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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.

User Bhadram
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1 Answer

3 votes

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.

Step-by-step explanation:

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


F_c=(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;


(mv_1^2)/(r_1)=(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{(v_1^2r_2)/(r_1)} .................(2)

By observing equation (2) carefully, the ratio
(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.

A centripetal force of 5.0 newtons is applied to a rubber stopper moving at a constant-example-1
User PatomaS
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