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
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
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