Question

If the radius of an object in circular motion is doubled, what change will occur in the centripetal force?

Answer 1

Answer:

The centripetal force will be 1/2 as big as it was. (option c)

Explanation:

Recall that centripetal force ($F_c$) is defined as: $F_c=m\,* \frac{v^2}{r}$ where "v" is the tangential velocity of the object in circular motion, "r" is the radius of rotation and "m" is the object's mass.

So if we start with such formula with a given mass, radius, and tangential velocity, and then we move to a situation where everything stays the same except for the radius which doubles, then the new centripetal force ($F'_c$) will be given by: $F'_c=m\,* \frac{v^2}{2r}$

and this is half (1/2) of the original force:

$F'_c=m\,* \frac{v^2}{2r}\\F'_c=m\,* \frac{v^2}{r}*\frac{1}{2} \\F'_c=F_c\,*\,\frac{1}{2}$

which is expressed by option "c" of the provided list.

Answer 2

Answer:

Answer:

The centripetal force will be 1/2 as big as it was. (option c)

Explanation:

Recall that centripetal force () is defined as:  where "v" is the tangential velocity of the object in circular motion, "r" is the radius of rotation and "m" is the object's mass.

So if we start with such formula with a given mass, radius, and tangential velocity, and then we move to a situation where everything stays the same except for the radius which doubles, then the new centripetal force () will be given by:

and this is half (1/2) of the original force:

which is expressed by option "c" of the provided list.

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Explanation: