Question

An object with a mass of `2 kg` is revolving around a point at a distance of `4 m`. If the object is making revolutions at a frequency of `4 Hz`, what is the centripetal force acting on the object?

Answer 1

`F_c~~5053N` radially inward.

Explanation:

The centripetal force is given in accordance with Newton's second law as:

`F_c=ma_c`

where `m` is the mass of the object and `a_c` is the centripetal acceleration experienced by the object

The centripetal acceleration is given by:

`a_c=v^2/r`

which is equivalent to `romega^2`. Therefore, we can write:

`F_c=mromega^2`

where `r` is the radius and `omega` is the angular velocity of the object

The angular velocity can also be expressed as:

`omega=2pif`

where `f` is the frequency of the revolution

And so our final expression becomes:

`color(blue)(F_c=mr(2pif)^2`

We are given:

• `|->"m"=2"kg"`

• `|->"r"=4"m"`

• `|->f=4"s"^-1`

Substituting these values into the equation we derived above:

`F_c=(2"kg")(4"m")(2pi(4"s"^-1))^2`

`=5053.237"N"`

`~~5053N` radially inward

This may also be expressed in scientific notation as `5xx10^3"N"` where significant figures are concerned.