Question

g A centripetal force F is applied to an object moving at a constant speed v in a horizontal circle of radius r. If the radius is quadrupled and the speed is doubled, what happens to the centripetal force

Answer 1

Answer:

The centripetal force remains constant.

Explanation:

The magnitude of centripetal force is given by the following formula:

F = mv²/r

where,

F = Centripetal Force

m = mass of the object

v = velocity of the object

r = radius of the circle

Now, if we quadruple the radius (r = 4r) and the speed is doubled (v = 2v). Then the centripetal force becomes:

F' = m(2 v)²/4r

F' = (4/4)(mv²/r)

F' = mv²/r

F' = F

Thus, the centripetal force remain constant.

Answer 2

When the radius of the circle is quadrupled and the speed is doubled, the centripetal force remains the same.

The given parameters:

• Centripetal force, = F
• Speed of the object, = v
• Radius of the circle, = r

The centripetal force applied to the object is calculated as follows;

$F = \frac{mv^2}{r} \\\\Fr = mv^2\\\\\frac{Fr}{v^2} = m\\\\\frac{F_1r_1}{v_1^2} = \frac{F_2r_2}{v_2^2} \\\\F_2 = \frac{F_1 r_1v_2^2}{r_2v_1^2}$

When the radius of the circle is quadrupled and the speed is doubled, the centripetal force becomes;

$F_2 = \frac{F_1 r_1v_2^2}{r_2v_1^2} \\\\F_2 = \frac{F_1 \times r_1 \times (2v_1)^2}{4r_1 \times v_1^2} \\\\F_2 = \frac{F_1 \times r_1 \times 4v_1 ^2 }{4r_1 \times v_1^2} \\\\F_2 = F_1$

Thus, when the radius of the circle is quadrupled and the speed is doubled, the centripetal force remains the same.

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