A model train, with a mass of $8 k g$ $8 kg$ , is moving on a circular track with a radius of $1 m$ $1 m$ . If the train's rate of revolution changes from $\frac{3}{8} H z$ $3/8 Hz$ to $\frac{5}{4} H z$ $5/4 Hz$ , by how much will the centripetal force applied by the tracks change by?

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A model train, with a mass of $8 k g$ $8 kg$ , is moving on a circular track with a radius of $1 m$ $1 m$ . If the train's rate of revolution changes from $\frac{3}{8} H z$ $3/8 Hz$ to $\frac{5}{4} H z$ $5/4 Hz$ , by how much will the centripetal force applied by the tracks change by?

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Answer 1 · 2018-03-10T10:13:14

We know that angular frequency $ω$ $omega$ and natural frequency $ν$ $nu$ are related as, $ω = 2 π ν$ $omega =2 pinu$

So,if the angular frequency changed from $ω_{1}$ $omega_1$ to $ω_{2}$ $omega_2$

we can say, $ω_{2} = 2 π ν_{2} = 2 π (\frac{5}{4}) = 7.854 r a d s^{- 1}$ $omega_2 =2pinu_2=2pi(5/4)=7.854 rads^-1$ and $ω_{1} = 2 π ν_{1} = 2 π (\frac{3}{8}) = 2.355 r a d s^{- 1}$ $omega_1=2pinu_1=2pi(3/8)=2.355 rads^-1$

Now,centripetal force is expressed as $m ω^{2} r$ $momega^2r$

So,change in centripetal force is $m r (ω_{2}^{2} - ω_{1}^{2}) = 8 \cdot 1 \cdot ({7.854}^{2} - {2.355}^{2}) = 449.113 N$ $mr(omega_2^2 - omega_1^2)=8*1*(7.854^2-2.355^2)=449.113 N$

Answer 2 · 2018-03-10T10:16:03

The change in centripetal force is $= 449.1 N$ $=449.1N$

Explanation:

The centripetal force is

$F = \frac{m v^{2}}{r} = m r ω^{2} N$ $F=(mv^2)/r=mromega^2N$

The mass of the train, $m = (8) k g$ $m=(8)kg$

The radius of the track, $r = (1) m$ $r=(1)m$

The frequencies are

$f_{1} = (\frac{3}{8}) H z$ $f_1=(3/8)Hz$

$f_{2} = (\frac{5}{4}) H z$ $f_2=(5/4)Hz$

The angular velocity is $ω = 2 π f$ $omega=2pif$

The variation in centripetal force is

$DeltaF=F_2-F_1$

$F_1=mromega_1^2=mr*(2pif_1)^2=8*1*(2pi*3/8)^2=44.4N$

$F_2=mromega_2^2=mr*(2pif_2)^2=8*1*(2pi*5/4)^2=493.5N$

$DeltaF=F_1-F_2=493.5-44.4=449.1N$

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A model train, with a mass of $8 k g$ $8 kg$ , is moving on a circular track with a radius of $1 m$ $1 m$ . If the train's rate of revolution changes from $\frac{3}{8} H z$ $3/8 Hz$ to $\frac{5}{4} H z$ $5/4 Hz$ , by how much will the centripetal force applied by the tracks change by?

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Explanation:

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2 Answers

Explanation:

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A model train, with a mass of $8 k g$ $8 kg$ , is moving on a circular track with a radius of $1 m$ $1 m$ . If the train's rate of revolution changes from $\frac{3}{8} H z$ $3/8 Hz$ to $\frac{5}{4} H z$ $5/4 Hz$ , by how much will the centripetal force applied by the tracks change by?