Question

A 400 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 5.21 N perpendicular to the tube. The maximum tension the tube can withstand without breaking is 40.0 N . Assume the coefficient of kinetic friction between steel block and steel table is 0.60.

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

Answer 1

here,

mass of block , m = 400 g = 0.4 kg

length of string , r = 1.2 m

force exerted , F= 5.21 N

for the tube to break

let the speed of block be v

maximum tension , T = m * v^2 /r

40 = 0.4 * v^2 /1.2

v = 10.95 m/s

the tangential acceleration , a = (F - uk * m * g)/m

a = ( 5.21 - 0.6 * 0.4 * 9.81) /0.4

a = 7.14 m/s^2

let the distance traveled be s

using third equation of motion

v^2 - u^2 = 2 * a * s

10.95^2 - 0 = 2 * 7.14 * s

s = 8.4 m

the number if revolution before the tube breaks , N = s /(2*pi*r)

N = 8.4 /(2*pi*1.2) = 1.12 rev