Question

A model train, with a mass of `4 kg`, is moving on a circular track with a radius of `7 m`. If the train's rate of revolution changes from `1/6 Hz` to `1/8 Hz`, by how much will the centripetal force applied by the tracks change by?

Answer 1

The centripetal force will decrease by `~~13"N"`

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 can be expressed in terms of the angular velocity (`omega`) as:

`a_c=romega^2`

Therefore, we can state:

`F_c=mromega^2`

The angular velocity can also be expressed in terms of the frequency of the motion as:

`omega=2pif`

Putting this all together, we have:

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

To find the change in centripetal force as the frequency changes, we're being asked for `DeltaF_c`, where:

`DeltaF_c=(F_c)_f-(F_c)_i`

`=mr(2pif_f)^2-mr(2pif_i)^2`

We can simplify this equation:

`=>mr(4pi^2f_f^2)-mr(4pi^2f_i^2)`

`=>color(purple)(DeltaF_c=4mrpi^2(f_f^2-f_i^2))`

We are provided with the following information:

• `->"m=4"kg"`

• `->"r"=7"m"`

• `->f_i=1/6"s"^-1`

• `->f_f=1/8"s"^-1`

Substituting these values into the equation we derived above:

`DeltaF_c=4(4"kg")(7"m")pi^2[(1/8/"s"^-1)^2-(1/6"s"^-1)^2]`

`~~color(red)(-13.43"N")`

Therefore, the centripetal force is decreased by `13.43"N"`.