A hollow cylinder, of radius \(R\) and mass \(M,\) rolls without slipping down a loop-the-loop track of radius \(r .\) The cylinder starts from rest at a height \(h\) above the horizontal section of track. What is the minimum value of \(h\) so that the cylinder remains on the track all the way around the loop? A hollow cylinder, of radius \(R\) and mass \(M,\) rolls without slipping down a loop-the- loop track of radius \(r .\) The cylinder starts from rest at a height \(h\) above the horizontal section of track. What is the minimum value of \(h\) so that the cylinder remains on the track all the way around the loop?

The variables we need to consider are: - \(h\): the initial height of the cylinder above the horizontal section of the track. - \(R\): the radius of the hollow cylinder. - \(r\): the radius of the loop-the-loop track. - \(M\): the mass of the hollow cylinder. - \(g\): the acceleration due to gravity. Initially, the cylinder is at rest and at a height \(h\). We need to find the minimum value of \(h\) for the cylinder to remain on the track all the way around the loop.

As the cylinder rolls down the track, its potential energy is converted into kinetic energy (both translational and rotational). The conservation of mechanical energy states that the initial mechanical energy is equal to the final mechanical energy: \(E_{initial} = E_{final}\) The initial mechanical energy is the potential energy due to its initial height, given by: \(E_{initial} = Mgh\) As the cylinder rolls without slipping, its final mechanical energy consists of both translational and rotational kinetic energy. The final mechanical energy is given by: \(E_{final} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2\), where \(v\) is the linear velocity, \(I\) is the moment of inertia, and \(\omega\) is the angular velocity of the cylinder. For a hollow cylinder of radius \(R\), the moment of inertia is given by: \(I = M R^2\) Since the cylinder rolls without slipping, the relationship between linear velocity and angular velocity is given by: \(v = R\omega\) Now, we can rewrite the final mechanical energy in terms of \(v\) as: \(E_{final} = \frac{1}{2}Mv^2 + \frac{1}{2}(MR^2)(\frac{v^2}{R^2})\) \(E_{final} = \frac{1}{2}Mv^2 + \frac{1}{2}Mv^2 = Mv^2\) By applying the conservation of mechanical energy, we get: \(Mgh = Mv^2\)

At the top of the loop (let's call this point A), the centripetal force acting on the cylinder is provided by the tension in the track and the gravitational force acting downward. Newton's second law states that: \(F_{net} = M a \) At the top of the loop, the net force acting is: \(F_{net} = T + Mg\) Since the acceleration at point A is centripetal acceleration, we have: \(M a_{centripetal} = T + Mg\) Centripetal acceleration is given by: \(a_{centripetal} = \frac{v^2}{r}\) Thus, we can rewrite the equation as: \(M\frac{v^2}{r}= T + Mg\)

To determine the minimum height, we must find the minimum tension in the track at the top of the loop for the cylinder to stay on the track. The minimum tension occurs when \(T = 0\), which gives us: \(M \frac{v^2}{r} = Mg\) From Step 2, we know that \(Mgh=Mv^2\). We can substitute this into the equation above: \(M \frac{Mgh}{r} = Mg\) Cancelling out \(M\) and \(g\) from both sides, we get: \(h = r\) Hence, the minimum value of \(h\) necessary for the hollow cylinder to remain on the track all the way around the loop is equal to the radius of the loop (\(r\)).