Question

A model train with a mass of `2 kg` is moving along a track at `6 (cm)/s`. If the curvature of the track changes from a radius of `3 cm` to `8 cm`, by how much must the centripetal force applied by the tracks change?

Answer 1

The centripetal force changes by `=0.15N`

Explanation:

The centripetal force is

`F=(mv^2)/r`

The mass, `m=(2)kg`

The speed, `v=(0.06)ms^-1`

The radius, `=(r) m`

The variation in centripetal force is

`DeltaF=F_2-F_1`

`F_1=mv^2/r_1=2*0.06^2/0.03=0.24N`

`F_2=mv^2/r_2=2*0.06^2/0.08=0.09N`

`DeltaF=0.24-0.09=0.15N`

Answer 2

The centripetal force is decreased by `0.15"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 velocity as:

`a_c=(v^2)/r`

Therefore, we can state:

`F_c=(mv^2)/r`

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

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

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

`=(mv^2)/r_f-(mv^2)/r_i`

We can simplify this equation:

`=>mv^2(1/r_f-1/r_i)`

`=>color(purple)(DeltaF_c=mv^2(1/r_f-1/r_i))`

We are provided with the following information:

• `->"m=2"kg"`

• `|->v=0.06"m"//"s"`

• `->"r_i"=0.03"m"`

• `->"r_f"=0.08"m"`

Substituting these values into the equation we derived above:

`DeltaF_c=(2"kg")(0.06"m"//"s")^2(1/0.08-1/0.03)`

`=>color(crimson)(-0.15"N")`

Therefore, the centripetal force is decreased by `0.15"N"`.