Question

A model train, with a mass of `3 kg`, is moving on a circular track with a radius of `2 m`. If the train's rate of revolution changes from `5/3 Hz` to `3/4 Hz`, by how much will the centripetal force applied by the tracks change by?

Answer 1

It is -524.62N.

Explanation:

The centripetal force is

`F=(mv^2)/r`.

We have both `m` and `r` we need to find `v` and we can calculate it.

The frequency tell us how many turns the train does in a second. A turn is long `2pir`, so the velocity is the frequency multiplied by the circumference. For the initial velocity we have

`v_i=5/3*2pir=5/3*2pi*2\approx20.94` m/s.

The initial centripetal force is then

`F_i=mv_i^2/r=3*20.94^2/2=657.72` N.

The final velocity is

`v_f=3/4*2pir=3/4*2pi*2\approx 9.42` m/s

and the final force is

`F_f=mv_f^2/r=3*9.42^2/2\approx133.1` N

The difference in force is then

`F_f-F_i=133.1-657.72=-524.62` N.

The negative sign is because the centripetal force decreases since the train is lowering the speed.

Answer 2

We have the following numbers at our disposal:

• `m = "3 kg"`, the mass of the train
• `r = "2 m"`, the radius of the train's path
• `omega_i = "5/3 rev"cdot"s"^(-1) xx (2pi " rad")/"rev"`, the initial angular velocity
• `omega_f = "3/4 rev"cdot"s"^(-1) xx (2pi " rad")/"rev"`, the final angular velocity

Recall that the sum of the centripetal "forces" is:

`\mathbf(sum F_c = (mv_T^2)/r)`

Hence, since we are looking for the change in the sum of the centripetal forces, `Delta(sum F_c)`, we are looking for `(mDeltav_T^2)/r`, where

`\mathbf(v_T = romega)`

is the tangential velocity in `"m/s"`, while the mass and radius are constant.

The rates of revolution were given in the question, but they are saying:

`"5/3 Hz"` `=` `"5/3 of a"` `\mathbf("full revolution")` `"per second"`

So, this angular velocity is actually

`omega = (5/3 cancel("rev"))/"s" xx (2pi "rad")/cancel("rev") = (10pi)/3 "rad/s"`.

Therefore, the change in the sum of the centripetal forces is:

`color(blue)(Delta(sum F_c))`

`= (mDeltav_T^2)/r`

`= (m(v_(Tf)^2 - v_(Ti)^2))/r`

`= (m((romega_f)^2 - (romega_i)^2))/r`

`= (("3 kg")[(("2 m")(3/4*2pi " rad/s"))^2 - (("2 m")(5/3*2pi " rad/s"))^2])/("2 m")`

`= (("3 kg")(9pi^2 "m"^2"/s"^2 - (400pi^2)/9 "m"^2"/s"^2))/("2 m")`

`=` `color(blue)(-"524.7 N")`

Thus, the sum of the centripetal forces has decreased by `"524.7 N"`.