Chapter 13.P, Problem 1P

A particle P moves with constant angular speed $\omega$ around a circle whose center is at the origin and whose radius is R. The particle is said to be in uniform circular motion. Assume that the motion is counterclockwise and that the particle is at the point $\left(R,0\right)$ when $t=0$. The position vector at time $t\ge 0$ is $r\left(t\right)=R\cos \omega ti\text{ + }R\sin \omega tj$.

(a) Find the velocity vector v and show that $v\cdot r=0$. Conclude that v is tangent to the circle and points in the direction of the motion.

(b) Show that the speed $\left|v\right|$ of the particle is the constant $\omega R$. The period T of the particle is the time required for one complete revolution. Conclude that

$T=\frac{2\pi R}{\left|v\right|}=\frac{2\pi }{\omega }$

(c) Find the acceleration vector a. Show that it is proportional to r and that it points toward the origin. An acceleration with this property is called a centripetal acceleration. Show that the magnitude of the acceleration vector is $\left|a\right|=R{\omega }^2$.

(d) Suppose that the particle has mass m. Show that the magnitude of the force F that is required to produce this motion, called a centripetal force, is

$\left|F\right|=\frac{m{\left|v\right|}^2}{R}$

To determine

a)

To find:

The velocity vector $v$ and show that $v·r=0.$ Conclude that $v$ is tangent to the circle and points in the direction of the motion.

Answer to Problem 1P

Solution:

$v= \omega R\left(-\sin \omega ti+\cos \omega tj\right)$

$v·r=0$

Explanation of Solution

1) Concept:

If $r\left(t\right)$ is the position vector of the particle at time $t$, the velocity vector at time $t$ is given by $v\left(t\right)=r\text{'}\left(t\right)$.

2) Given:

$r\left(t\right)=R\cos \omega ti+R\sin \omega t j$

3) Calculation:

Consider $r\left(t\right)=R\cos \omega ti+R\sin \omega t j$

The velocity vector at time $t$ is given by $v\left(t\right)=r\text{'}\left(t\right)$.

Therefore, differentiating $r\left(t\right)$ with respect to $t$,

$v\left(t\right)={\left(R\cos \omega t\right)}^{\text{'}}i+{\left(R\sin \omega t\right)}^{\text{'}}j$

$v\left(t\right)= -\omega R\sin \omega ti+\omega R\cos \omega tj$

Now $v·r=\left(-\omega R\sin \omega ti+\omega R\cos \omega tj\right)·\left(R\cos \omega ti+R\sin \omega t j\right)$

$=\left(-\omega R\sin \omega t\right)\left(R\cos \omega t\right)+\left(\omega R\cos \omega t\right)\left(R\sin \omega t\right)$

$= -\omega R^2\sin \omega t\cos \omega t+\omega R^2\sin \omega t\cos \omega t$

$v·r=0$

As $v·r=0$, $v$ and $r$ are perpendicular to each other.

$r$ is along the radius of the circle and $v$ is perpendicular to $r;$ therefore, $v$ is tangent to the circle.

Since $v$ is a velocity vector, $v$ points in the direction of motion.

Conclusion:

$v= \omega R\left(-\sin \omega ti+\cos \omega tj\right).$

$v·r=0$

$v$ is tangent to the circle and points in the direction of the motion.

To determine

b)

To show:

The speed of particle $\left|v\right|$ is constant $\omega R$ and conclude the period of particle is $T=\frac{2\pi R}{ \omega R}=\frac{2\pi }{\omega }$.

Answer to Problem 1P

Solution:

$\left|v\right|=\omega R$

$T=\frac{2\pi }{\omega }$

Explanation of Solution

1)  Concept:

Speed of a particle is the magnitude of velocity vector $\left|v\left(t\right)\right|$.

$Time =\frac{distance}{speed}$

2) Given:

$r\left(t\right)=R\cos \omega ti+R\sin \omega t j$

3) Calculation:

From part (a),

$v= -\omega R\sin \omega ti+\omega R\cos \omega tj$

Therefore, as the speed of a particle is the magnitude of velocity vector $\left|v\left(t\right)\right|,$

$\left|v\left(t\right)\right|=\sqrt{(-\omega R\sin \omega t{)}^2+{\left(\omega R\cos \omega t\right)}^2}$

$=\sqrt{{\omega }^2{R}^2{\sin }^2\omega t+{\omega }^2{R}^2{\cos }^2\omega t}$

$= \sqrt{{\omega }^2{R}^2({\sin }^2\omega t+{\cos }^2\omega t)}$

$\left|v\left(t\right)\right|=\omega R$

The period $T$ of particle is the time required to complete one revolution.

$Time =\frac{distance}{speed}$

At the speed $\omega R,$ the particle completes one revolution of distance $2\pi R$.

Therefore, period is $T=\frac{2\pi R}{\omega R}=\frac{2\pi }{\omega }$

Conclusion:

The speed of particle $\left|v\right|$ is constant $\omega R,$ and the particle completes one revolution in period $T=\frac{2\pi R}{\omega R}=\frac{2\pi }{\omega }.$

To determine

c)

To find:

Acceleration vector $a$ and to show it is proportional to $r$ and it points towards the origin. Also show that the magnitude of acceleration vector is $\left|a\right|=R{\omega }^2$.

Answer to Problem 1P

Solution:

$a=-{\omega }^{2}r$

Explanation of Solution

1) Concept:

The acceleration of a particle is $a\left(t\right)=v^{\text{'}}\left(t\right)=r\text{'}\text{'}\left(t\right)$.

2) Calculation:

From part (a),$v= -\omega R\sin \omega ti+\omega R\cos \omega tj$

The acceleration of particle is $a\left(t\right)=v^{\text{'}}\left(t\right)=r\text{'}\text{'}\left(t\right)$

Therefore, differentiating $v\left(t\right)$ with respect to $t$,

$a\left(t\right)={\left(-\omega R\sin \omega t\right)}^{\text{'}} i+{\left(\omega R\cos \omega t\right)}^{\text{'}}j$

$=- {\omega }^{2}R\cos \omega ti-{\omega }^{2}R\sin \omega tj$

$=- {\omega }^{2}R( \cos \omega ti+\sin \omega tj)$

$= - {\omega }^{2}r$

This concludes that acceleration is proportional to $r,$ and as it has a negative sign, it points in the opposite direction of $r,$ that is, towards the origin.

The acceleration with this property is called centripetal acceleration.

Now, the magnitude of acceleration vector is given by

$\left|a\right|=\left|- {\omega }^{2}r\right|$

$={\omega }^2\left|R\left(\cos \omega ti+\sin \omega tj\right)\right|$

$= {\omega }^{2}R\left|\cos \omega ti+\sin \omega tj\right|$

$\left|a\right|={\omega }^{2}R$.

Conclusion:

The acceleration of particle is $a\left(t\right)= - {\omega }^{2}r,$ and it is proportional to $r.$ As it has a negative sign, it points in the opposite direction of $r,$ that is, towards the origin.

Magnitude of acceleration vector is given by $\left|a\right|={\omega }^{2}R$.

To determine

d)

To show:

The magnitude of force required to produce this motion is $F=\frac{m{\left|v\right|}^2}{R}$

Answer to Problem 1P

Solution:

$\left|F\right|=\frac{m{\left|v\right|}^2}{R}$

Explanation of Solution

1) Concept:

Here use Newton’s Second Law of motion.

Newton’s Second Law:

If at any time $t$, if a force $F\left(t\right)$ acts on an object of mass $m$ producing an acceleration $a\left(t\right)$, then

$F\left(t\right)=ma\left(t\right)$

2) Calculation:

By using Newton’s Second Law of motion, $F\left(t\right)=ma\left(t\right)$

Therefore, magnitude of force required to produce this motion is

$\left|F\right|=m\left|a\right|$

From part (c),$\left|a\right|={\omega }^{2}R$

Hence, $\left|F\right|=m{\omega }^{2}R$

$= \frac{m{\omega }^2{R}^2}{R}$

$= \frac{m{\left(\omega R\right)}^2}{R}$

$\left|F\right|=\frac{m{\left|v\right|}^2}{R}$

Conclusion:

The magnitude of force required to produce this motion is $\left|F\right|=\frac{m{\left|v\right|}^2}{R}$