Question

A 600 g steel block rotates on a steel table while attached to a 1.20 m -long hollow tube. Compressed air fed through the tube and ejected from a nozzle on the back of the block exerts a thrust force of 4.71 N perpendicular to the tube. The maximum tension the tube can withstand without breaking is 50.0 N. Assume the coefficient of kinetic friction between steel block and steel table is 0.60. (Figure 1) 

Part A 

If the block starts from rest, how many revolutions does it make before the tube breaks? 

Answer 1

given
m = 600 g = 0.6 kg
R = 1.20 m
F_thrust = 4.71 N
T_max = 50.0 N
mue_k = 0.6

let v is the maximum speed of the block

T_max = m*v^2/r

v = sqrt(T_max*r/m)

= sqrt(50*1.2/0.6)

= 10 m/s

let a is the tangential acceleration of the block

net tangential force acting on the block, Fnet = F_thrust - fk

m*a = F_thrust - mue_k*N

m*a = F_thrust - mue_k*m*g

a = (F_thrust - mue_k*m*g)/m

= (4.71 - 0.6*0.6*9.8)/0.6

= 1.97 m/s^2

let d is the total distance traveled before tube breaks.

use, d = (v^2 - u^2)/(2*a)

= (10^2 - 0^2)/(2*1.97)

= 25.38

number of revolutions made by the block before tube breaks,

N = d/(2*pi*R)

= 25.38/(2*pi*1.2)

= 3.37 <<<<<<<<<<-------------Answer