Question

Question: A set of keys on the end of a string is swung steadily in a horizontal circle. In one trial, it moves at speed v in a circle of radius In a second trial, it moves at a higher speed 4v in a circle of radius In the second trial, how does the period of its motion compare with its period in the first trial?

(a) It is the same as in the first trial.

(b) It is 4 times larger.

(c) It is one-fourth as large.

(d) It is 16 times larger.

(e) It is one-sixteenth as large.

Short answer

Time period is the time taken by the ball to complete one revolution.

It is denoted by If ‘r’ is the radius of the circle of motion, then in time ‘T’ our ball covers a distance

Step-by-step solution

Step 1: Define Time period of circular motion.

Time period is the time taken by the ball to complete one revolution.

It is denoted by If ‘r’ is the radius of the circle of motion, then in time ‘T’ our ball covers a distance

Explanation for correct option

In two trials, a set of keys on the end of a string is swung steadily in a horizontal circle. The two trials are carried out at different speeds and with different radiuses. The duration of the second trial should be compared to the duration of the first trial. This problem is conceptualised using the link between time period, velocity, and radius of a circular path. The time period of a circularly moving particle is given by
$T=\frac{\text{2}\pi r}{v}$

Second trial: Speed

Radius of the circular path is

Time period of the particle in the first trial is
$$\begin{gathered} T_2=\frac{\text{2}\pi \left(4r\right)}{\left(4v\right)} \\ =\frac{\text{2}\pi r}{v} \\ =T_1 \end{gathered}$$
The time period in the second trial is same as that of time period in the first trial. So, option (a) is true.