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

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Answer 1 · 2017-08-09T21:24:53

The centripetal force will decrease by $\approx 13 N$ $~~13"N"$

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

$F_{c} = m a_{c}$ $F_c=ma_c$

where $m$ $m$ is the mass of the object and $a_{c}$ $a_c$ is the centripetal acceleration experienced by the object

The centripetal acceleration can be expressed in terms of the angular velocity ( $ω$ $omega$ ) as:

$a_{c} = r ω^{2}$ $a_c=romega^2$

Therefore, we can state:

$F_{c} = m r ω^{2}$ $F_c=mromega^2$

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

$ω = 2 π f$ $omega=2pif$

Putting this all together, we have:

$F_{c} = m r {(2 π f)}^{2}$ $color(blue)(F_c=mr(2pif)^2$

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:

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 $13.43"N"$ .

A model train, with a mass of 4 kg4kg4 kg, is moving on a circular track with a radius of 7 m7m7 m. If the train's rate of revolution changes from 1/6 Hz16Hz1/6 Hz to 1/8 Hz18Hz1/8 Hz, by how much will the centripetal force applied by the tracks change by?