Question 3.4.1: A solid cylinder of mass m and radius r starts from rest and......
A solid cylinder of mass m and radius r starts from rest and rolls down the incline at an angle θ (Figure 3.4.2). The static friction coefficient is \mu_{S}. Determine the acceleration of the center of mass a_{G_{x}} and the angular acceleration α. Assume that the cylinder rolls without bouncing, so that a_{G_{y}}=0. Assume also that the cylinder rolls without slipping. Use two approaches to solve the problem: (a) Use the moment equation about G, and (b) use the moment equation about P. (c) Obtain the frictional condition required for the cylinder to roll without slipping.
The free body diagram is shown in the figure. The friction force is F and the normal force is N. The force equation in the x direction is
m a_{G_{x}}=f_{x}=m g\sin\theta-F (1)
In the y direction,
m a_{G_{Y}}=f_{y}=N-m g\cos\theta (2)
If the cylinder does not bounce, then a_{G_{y}}=0 and thus from (2),
N=m g\cos\theta (3)
a. The moment equation about the center of mass gives
I_{G}\alpha=M_{G}=F r (4)
Solve for F and substitute it into (1):
m a_{G_{x}}=m g\sin\theta-{\frac{I_{G}\alpha}{r}} (5)
If the cylinder does not slip, then
a_{G_{x}}=r\alpha (6)
Solve this for α and substitute into (5):
m a_{G_{x}}=m g\sin\theta-{\frac{I_{G}a_{G_{x}}}{r^{2}}}Solve this for a_{G_{x}}{:}
a_{G_{x}}=\frac{m g r^{2}\sin\theta}{m r^{2}+I_{G}} (7)
For a solid cylinder, I_{G}=m r^{2}/2, and the last expression reduces to
a_{G_{x}}={\frac{2}{3}}g\sin\theta (8)
b. We could have used instead the moment equation (2.4.5) about the point P, which is accelerating. This equation is
x(t)=C_{1}e^{-r_{1}t}+C_{2}e^{-r_{2}t}+\cdot\cdot\cdot+C_{n}e^{-r_{n}t} (2.4.5)
M_{P}=I_{G}\alpha+m a_{G_{x}}dwhere d =r and M_{P} = (mg sin θ)r. Thus
m g r\sin\theta=I_{G}{\frac{a_{G_{x}}}{r}}+m a_{G_{x}}rSolving this for a_{G_{x}} we obtain the same expression as (7). The angular acceleration is found from (6).
c. The maximum possible friction force is F_{\mathrm{max}}=\mu_{s}N=\mu_{s}m g\cos\theta. From (4), (6), and (7),
F={\frac{I_{G}\alpha}{r}}={\frac{I_{G}a_{G_{x}}}{r^{2}}}={\frac{I_{G}m g\sin\theta}{I_{G}+m r^{2}}}If F_{\mathrm{max}}\gt F, the cylinder will not slip. The condition of no slip is therefore given by
\mu_{s}\cos\theta\gt \frac{I_{G}\sin\theta}{I_{G}+m r^{2}} (9)
For a solid cylinder, this reduces to
\mu_{s}\cos\theta\gt {\frac{1}{3}}\sin\theta (10)