Question

A hollow spherical shell with mass 2.45 kg rolls without slipping down a slope that makes an angle of 39.0 degrees with the horizontal.

Answer 1

Moment of inertia of spherical shell I = ( 2/3) m r ²Mass m = 2.45 kgAngle θ = 39 degrees(A). The magnitude of the acceleration a_cm of the center of mass of the spherical shellis a = g sin θ / [I / mr ² ]= g sin θ / [(2/3)]= ( 3/2) g sin θ= (3/2) x 9.8 x sin 39= 9.25 m / s ²B)The magnitude of the frictional force acting on the spherical shell F = μ mg cos θWhere μ = Coefficient of friction.If the object not in motion ,then μ = coeffcient of static friction=tan 39= 0.8097So, magnitude of static frctional force F = 15.1 NIf the object in motion then μ = coeffcient of rooling friction