A model train, with a mass of #3 kg#, is moving on a circular track with a radius of #2 m#. If the train's rate of revolution changes from #5/3 Hz# to #3/4 Hz#, by how much will the centripetal force applied by the tracks change by?

2 Answers
May 31, 2016

Answer:

It is -524.62N.

Explanation:

The centripetal force is

#F=(mv^2)/r#.

We have both #m# and #r# we need to find #v# and we can calculate it.

The frequency tell us how many turns the train does in a second. A turn is long #2pir#, so the velocity is the frequency multiplied by the circumference. For the initial velocity we have

#v_i=5/3*2pir=5/3*2pi*2\approx20.94# m/s.

The initial centripetal force is then

#F_i=mv_i^2/r=3*20.94^2/2=657.72# N.

The final velocity is

#v_f=3/4*2pir=3/4*2pi*2\approx 9.42# m/s

and the final force is

#F_f=mv_f^2/r=3*9.42^2/2\approx133.1# N

The difference in force is then

#F_f-F_i=133.1-657.72=-524.62# N.

The negative sign is because the centripetal force decreases since the train is lowering the speed.

May 31, 2016

We have the following numbers at our disposal:

  • #m = "3 kg"#, the mass of the train
  • #r = "2 m"#, the radius of the train's path
  • #omega_i = "5/3 rev"cdot"s"^(-1) xx (2pi " rad")/"rev"#, the initial angular velocity
  • #omega_f = "3/4 rev"cdot"s"^(-1) xx (2pi " rad")/"rev"#, the final angular velocity

Recall that the sum of the centripetal "forces" is:

#\mathbf(sum F_c = (mv_T^2)/r)#

Hence, since we are looking for the change in the sum of the centripetal forces, #Delta(sum F_c)#, we are looking for #(mDeltav_T^2)/r#, where

#\mathbf(v_T = romega)#

is the tangential velocity in #"m/s"#, while the mass and radius are constant.

The rates of revolution were given in the question, but they are saying:

#"5/3 Hz"# #=# #"5/3 of a"# #\mathbf("full revolution")# #"per second"#

So, this angular velocity is actually

#omega = (5/3 cancel("rev"))/"s" xx (2pi "rad")/cancel("rev") = (10pi)/3 "rad/s"#.

Therefore, the change in the sum of the centripetal forces is:

#color(blue)(Delta(sum F_c))#

#= (mDeltav_T^2)/r#

#= (m(v_(Tf)^2 - v_(Ti)^2))/r#

#= (m((romega_f)^2 - (romega_i)^2))/r#

#= (("3 kg")[(("2 m")(3/4*2pi " rad/s"))^2 - (("2 m")(5/3*2pi " rad/s"))^2])/("2 m")#

#= (("3 kg")(9pi^2 "m"^2"/s"^2 - (400pi^2)/9 "m"^2"/s"^2))/("2 m")#

#=# #color(blue)(-"524.7 N")#

Thus, the sum of the centripetal forces has decreased by #"524.7 N"#.