = 50g= 0.05kg force= 50 x 10= 500N extension= 7.8cm = 0.078m using the formula Force= Ke K = force/extension 500/.078 = 6410.25N/m
Sampson
1.2 Decrease in length= -4.7cm =-0.047m mass=? acceleration due to gravity= 10 force = K x e force= mass x acceleration m x a = K x e mass = K x e/acceleration = 6410.25 x 0.047/10 = 30.13kg
Sampson
1.1 6.28Nm-¹
Anita
1.2 0.03kg or 30g
Anita
I used g=9.8ms-²
Anita
you should explain how yoy got the answer Anita
Grant
ok
Anita
with the fomular F=mg I got the value for force because now the force acting on the spring is the weight of the object and also you have to convert from grams to kilograms and cm to meter
Anita
so the spring constant K=F/e where F is force and e is extension
Anita
In this first example why didn't we use P=P° + ¶hg where ¶ is density
Density = force applied x area p=fA =p = mga, then a=h therefore substitute =p =mgh
Hlehle
Hlehle
sorry I had a little typo in my question
Anita
Density = m/v (mass/volume) simple as that
Augustine
Hlehle vilakazi how density is equal to force * area and you also wrote p= mgh which is machenical potential energy ? how ?
Manorama
what is wave
who can state the third equation of motion
Alfred
wave is a distrubance that travelled in medium from one point to another with carry energy .
Manorama
wave is a periodic disturbance that carries energy from one medium to another..
Augustine
what exactly is a transverse wave then?
Dharmee
two particles rotate in a rigid body then acceleration will be ?
same acceleration for all particles because all prticles will be moving with same angular velocity.so at any time interval u find same acceleration of all the prticles
Zaheer
what is electromagnetism
It is the study of the electromagnetic force, one of the four fundamental forces of nature. ... It includes the electric force, which pushes all charged particles, and the magnetic force, which only pushes moving charges.
Energy
what is units?
units as in how
praise
What is th formular for force
F = m x a
Santos
State newton's second law of motion
can u tell me I cant remember
Indigo
force is equal to mass times acceleration
Santos
The acceleration of a system is directly proportional to the and in the same direction as the external force acting on the system and inversely proportional to its mass that is f=ma
David
The rate of change of momentum of a body is directly proportional to the force exerted on that body.
Rani