Problem 3/348 (Curvilinear Motion)
The 26-in. drum rotates about a horizontal axis
with a constant angular velocity Ω = 7.5 rad/sec.
The small block A has no motion relative to the
drum surface as it passes the bottom position θ =
0. Determine the coefficient of static friction μs
that would result in block slippage at an angular
position θ; plot your expression for 0 ≤ θ ≤ 180°.
Determine the minimum required coefficient
value μmin that would allow the block to remain
fixed relative to the drum throughout a full
revolution. For a friction coefficient slightly less
than μmin, at what angular position θ would
slippage occur?
Problem Formulation
From the free body and mass acceleration diagrams,
[ΣFn = ma n ]
N − mg cos θ = mrΩ 2
[ΣFt
F − mg sin θ = 0
= ma t ]
For impending slip we have F = μ s N . Substituting F into the above and solving
gives,
μs =
g sin θ
sin θ
=
2
1.8925 + cos θ
g cos θ + rΩ
The last two questions can be answered only after plotting μs as a function of θ.
MATLAB Worksheet and Scripts
%%%%%%%%%%%%%%%% Script #1
%%%%%%%%%%%%%%%%%%%%
% This script solves our two equations
symbolically
% for mu_s and N
% O = Omega
% x = mu_s
% y = N
syms theta O g m r x y
eqn1 = y-m*g*cos(theta)-m*r*O^2;
KINETICS OF PARTICLES 2
eqn2 = x*y-m*g*sin(theta);
% remember that we write our equations in the form
% expression = 0 and then omit the "=0"
[x,y] = solve(eqn1,eqn2)
%%%%%%%%%%%%%%end of script %%%%%%%%%%%%%%%%%%%%
Output of script #1
x=
g*sin(theta)/(g*cos(theta)+r*O^2)
y=
m*g*cos(theta)+m*r*O^2
%%%%%%%%%%%%%%%% Script #2 %%%%%%%%%%%%%%%%%%%%
% This script plots mu_s as a function of theta
theta = 0:0.02:pi;
mu_s = sin(theta)./(1.8925+cos(theta));
plot(theta*180/pi, mu_s)
xlabel('theta (degrees)')
title('coefficient of static friction')
%%%%%%%%%%%%%%end of script %%%%%%%%%%%%%%%%%%%%
KINETICS OF PARTICLES 3
If the block is not to slip at any angle θ, the coefficient of friction must be greater
than or equal to any value shown on the plot above. Thus, the minimum required
coefficient value μmin that would allow the block to remain fixed relative to the
drum throughout a full revolution is equal to the maximum value in the plot
above. The location where this maximum occurs can be found by solving the
equation dμ s / dθ = 0 for θ. This θ can then be substituted into μs to yield the
required value for μmin. Here’s how we do this with MATLAB.
EDU» syms theta
EDU» mu_s = sin(theta)./(1.8925+cos(theta));
EDU» dmu = diff(mu_s,theta)
dmu = cos(theta)/(757/400+cos(theta))+sin(theta)^2/(757/400+cos(theta))^2
EDU» theta_m = solve(dmu,theta)
theta_m =
-atan(1/302800*236697316401^(1/2))+pi
atan(1/302800*236697316401^(1/2))-pi
EDU» eval(theta_m)
ans =
2.1275
-2.1275
EDU» mu_min = subs(mu_s,theta,2.1275)
mu_min = 0.6224
From the above we see that μmin = 0.622. If μs is slightly less than this value, the
block will slip when θ = 2.128 rads (121.9°)
KINETICS OF PARTICLES 4