Numerical Problems and Solutions of Centripetal Force and Centripetal Acceleration

Solutions to Numerical Problems of the Centripetal Force and Centripetal Acceleration

Numerical Problems and Solutions

Q.1 An object of mass $0.20 \: Kg$ is being revolved in a circular path of diameter $2.0 \: m$ on a frictionless horizontal plane by means of a string. It performs $10$ revolutions in $3.14 \: s$. Find the centripetal force acting on an object.

Solution:
Given that:
Mass of body $(m)=0.20 \: Kg$
Diameter of Circular path $(d)=2.0: m$
Number of revolutions $(n)=10$
Time taken to complete $(10)$ revolution$(t)=3.14 s$
The centripetal force acting on a body $(F)=?$
Now the centripetal force:
$F=m r \omega^2$
$F=m r \left( \frac{2 \pi n}{t} \right)^2$
Now Substitute the given values in the centripetal force formula:
$F=0.20 \times 1 \left( \frac{2 \times 3.14 \times 10}{3.14} \right)^2$
$F=0.8 N$

Q.2 A car of mass $1200 Kg$ is moving with a speed of $10.5 ms^{-1}$ on a circular path of radius $20 m$ on a level road. What will be the frictional force between the car and the road so that the car does not slip?

Solution:
Given that:
Mass of car $(m) = 1200 \: Kg$
Speed of car $(v) = 10.5: ms^{-1}$
The radius of circular path $(r)= 20 m$
The frictional force $(F)=?$
The frictional force $(F)$ should be equal to the required centripetal force i.e
$F=\frac{mv^{2}}{r}$
Now Substitute the given values in the centripetal force formula:
$F=\frac{1200 \times (10.5)^{2}}{20}$
$F=6.615 \times 10^{3} N$

Q.3 A string can bear a maximum tension of $50 N$ without breaking. A body of mass $1 kg$ is tied to one end of $2 m$ long piece of the string and rotated in a horizontal plane. Find the maximum linear velocity with which the string would not break.

Solution:
Given that:
Maximum tension on the string without breaking $(F)=50 N$
mass of the a body $(m)=1 Kg$
length of the string $(l)=2 m$
The maximum velocity with which the string would not break $(v)=?$
We know that the centripetal force:
$F=\frac{mv^{2}}{r}$
$v=\sqrt{\frac{F \: r}{m}}$
Now Substitute the given values in the centripetal force formula:
$v=\sqrt{\frac{50 \times 2 2}{1}}$
$v=10 ms^{-1}$

Q.4 The moon revolves around the earth in $2.36 \times 10^{6} s$ in a circular orbit of radius $3.85 \times 10^{5} Km$. Calculate the centripetal acceleration produced in the motion of the moon.

Solution:
Given that:
The radius of circular orbit of the moon $(r)=3.85 \times 10^{5} Km = 3.85 \times 10^{8} m$
Time taken to complete one revolution $(T)=2.36 \times 10^{6} s$
Centripetal acceleration produced in the motion of the moon $=?$
The centripetal acceleration:
$a= \omega^{2} r $
$a= \left( \frac{2 pi}{T} \right) r \qquad \left( \because \omega= \frac{2 pi}{T} \right) $
Now Substitute the given values in the centripetal acceleration formula:
$a= \left( \frac{2 \times 3.14}{2.36 \times 10^{6}} \right) \times 3.85 \times 10^{8} $
$a=2.73 \times 10^{-3} ms^{-2} N$

Q.5 Calculate the centripetal acceleration at a point on the equator of the earth. The radius of the earth is $6.4 \times 10^{6} m$ and it completes one rotation per day about its axis.

Solution:
Given that:
Radius of earth $(r)=6.4 \times 10 ^{6} m$
Time taken to complete one rotation per day about its axis$(T)=24 \times 60 \times 60 s$
centripetal acceleration at a point on equator $=?$
The centripetal acceleration:
$a= \omega^{2} r $
$a= \left( \frac{2 pi}{T} \right) r \qquad \left( \because \omega= \frac{2 pi}{T} \right) $
Now Substitute the given values in the centripetal acceleration formula:
$a= \left( \frac{2 \times 3.14}{24 \times 60 \times 60} \right) \times 6.4 \times 10^{6} $
$a=3.37 \times 10^{-2} ms^{-2} N$

Numerical Aperture and Acceptance Angle of the Optical Fibre

Angle of Acceptance → If incident angle of light on the core for which the incident angle on the core-cladding interface equals the critical angle then incident angle of light on the core is called the "Angle of Acceptance. Transmission of light when the incident angle is equal to the acceptance angle If the incident angle is greater than the acceptance angle i.e. $\theta_{i}>\theta_{0}$ then the angle of incidence on the core-cladding interface will be less than the critical angle due to which part of the incident light is transmitted into cladding as shown in the figure below Transmission of light when the incident angle is greater than the acceptance angle If the incident angle is less than the acceptance angle i.e. $\theta_{i}<\theta_{0}$ then the angle of incidence on the core-cladding interface will be greater than the critical angle for which total internal reflection takes place inside the core. As shown in the figure below Transmission of lig

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Fraunhofer diffraction due to a single slit

Let $S$ be a point monochromatic source of light of wavelength $\lambda$ placed at the focus of collimating lens $L_{1}$. The light beam is incident normally from $S$ on a narrow slit $AB$ of width $e$ and is diffracted from it. The diffracted beam is focused at the screen $XY$ by another converging lens $L_{2}$. The diffraction pattern having a central bright band followed by an alternative dark and bright band of decreasing intensity on both sides is obtained. Analytical Explanation: The light from the source $S$ is incident as a plane wavefront on the slit $AB$. According to Huygens's wave theory, every point in $AB$ sends out secondary waves in all directions. The undeviated ray from $AB$ is focused at $C$ on the screen by the lens $L_{2}$ while the rays diffracted through an angle $\theta$ are focussed at point $p$ on the screen. The rays from the ends $A$ and $B$ reach $C$ in the same phase and hence the intensity is maximum. Fraunhofer diffraction due to

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Particle in one dimensional box (Infinite Potential Well)

Let us consider a particle of mass $m$ that is confined to one-dimensional region $0 \leq x \leq L$ or the particle is restricted to move along the $x$-axis between $x=0$ and $x=L$. Let the particle can move freely in either direction, between $x=0$ and $x=L$. The endpoints of the region behave as ideally reflecting barriers so that the particle can not leave the region. A potential energy function $V(x)$ for this situation is shown in the figure below. Particle in One-Dimensional Box(Infinite Potential Well) The potential energy inside the one -dimensional box can be represented as $\begin{Bmatrix} V(x)=0 &for \: 0\leq x \leq L \\ V(x)=\infty & for \: 0> x > L \\ \end{Bmatrix}$ $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2m}{\hbar^{2}}(E-V)\psi(x)=0 \qquad(1)$ If the particle is free in a one-dimensional box, Schrodinger's wave equation can be written as: $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2mE}{\hbar^{2}}\psi(x)=0$ $\frac{d^{2} \psi(x)}{d x

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