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} $

$a=3.37 \times 10^{-2} ms^{-2} N$