Question

A stone is whirled with uniform speed in a horizontal circle of radius 15cm .It takes the stone 10 seconds to describe an arc of length 4cm . Calculate . a.Angular velocity b. linear velocity c. periodic time​

Answer 1

a) The angular velocity is 0.004 rad/s

b) The linear velocity is $6\cdot 10^{-4} m/s$

c) The period of motion is 235 s

Explanation:

a)

The angular velocity of an object in circular motion is the rate of change of the angular position of the object. It can be calculated as follows:

$\omega=\frac{s}{t}$

where

s is the arc of length covered by the object

t is the time elapsed

For the stone moving in the horizontal circle in the problem, we have

s = 4 cm = 0.04 m is the length of the arc covered

t = 10 s is the time elapsed

Substituting, we find the angular velocity:

$\omega=\frac{0.04}{10}=0.004 rad/s$

b)

In a circular motion, the linear velocity is related to the angular velocity by the equation

$v=\omega r$

where

$v$ is the linear velocity

$\omega$ is the angular velocity

r is the radius of the circle

For the stone in this problem, we have:

$\omega=0.004 rad/s$ (angular velocity, calculated in part a)

r = 15 cm = 0.15 m (radius of the circle)

Substituting, we find

$v=(0.004)(0.15)=6\cdot 10^{-4} m/s$

c)

In a circular motion, the period is the time taken for the object to complete exactly one revolution.

The circle in this problem has a radius of

r = 0.15 m

So, the circumference is

$L=2\pi r = 2\pi(0.15)=0.94 m$

We know that the stone covers an arc length of s = 0.04 m in 10 s, therefore we can use the following proportion to find the time taken for the stone to cover one circumference (and this time is the period of the motion):

$0.04 m: 10 s = 0.94 m : T\\T=\frac{(10s)(0.94m)}{0.04m}=235 s$

Learn more about circular motion:

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