Question

Chapter 3, Practice Problem 3/115 The quarter-circular hollow tube of circular cross section starts from rest at time t = 0 and rotates about point o in a horizontal plane with a constant counterclockwise angular acceleration 7 = 2.9 rad/s2. At what time t will the 0.30-kg particle P slip relative to the tube? The coefficient of static friction between the particle and the tube is us = 0.53. 0.58 m 51 Answer: t = the tolerance is +/-2%

Answer 1

The centripetal force due to quarter circle is given as, F=mr$\omega$²

where m=mass of the particle=0.3 kg

r=radius of the quarter circle=0.58 m

If $\omega$ =angular frequency of the system

From the laws of circular motion, $\omega$ =$\omega$₀+$\alpha$t

where $\alpha$ =angular acceleration=2.9 rad/sec²

Since the particle starts from rest, $\omega$ ₀=0

Thus, $\omega$ =$\alpha$t

t= time taken by the particle to slip relative to the tube due to rotation

Also, F=$\mu$N

where $\mu$ =coefficient of static friction=0.53(given)

N=normal force =mg

Equating the centripetal and frictional forces,

mr($\alpha$t)²=$\mu$mg

Thus, time taken for the particle to slid whencentripetal force exceeeds the frictional force=t=($\mu$g/r$\alpha$²)^1/2

Thus, t=(0.53*0.58/9.8*2.9²)^1/2=0.061 sec= 61 ms