A hollow spherical shell with mass 2.00 kg rolls without slipping down a 38 degree slope.?

Question

1. find the acceleration of the shell. 2. find the friction force on it. 3. if the mass increased to 4 kg find the acceleration of the shell and the friction force on it.

Answer 1

Moment of inertia for a hollow a sphere is
I=(2/3) m r^2
T=I A
a=Ar
T=rF
F=mg sin(38)
T=r mg sin(38)
finally a= (T/I) r
a=(r mg sin(38)/I)r
a=(r mg sin(38)/[(2/3) m r^2] r
a=g sin(38)/(2/3) =
a=3 g sin(38)/2=
a= 3 x 9.8 x sin(38)/2=9.0m/s^2

The acceleration will be the same for 4 kg as it is for 2kg.

Since it roles not slides mass has nothing to do with friction .

Answer 2

a(cm) ability acceleration on the midsection of mass. to discover the Frictional stress, mg sin(40) - Friction = ma Friction = mg sin(40) - ma to discover the minimum coefficient of friction uk, Friction = uk * general stress uk = Friction / general stress uk = friction / mg cos (40)