A model train with a mass of #2 kg# is moving along a track at #9 (cm)/s#. If the curvature of the track changes from a radius of #6 cm# to #25 cm#, by how much must the centripetal force applied by the tracks change?

2 Answers
Apr 19, 2018

Answer:

The change in centripetal force is #=0.2052N#

Explanation:

The centripetal force is

#vecF_C=(mv^2)/r*vecr#

The mass is of the train #m=2kg#

The velocity of the train is #v=0.09ms^-1#

The radii of the tracks are

#r_1=0.06m#

and

#r_2=0.25m#

The variation in the centripetal force is

#DeltaF=F_2-F_1#

The centripetal forces are

#||F_1||=2*0.09^2/0.06=0.27N#

#||F_2||=2*0.09^2/0.25=0.0648N#

#DeltaF=F_1-F_2=0.27-0.0648=0.2052N#

Apr 19, 2018

Answer:

#0.378" N"#

Explanation:

Consider a diagram:

enter image source here

We know that centripetal acceleration, towards the centre, is #a=v^2/r#

#v=9" cm s"^-1#
#=0.09" m s"^-1#

Now, use the formula to find the acceleration in each case for each radius.

Let #a_1# be the acceleration for #6"cm"# and #a_2# be the acceleration for #25"cm"#

#a_1=0.09^2/0.06#
#=0.135" m s"^-2#

#a_2=0.09^2/0.25#
#=0.324" m s"^-2#

Now, since the acceleration will be towards the centre of the circle, resolve forces towards the centre using Newton's Second Law

#F_1=2xx0.135#
#=0.27" N"#

#F_2=2xx0.324#
#=0.648" N"#

So the increase in centripetal force is:

#DeltaF=0.648-0.27#
#=0.378" N"#