A hollow sphere of mass M and radius R is released from rest on an inclined plane of inclination θ with horizontal. The minimum coefficient of friction between sphere and plane so that sphere rolls without slipping, is

The sphere will roll without slipping if the frictional force is enough to prevent it from sliding. The frictional force is given by f = μN, where μ is the coefficient of friction and N is the normal force.

Step 1: Calculate the normal force (N) The normal force is the component of the weight of the sphere perpendicular to the inclined plane. It can be calculated as: N = Mgcos(θ)

Step 2: Calculate the frictional force (f) The frictional force is the component of the weight of the sphere parallel to the inclined plane. It can be calculated as: f = Mgsin(θ)

Step 3: Set up the equation for rolling without slipping For the sphere to roll without slipping, the frictional force must be equal to the torque due to the frictional force. The torque (τ) can be calculated as: τ = fR = Iα where I is the moment of inertia of the sphere and α is the angular acceleration.

For a hollow sphere, I = 2/3MR^2. And α can be related to the linear acceleration (a) as α = a/R.

Substituting these values in, we get: fR = (2/3MR^2)(a/R) Solving for a, we get: a = 3f/(2MR)

Step 4: Set up the equation for the acceleration (a) The acceleration of the sphere down the inclined plane can also be calculated from the forces acting on it. It is given by: a = g*sin(θ) - f/M

Step 5: Solve for the coefficient of friction (μ) Setting the two expressions for a equal to each other, we get: 3f/(2MR) = gsin(θ) - f/M Solving for f, we get: f = (2MRgsin(θ))/(3+2R) And finally, solving for μ, we get: μ = f/N = (2Rgsin(θ))/(3gcos(θ)+2R)

So, the minimum coefficient of friction for the sphere to roll without slipping is μ = (2Rgsin(θ))/(3gcos(θ)+2R).