4.11 Uniform circular motion

UNIFORM CIRCULAR MOTION

When an object follows a circular path at a constant speed, the motion of the object is called uniform circular motion. The word “uniform” refers to the speed, which is uniform (constant) throughout the motion. Let’s assume an object is moving with uniform speed \(v\) in a circle of radius \(r\). Since the velocity of the object is changing continuously in direction, the object undergoes acceleration. Let us find the magnitude and the direction of this acceleration.

Centripetal acceleration

The acceleration of an object moving with speed \(v\) in a circle of radius \(r\) has a magnitude \(v^{2} / r\) and is always directed towards the centre. This is why this acceleration is called centripetal acceleration. 

\(a_{\mathrm{c}}=\left(\frac{v}{r}\right) v=v^{2} / r\)

Angular Velocity

Figure 4o

Suppose a object \(P\) is moving in a circle of radius \(r\) (Figure 4o). Let \(O\) be the centre of the circle. Let \(O\) be the origin and \(O X\) the \(X\)-axis. We call \(\theta\) the angular position of the object \(P\). As the object moves on the circle, from position \(P\) to \(P^{\prime}\) in time \(\Delta t\) the line \( O P\) turns through an angle \(\Delta \theta\) as shown in the figure 4o. \(\Delta \theta\) is called the angular distance. 

The rate of change of angular position (displacement) is called angular velocity. Thus, the angular velocity is

\(\omega=\lim _{\Delta t \rightarrow 0} \frac{\Delta \theta}{\Delta t}=\frac{d \theta}{d t}\)

Now, if the distance travelled by the object during the time \(\Delta t\) is \(\Delta \mathrm{s}\), i.e. \(P P^{\prime}\) is \(\Delta \mathrm{s}\), then speed of the object:
\(
v=\frac{\Delta s}{\Delta t}
\)
but \(\Delta s=r \Delta \theta\). Therefore :
\(
\begin{gathered}
v=r \frac{\Delta \theta}{\Delta t}=\mathrm{r} \omega \\
v=r \omega
\end{gathered}
\)
We can express centripetal acceleration \(a_{c}\) in terms of angular speed as

\(\begin{aligned}
&a_{c}=\frac{v^{2}}{r}=\frac{\omega^{2} R^{2}}{r}=\omega^{2} r \\
&a_{c}=\omega^{2} r
\end{aligned}\)

The time taken by an object to make one revolution is known as its time period \(T\) and the number of revolution made in one second is called its frequency \(f(=1 / T)\). However, during this time the distance moved by the object is \(s=2 \pi r\).
Therefore, \(v=2 \pi r / T=2 \pi r f\)
In terms of frequency \(v\), we have
\(
\begin{gathered}
\omega=2 \pi f \\
v=2 \pi r f \\
a_{c}=4 \pi^{2} f^{2} r
\end{gathered}
\)

Angular acceleration

The rate of change of angular velocity is called angular acceleration. Thus, the angular acceleration is
\(
\alpha=\frac{d \omega}{d t}=\frac{d^{2} \theta}{d t^{2}}
\)
If the angular acceleration \(\alpha\) is constant, we have
\(
\begin{aligned}
&\theta=\omega_{0} t+\frac{1}{2} \alpha t^{2} \\
&\omega=\omega_{0}+\alpha t
\end{aligned}
\)
and
\(\omega^{2}=\omega_{0}^{2}+2 \alpha \theta\)
where \(\omega_{0}\) and \(\omega\) are the angular velocities at \(t=0\) and at time \(t\) and \(\theta\) is the angular position at time \(t\).

Tangential acceleration

The direction of velocity \(v\) is along the tangent to the circle at every point. We know that the \(v\) is the linear speed of the object and is given by

\(v=r \omega\)

Differentiating equation the above equation with respect to time, the rate of change of speed is
\(
\begin{aligned}
\text { or, } \quad & a_{t}=\frac{d v}{d t}=r \frac{d \omega}{d t} \\
a_{t} &=r \alpha .
\end{aligned}
\)
Remember that \(a_{t}=\frac{d v}{d t}\) is the rate of change of speed and is not the rate of the change of velocity. It is, therefore, not equal to the net acceleration.

We should remember that \(a_{t}\) is called the tangential acceleration along the tangent and hence we have used the suffix \(t\). 

Example: 4.8

A particle moves in a circle of radius \(20 \mathrm{~cm}\) with a linear speed of \(10 \mathrm{~m} / \mathrm{s}\). Find the angular velocity.
Solution: The angular velocity is
\(
\omega=\frac{v}{r}=\frac{10 \mathrm{~m} / \mathrm{s}}{20 \mathrm{~cm}}=50 \mathrm{rad} / \mathrm{s}
\)

Example: 4.9

A particle travels in a circle of radius \(20 \mathrm{~cm}\) at a speed that uniformly increases. If the speed changes from \(5.0 \mathrm{~m} / \mathrm{s}\) to \(6.0 \mathrm{~m} / \mathrm{s}\) in \(2.0 \mathrm{~s}\), find the angular acceleration.
Solution: The tangential acceleration is given by
\(
\begin{aligned}
a_{t} &=\frac{d v}{d t}=\frac{v_{2}-v_{1}}{t_{2}-t_{1}} \\
&=\frac{6 \cdot 0-5 \cdot 0}{2 \cdot 0} \mathrm{~m} / \mathrm{s}^{2}=0.5 \mathrm{~m} / \mathrm{s}^{2} .
\end{aligned}
\)
The angular acceleration is \(\alpha=a_{t} / r\)
\(
=\frac{0.5 \mathrm{~m} / \mathrm{s}^{2}}{20 \mathrm{~cm}}=2.5 \mathrm{rad} / \mathrm{s}^{2} \text {. }
\)

UNIT VECTORS ALONG THE RADIUS AND THE TANGENT

Let’s calculate the \(e_{r}\) known as the radial unit vector and \(\overrightarrow{e_{t}}\) the tangential unit vector along the radius of a circle. Consider an object moving in a circle with the origin at \(O\) and, the line \(O X\) as the \(X\)-axis and a perpendicular radius \(O Y\) as the \(Y\)-axis. The angular position of the object at the point \(P\) is \(\theta\).

Figure 4p

 

Draw a unit vector \(\overrightarrow{P A}=\overrightarrow{e_{r}}\) along the outward radius and a unit vector \(\overrightarrow{P B}=\overrightarrow{e_{t}}\) along the tangent in the direction of increasing \(\theta\). We call \(\overrightarrow{e_{r}}\) the radial unit vector and \(\overrightarrow{e_{t}}\) the tangential unit vector. Draw \(P X^{\prime}\) parallel to the \(X\)-axis and \(P Y^{\prime}\) parallel to the \(Y\)-axis. From the figure 4p,

\(\overrightarrow{P A}=\vec{i} P A \cos \theta+\vec{j} P A \sin \theta\)
or, \(\quad \overrightarrow{e_{r}}=\vec{i} \cos \theta+\vec{j} \sin \theta\). …………….. (4.1)

where \(\vec{i}\) and \(\vec{j}\) are the unit vectors along the \(X\) and \(Y\) axes respectively. Similarly,
\(\overrightarrow{P B}=-\vec{i} P B \sin \theta+\vec{j} P B \cos \theta\)
or, \(\quad \overrightarrow{e_{t}}=-\vec{i} \sin \theta+\vec{j} \cos \theta\). …………….. (4.2)

ACCELERATION IN CIRCULAR MOTION

From Figure 4p, It is clear that the position vector of the object at time \(t\) is
\(
\begin{aligned}
\vec{r} &=\overrightarrow{O P}=O P \overrightarrow{e_{r}} \\
&=r(\vec{i} \cos \theta+\vec{j} \sin \theta) \ldots \ldots \text { (4.3) }
\end{aligned} \)
Differentiating equation (4.3) with respect to time, the velocity of the object at time \(t\) is
\(
\begin{aligned}
\vec{v}=\frac{d \vec{r}}{d t} &=\frac{d}{d t}[r(\vec{i} \cos \theta+\vec{j} \sin \theta)] \\
&=r\left[\vec{i}\left(-\sin \theta \frac{d \theta}{d t}\right)+\vec{j}\left(\cos \theta \frac{d \theta}{d t}\right)\right] \\
&=r \omega[-\vec{i} \sin \theta+\vec{j} \cos \theta] \ldots \ldots \text { (4.4) }
\end{aligned} \)

The term \(r \omega\) is the speed of the object at time \(t\) (\(v=r \omega\)) and the vector in the square bracket is the unit vector \(\overrightarrow{e_{t}}\) along the tangent. Thus, the velocity of the object at any instant is along the tangent to the circle and its magnitude is \(v=r \omega\).

The acceleration of the object at time \(t\) is \(\vec{a}=\frac{d \vec{v}}{d t} \cdot\) From (equation 4.4),
\(
\begin{aligned}
\vec{a} &=r\left[\omega \frac{d}{d t}[-\vec{i} \sin \theta+\vec{j} \cos \theta]+\frac{d \omega}{d t}[-\vec{i} \sin \theta+\vec{j} \cos \theta]\right] \\
&=\omega r\left[-\vec{i} \cos \theta \frac{d \theta}{d t}-\vec{j} \sin \theta \frac{d \theta}{d t}\right]+r \frac{d \omega}{d t} \overrightarrow{e_{t}} \\
&=-\omega^{2} r[\vec{i} \cos \theta+\vec{j} \sin \theta]+r \frac{d \omega}{d t} \overrightarrow{e_{t}} \\
&=-\omega^{2} r \overrightarrow{e_{r}}+\frac{d v}{d t} \overrightarrow{e_{t}} \ldots \ldots \text { (4.5) }
\end{aligned}\)
where \(\overrightarrow{e_{r}}\) and \(\overrightarrow{e_{t}}\) are the unit vectors along the radial and tangential directions respectively and \(v\) is the speed of the objectat time \(t\). We have used
\(
r \frac{d \omega}{d t}=\frac{d}{d t}(r \omega)=\frac{d v}{d t} .
\)

Example: 4.10

Prove that if an object moves in a circle of radius \(r\) with a constant speed \(v\), its acceleration is \(v^{2} / r\) directed towards the centre. 

Solution: If a particle moves in the circle with a uniform speed, we call it a uniform circular motion. In this case, \(\frac{d v}{d t}=0\) and equation (4.5) gives
\(\vec{a}=-\omega^{2} r \overrightarrow{e_{r}}\)
Thus, the acceleration of the particle is in the direction of \(-\overrightarrow{e_{r}}\), that is, towards the centre. The magnitude of the acceleration is
\(
\begin{aligned}
a_{r} &=\omega^{2} r \\
&=\frac{v^{2}}{r^{2}} r=\frac{v^{2}}{r}
\end{aligned}\)

[ Note: some books use the notation \(a_{c}\) instead of \(a_{r}\) ]

Thus, if a particle moves in a circle of radius \(r\) with a constant speed \(v\), its acceleration is \(v^{2} / r\) directed towards the centre known as the centripetal acceleration. Note that the speed remains constant, the direction continuously changes, and hence the “velocity” changes and there is an acceleration during the motion.

Example: 4.11

Find the magnitude of the linear acceleration of a particle moving in a circle of radius \(10 \mathrm{~cm}\) with uniform speed completing the circle in \(4 \mathrm{~s}\).

Solution: The distance covered in completing the circle is \(2 \pi r=2 \pi \times 10 \mathrm{~cm}\). The linear speed is \(v=2 \pi r / t=\frac{2 \pi \times 10 \mathrm{~cm}}{4 \mathrm{~s}}=5 \pi \mathrm{cm} / \mathrm{s}.\)
The linear acceleration is
\(
a=\frac{v^{2}}{r}=\frac{(5 \pi \mathrm{cm} / \mathrm{s})^{2}}{10 \mathrm{~cm}}=2.5 \pi^{2} \mathrm{~cm} / \mathrm{s}^{2} \text {. }
\)
This acceleration is directed towards the centre of the circle.

Nonuniform Circular Motion

If the speed of the object moving in a circle is not constant, the acceleration has both the radial and the tangential components. According to equation (4.5), the radial and the tangential accelerations are
and \(\quad a_{r}=-\omega^{2} r=-v^{2} / r \text { and } \quad a_{t}=\frac{d v}{d t} \ldots \ldots \text { (4.6) }\) 
Thus, the component of the acceleration towards the centre is \(\omega^{2} r=v^{2} / r\) and the component along the tangent (along the direction of motion) is \(d v / d t\). The magnitude of the acceleration is
\(
a=\sqrt{a_{r}^{2}+a_{t}^{2}}=\sqrt{\left(\frac{v^{2}}{r}\right)^{2}+\left(\frac{d v}{d t}\right)^{2}}
\)

Figure 4q

The direction of this resultant acceleration makes an angle \(\alpha\) with the radius (figure 4q) where
\(
\tan \alpha=\left(\frac{d v}{d t}\right) /\left(\frac{v^{2}}{r}\right) .
\)

Example: 4.12

An object moves in a circle of radius \(20 \mathrm{~cm}\). It’s linear speed is given by \(v=2 t\), where \(t\) is in second and \(v\) in metre/second. Find the radial and tangential acceleration at \(t=3 \mathrm{~s}\).

Solution: The linear speed at \(t=3 \mathrm{~s}\) is
\(
v=2 t=6 \mathrm{~m} / \mathrm{s} \text {. }
\)
The radial acceleration at \(t=3 \mathrm{~s}\) is
\(
a_{r}=v^{2} / r=\frac{36 \mathrm{~m}^{2} / \mathrm{s}^{2}}{0 \cdot 20 \mathrm{~m}}=180 \mathrm{~m} / \mathrm{s}^{2} \text {. }
\)

The tangential acceleration is \(a_{t}=\frac{d v}{d t}=\frac{d(2 t)}{d t}=2 \mathrm{~m} / \mathrm{s}^{2} .\)

Centripetal Force

If an object moves in a circle as seen from an inertial frame, a resultant nonzero force must act on the particle. That is because an object moving in a circle is accelerated and acceleration can be produced in an inertial frame only if a resultant force acts on it. If the speed of the object remains constant, the acceleration of the object is towards the centre and its magnitude is \(v^{2} / r\). Here \(v\) is the speed of the object and \(r\) is the radius of the circle. The resultant force must act towards the centre and its magnitude \(F\) must satisfy
\(a=\frac{F}{m}\)
\(\frac{v^{2}}{r}=\frac{F}{m}\)
\(F=\frac{m v^{2}}{r} \ldots \ldots \text { (4.7) }\).
Since this resultant force is directed towards the centre, it is called centripetal force. Thus, a centripetal force of magnitude \(m v^{2} / r\) is needed to keep the object in a uniform circular motion.

CIRCULAR TURNINGS AND BANKING OF ROADS

When there is no friction the vertical component of the road acts as a normal force due to which the weight of the vehicle is balanced and the horizontal component gives in to the centripetal force in the direction of the centre of the curvature of the road. The inclination occurs at the horizontal and longitudinal axis. 

Figure 4r

Consider a car of mass \(m\) moving at a speed \(v\) is making a turn on the circular path of radius \(r\). Let \(\theta\) be the banking angle and \(f\) be the frictional force acting between the road and the car’s tires. The external forces acting on the vehicle are
(i) weight \(mg\)
(ii) Normal contact force \(N\) and
(iii) friction \(f\)

If the road is horizontal, the normal force \(N\) is vertically upward. The only horizontal force that can act towards the centre is the friction \(f\). The tires get a tendency to skid outward and the frictional force which opposes this skidding acts towards the centre. Thus, for a safe turn, we must have
\(\frac{v^{2}}{r}=\frac{f}{m}\)
or, \(\quad f=\frac{m v^{2}}{r}\)

However, there is a limit to the magnitude of the frictional force. If \(\mu_{s}\) is the coefficient of static friction between the tires and the road, the magnitude of friction \(f\) cannot exceed \(\mu_{s} N\). For vertical equilibrium \(N=mg\), so that
\(f \leq \mu_{s} mg\)
Thus, for a safe turn
\(\frac{m v^{2}}{r} \leq \mu_{s} mg\)

Therefore, \(\quad \mu_{s} \geq \frac{v^{2}}{r g}\).

Friction is not always reliable at circular turns if high speeds and sharp turns are involved. To avoid dependence on friction, the roads are banked at the turn so that the outer part of the road is somewhat lifted up as compared to the inner part (figure 4r). 

The surface of the road makes an angle \(\theta\) with the horizontal throughout the turn. The normal force \(N\) makes an angle \(\theta\) with the vertical. At the correct speed, the horizontal component of \(N\) is sufficient to produce the acceleration towards the centre and the self-adjustable frictional force keeps its value zero.

\(N \cos \theta \rightarrow\) Normal reaction along the vertical axis

\(\mathrm{f} \sin \theta \rightarrow\) Frictional force along the vertical axis

\(\mathrm{mg} \rightarrow\) Weight of the vehicle acting downwards

Applying Newton’s second law along the radius and the first law in the vertical direction,
\(\Rightarrow N \sin \theta=\frac{m v^{2}}{r}\)
and, \(N \cos \theta=mg \)
These equations give, \(\tan \theta=\frac{v^{2}}{r g}\); \(\theta=\tan ^{-1} \frac{v^{2}}{r g} \ldots \ldots \text { (4.8) }\)

The angle \(\theta\) depends on the speed of the vehicle as well as on the radius of the turn. If the speed of the car is a little less or a little more than the correct speed, the self-adjustable static friction operates between the tires and the road, and the vehicle does not skid or slip. If the speed is too different from that given by equation (4.8), even the maximum friction cannot prevent a skid or a slip.

Example: 4.13

The road at a circular turn of radius \(10 \mathrm{~m}\) is banked by an angle of \(10^{\circ}\). With what speed should a vehicle move on the turn so that the normal contact force is able to provide the necessary centripetal force?
Solution:  If \(v\) is the correct speed,
\(\tan \theta=\frac{v^{2}}{r g}\)

\(\quad \begin{aligned}
v &=\sqrt{r g \tan \theta} \\
&=\sqrt{(10 \mathrm{~m})\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)\left(\tan 10^{\circ}\right)}=4 \cdot 2 \mathrm{~m} / \mathrm{s}
\end{aligned}\)

CENTRIFUGAL FORCE

Let us assume an observer is sitting in a closed cabin which is made to rotate about the vertical \(Z\)-axis at a uniform angular velocity \(\omega\) (figure 4s). The \(X\) and \(Y\) axes are fixed in the cabin. Consider a heavy box of mass \(m\) kept on the floor at a distance \(r\) from the \(Z\)-axis. Suppose the floor and the box are rough and the box does not slip on the floor as the cabin rotates. The box is at rest with respect to the cabin and hence is rotating with respect to the ground at an angular velocity \(\omega\). Let us first analyze the motion of the box from the ground frame. In this frame (which is inertial) the box is moving in a circle of radius \(r\). It, therefore, has an acceleration \(v^{2} / r=\omega^{2} r\) towards the centre. The resultant force on the box must be towards the centre and its magnitude must be \(m \omega^{2} r\). The forces on the box are
(i) weight \(mg\)
(ii) normal force\(N\) by the floor
(iii) friction \(f\) by the floor.
Figure (4s) shows the free-body diagram for the box. Since the resultant is towards the centre and its magnitude is \(m \omega^{2} r\), we should have
\(f=m \omega^{2} r\)

The floor exerts a force of static friction \(f=m \omega^{2} r\) towards the origin.

When we observe the same box from the frame of the rotating cabin. The observer there finds that the box is at rest. If we apply Newton’s laws, the resultant force on the box should be zero. The weight and the normal contact force balance each other but the frictional force \(f=m \omega^{2} r\) acts on the box towards the origin. To make the resultant zero, a pseudo force must be assumed which acts on the box away from the centre (radially outward) and has a magnitude \(m \omega^{2} r\). This pseudo force is called the centrifugal force. Therefore, the Centrifugal force = \(m \omega^{2} r\).
The free-body diagram is shown in Figure 4s-c. As the box is at rest, Newton’s first law gives
\(f=m \omega^{2} r\)

EFFECT OF EARTH’S ROTATION ON APPARENT WEIGHT

The earth rotates about its axis at an angular speed of one revolution in 24 hours. The line joining the north and the south poles is the axis of rotation. Every point on the earth moves in a circle. A point at the equator moves in a circle of radius equal to the radius of the earth and the centre of the circle is same as the centre of the earth. Consider a place \(P\) on the earth (figure 4t).

 

Drop a perpendicular \(P C\) from \(P\) to the axis \(S N\). The place \(P\) rotates in a circle with the centre at \(C\). The radius of this circle is \(C P(r)\). The angle between the axis \(S N\) and the radius \(O P\) through \(P\) is called the colatitude of the place \(P\). We have
\(
\begin{aligned}
C P &=O P \sin \theta \\
r &=R \sin \theta
\end{aligned}
\)
\(
\text { or, } \quad r=R \sin \theta
\); where \(R\) is the radius of the earth.
If we work from the frame of reference of the earth, we shall have to assume the existence of the pseudo forces. In particular, a centrifugal force \(m \omega^{2} r\) has to be assumed on any particle of mass \(m\) placed at \(P\). Here \(\omega\) is the angular speed of the earth. If we discuss the equilibrium of bodies at rest in the earth’s frame, no other pseudo-force is needed.

Consider a heavy particle of mass \(m\) suspended through a string from the ceiling of a laboratory at colatitude \(\theta\) (figure 4u). Looking from the earth’s frame the particle is in equilibrium and the forces on it are
(a) gravitational attraction \(mg\) towards the centre of the earth, i.e., vertically downward.

(b) centrifugal force \(m \omega^{2} r\) towards \(C P\) and

(c) the tension in the string \(T\) along the string.

As the particle is in equilibrium (in the frame of earth), the three forces on the particle should add up to zero.
The resultant of \(mg\) and \(m \omega^{2} r\)
\(
\begin{aligned}
&=\sqrt{(m g)^{2}+\left(m \omega^{2} r\right)^{2}+2(m g)\left(m \omega^{2} r\right) \cos \left(90^{\circ}+\theta\right)} \\
&=m \sqrt{g^{2}+\omega^{4} R{ }^{2} \sin ^{2} \theta-2 g \omega^{2} R \sin ^{2} \theta} \\
&=m g^{\prime}
\end{aligned}
\)

where \(g^{\prime}=\sqrt{g^{2}-\omega^{2} R \sin ^{2} \theta\left(2 g-\omega^{2} R\right)}\ldots(4.9)\)
Also, the direction of this resultant makes an angle \(\alpha\) with the vertical \(O P\), where
\(
\begin{aligned}
\tan \alpha &=\frac{m \omega^{2} r \sin \left(90^{\circ}+\theta\right)}{m g+m \omega^{2} r \cos \left(90^{\circ}+\theta\right)} \\
&=\frac{\omega^{2} R \sin \theta \cos \theta}{g-\omega^{2} R \sin ^{2} \theta}\end{aligned} \ldots(4.10)\)

As the three forces acting on the particle must add up to zero, the force of tension must be equal and opposite to the resultant of the rest two. Thus, the magnitude of the tension in the string must be \(m g^{\prime}\) and the direction of the string should make an angle \(\alpha\) with the true vertical.

The magnitude of \(g^{\prime}\) is also different from \(g\). As \(2 g>\omega^{2} R\), it is clear from equation (4.9) that \(g^{\prime}<g\). One way of measuring the weight of a body is to suspend it by a string and find the tension in the string. The tension itself is taken as a measure of the weight. As \(T=m g^{\prime}\), the weight so observed is less than the true weight mg. This is known as the apparent weight. Similarly, if a person stands on the platform of a weighing machine, the platform exerts a normal force \(N\) which is equal to \(m g^{\prime}\). The reading of the machine responds to the force exerted on it and hence the weight recorded is the apparent weight \(m g^{\prime}\).
At equator, \(\theta=90^{\circ}\) and equation (4.9) gives
\(
\begin{aligned}
g^{\prime} &=\sqrt{g^{2}-2 g \omega{ }^{2} R+\omega{ }^{4} R{ }^{2}} \\
&=g-\omega{ }^{2} R \\
\text { or, } \quad m g^{\prime} &=m g-m \omega{ }^{2} R .
\end{aligned}
\)
This can be obtained in a more straightforward way. At the equator, \(m \omega{ }^{2} R\) is directly opposite to \(mg\) and the resultant is simply \(mg-m \omega^{2} R\). Also, this resultant is towards the centre of the earth so that at the equator the plumb line stands along the true vertical.

At poles, \(\theta=0\) and equation (4.9) gives \(g^{\prime}=g\) and equation (4.10) shows that \(\alpha=0\). Thus, there is no apparent change in \(g\) at the poles. This is because the poles themselves do not rotate and hence the effect of the earth’s rotation is not felt there.

Example: 4.14

A body weighs \(98 \mathrm{~N}\) on a spring balance at the north pole. What will be its weight recorded on the same scale if it is shifted to the equator? Use \(g=G M / R^{2}=9.8 \mathrm{~m} / \mathrm{s}^{2}\) and the radius of the earth \(R=6400 \mathrm{~km}\).

Solution: At the poles, the apparent weight is the same as the true weight. Thus,
\(
\begin{aligned}
& 98 \mathrm{~N} &=m g=m\left(9 \cdot 8 \mathrm{~m} / \mathrm{s}^{2}\right) \\
\text { or, } & m &=10 \mathrm{~kg} .
\end{aligned}
\)
At the equator, the apparent weight is
\(m g^{\prime}=m g-m \omega{ }^{2} R \text {. }\)
The radius of the earth is \(6400 \mathrm{~km}\) and the angular speed is
\(
\omega=\frac{2 \pi \mathrm{rad}}{24 \times 60 \times 60 \mathrm{~s}}=7 \cdot 27 \times 10^{-5} \mathrm{rad} / \mathrm{s}
\)
Thus,
\(
\begin{aligned}
m g^{\prime} &=98 \mathrm{~N}-(10 \mathrm{~kg})\left(7 \cdot 27 \times 10^{-5} \mathrm{~s}^{-1}\right)^{2}(6400 \mathrm{~km}) \\
&=97 \cdot 66 \mathrm{~N} .
\end{aligned}
\)

Example 4.15: An insect trapped in a circular groove of radius \(12 \mathrm{~cm}\) moves along the groove steadily and completes 7 revolutions in \(100 \mathrm{~s}\). (a) What is the angular speed, and the linear speed of the motion? (b) Is the acceleration vector a constant vector? What is its magnitude?

Answer: This is an example of uniform circular motion. Here \(R=12 \mathrm{~cm}\). The angular speed \(\omega\) is given by
\(
\omega=2 \pi / T=2 \pi \times 7 / 100=0.44 \mathrm{rad} / \mathrm{s}
\)
The linear speed \(v\) is :
\(
v=\omega R=0.44 \mathrm{~s}^{-1} \times 12 \mathrm{~cm}=5.3 \mathrm{~cm} \mathrm{~s} \mathrm{~s}^{-1}
\)
The direction of velocity \(v\) is along the tangent to the circle at every point. The acceleration is directed towards the centre of the circle. Since this direction changes continuously, acceleration here is not a constant vector. However, the magnitude of acceleration is constant:
\(
\begin{aligned}
& a=\omega^2 R=\left(0.44 \mathrm{~s}^{-1}\right)^2(12 \mathrm{~cm}) \\
& =2.3 \mathrm{~cm} \mathrm{~s}^{-2}
\end{aligned}
\)