# 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

#### Answer:

It is -524.62N.

#### Explanation:

The centripetal force is

We have both

The frequency tell us how many turns the train does in a second. A turn is long

The initial centripetal force is then

The final velocity is

and the final force is

The difference in force is then

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

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**,

#\mathbf(v_T = romega)#

is the **tangential velocity** in *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