Question

A particle is executing uniform circular motion in a path of radius 3 m with a frequency of 1 revolution per minute. By how much does the total distance traveled by the particle in 70 seconds exceed the magnitude of its displacement over the same time interval?

• A. 0
• B. $(\pi -3)m$
• C. $7\pi m$
• D. $(7\pi -3)m$

Answer 1

The correct option is D

$(7\pi -3)m$

Step 1: Given data

The radius of the circle R = 3 m

Frequency of rotation f = 1 revolution per minute= 1 rotation in 60 seconds.

Time t= 70 sec

Step 2: Finding the distance traveled in 70 seconds

Distance:

1. Distance is defined as the total path covered by an object from the initial position to the final position.
2. When an object moves in the circular path and completes one rotation.
3. Then, the distance traveled will be equal to the circumference of the circle.

So, Distance traveled in rotation = circumference of the circle

$$\begin{gathered} =2\pi r \\ =2\pi \times 3 \\ =6\pi m \end{gathered}$$

1. The frequency of rotation is defined as the number of rotations of the particle in one sec.
2. Given, The particle completes 1 rotation in 1 minute.
3. So, frequency f= 1 rotation in 1 min
4. Converting to sec, there will be $\frac{1}{60}$ rotation in 1 sec
5. So, the number of rotations in 70 sec is $\frac{70}{60}$

The total distance traveled by a particle in 70 seconds is,

Distance= $Dis\tan cetraveledinonerotation\times Numberofrotationin70sec$

=$6\pi \times \frac{70}{60}$

=$7\pi m$

Step 3: Finding the displacement traveled in 70 seconds

Displacement:

1. Displacement is defined as the shortest distance between the initial point and the final point.
2. When the object completes one complete circle then, the initial point is equal to the final point.
3. In 60 seconds the particle reaches the initial position.
4. We need to find the displacement of a particle in 70 sec (ie; 10 sec after the completion of one circle)
5. In 10 sec, the angle subtended at the center is $\frac{2\pi }{6}$=$\frac{\pi }{3}$ ie; $60°$
6. So, In 70 sec the displacement is the chord of an equilateral triangle
7. As it is an equilateral triangle, the sides of the triangle will be equal to the radius of the circular path.

So, the displacement in 70 sec is $3m$

Now, the total distance traveled by the particle is greater than the value of displacement.

Distance traveled in 70 sec is $7\pi meters$

Displacement in 70 sec is $3m$

So, distance exceeds the displacement by $(7\pi -3)m$

Hence,

Option (d) is correct.