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?

2 Answers
Aug 12, 2017

Answer:

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#

Aug 12, 2017

Answer:

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"#.