Question

Question: A rubber stopper 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 3v in a circle of radius In this second trial, is its acceleration

(a) The same as in the first trial,

(b) Three times larger,

(c) One-third as large,

(d) Nine times larger, or

(e) One-ninth as large?

Short answer

The centripetal acceleration in the second trail is three times that of the first trail.

So, option (b) is correct

Step-by-step solution

Step 1: Define centripetal force

The force operating on an object in curvilinear motion directed toward the axis of rotation or centre of curvature is known as centripetal force. It is always perpendicular to the velocity of the body moving on the circular path. The centripetal force of an item moving on a circular path is given as-
$\mathbf{F}\mathbf{}\mathbf{=}\frac{{\mathbf{mv}}^{\mathbf{2}}}{\mathbf{r}}$

Explanation for correct option

The magnitude of a particle's centripetal acceleration is given by if it moves in a circular route of radius r at a constant speed v.
$\mathrm{a}=\frac{{\mathrm{v}}^2}{\mathrm{r}}$
It moves at a speed of v in a circle with a radius of r in one trial. It moves at a fast speed of 3 v in a circle with a radius of 3 r in the second trial.

In first trail, the centripetal acceleration of rubber stopper is $a_1=\frac{v^2}{r}$
In the second trial, the centripetal acceleration of rubber stopper is
$$\begin{gathered} a_2=\frac{{\left(3v\right)}^2}{3r} \\ =\frac{9v^2}{3r} \\ =3\left(\frac{v^2}{r}\right) \\ =3a_1 \end{gathered}$$
As a result, the centripetal acceleration in the second trail is three times that of the first trail.

So, option (b) is correct