Posted by Karla on .
A hollow, spherical shell with mass 2.00 kg rolls without slipping down a slope angled at 38.0 degrees. (a) Find the acceleration, the friction force, and the minimum coefficient of fricition needed to prevent slipping.
You will find the moment of inertia (I) of a hollow sphere of radius R derived here: (Broken Link Removed)
I = (2/3) MR^2
If it rolls wthout slippng, the increase in Kinetic Energy equals the decrease in Potential Energy. The kinetic energy is in two parts, rotation and translation.
M g H sin 38 = (1/2) M V^2 + (1/2) I (V/R)^2
= [(1/2) + (1/3)] M V^2
The M's and R's cancel and
gH sin 38 = (5/6) V^2
V = sqrt (1.2 g H sin 38) = sqrt (2 a H)
where a is the acceleration
a = 0.6 g sin 38
Use that acceleration and Newton's second law t compute the actual friction force, F.
M g sin 38 - F = M a
Mg sin 38 - M*0.6 g sin 38 = F
F = 0.4 M g sin 38
To provide this amount of friction, the static coefficient of friction mu,s must equal or exceed a value given by
M g cos 38 * mu,s = 0.4 M g sin 38
mu,s - 0.4 tan 38
Higher values of mu,s will also prevent slipping
The V/R term is the angular rtation velocity.