Question

A centripetal force of 5.0 newtons is applied to a rubber stopper moving at a constant speed in a horizontal circle. If the same force is applied, but the radius is made smaller, what happens to the speed, v, and the frequency, f, of the stopper.

Answer 1

Answer:

Both the frequency f and velocity v will increase.

When the radius reduces, the circumference of the circular path becomes smaller which means that more number of revolutions can be made per unit time as long as the force is kept constant; this is an increase in frequency.

Explanation:

The centripetal force acting on a mass in circular motion is given by equation (1);

$F_c=\frac{mv^2}{r}....................(1)$

where m is the mass of the object and r is radius of the circle. From equation one we see that the centripetal force is directly proportional to the square of the velocity and inversely proportional to the radius of the circular path.

However, according to the problem, the force is constant while the radius and the velocity changes. Therefore we can write the following equation;

$\frac{mv_1^2}{r_1}=\frac{mv_2^2}{r_2}......................(2)$

Also recall that m is constant so it cancels out from both sides of equation (2). Therefore from equation we can write the following;

$v_2=\sqrt{\frac{v_1^2r_2}{r_1}} .................(2)$

By observing equation (2) carefully, the ratio $\frac{r_2}{r_1}$ will with the square root increase $v_1$ since $r_2$ is lesser than $r_1$.

Hence by implication, the value of $v_2$ will be greater than $v_1$.

As the radius changes from $r_1$ to $r_2$, the velocity also changes from $v_1$ to $v_2$.