Solutions are: a) 3.0m/s b) 36m/s^2 c) the normal force between the drum and the

Question

Solutions are:

a) 3.0m/s

b) 36m/s^2

c) the normal force between the drum and the pebble

d)1.8N directed inward

e) 0.50N directed upward

f) 0.28

g) the value of ? will not change

please I need step by step process to the soluton. Thanks!

Pebble inside a rotating drum. Consider a hollow drum of radius r = 25 cm rotating about some vertical axis with a constant angular velocity omega = 12^rad/s as shown in Fig. 1. The magnitude of the angular velocity is just the rate of rotation, often expressed in radians per second (or alternatively, the number of revolutions divided by 2pi per second.) The speed nu of a particle at a distance of r from the axis of rotation can be found from the relation nu = romega. A pebble of mass m = 50 g placed inside the drum is observed to rotate with the cylinder and not fall. (a) What is the speed of the pebble? (b) What is the magnitude of its radial acceleration? (c) What force is acting as the centripetal force in this problem? (d) What is the normal force between the pebble and the drum? (e) What is the force of friction between the pebble and the drum? (f) What, then, is the minimum coefficient of static friction between the drum and pebble? (g) How would your answer for (f) change if the mass of the pebble were doubled?

Explanation / Answer

a. Speed of pebble:

v = r*w =0.25*12 = 3.0 m/s.

b. Radial acceleration, a =v^2/r = 3^2 /0.25 =36 m/s^2

c. normal force between drum and pebble,

N =m*a = 50*10^-3*36 =1.8 N

d. Centripetal force, f= m*a =1.8 N

e. Frictional force,

Fr =mu*N =m*g =50*10^-3*9.81 =0.4905

f. Minimum co.eff. of friction,

mu = 0.4905 / N =0.4905 /1.8 =0.2725

g. if m' = 2*50 = 100 grams

mu=0.4905/N =0.4905 / 2*m*a =0.13625.

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